We will examine objects that are familiar and concrete, but we will sometimes pose questions about them that are not easy to answer.
The purpose is to introduce the themes that will be studied in the rest of the text, but also to begin looking at concrete situations.

We look into the idea of symmetry.

We may find it hard to give an adequate answer if we ask ourselves what we mean by symmetry.
When we see that symmetry can be given an operational definition, we will be able to associate a structure with each object.

The goal of the book is to encourage you to think for yourself.

Take some time to think about the questions.
Start with a list of objects that are symmetrical: a sphere, a circle, a cube, a square, a rectangular box, etc.

Take some time to think about the questions you're reading about.

A rectangular card is an example of a symmetric object.

The card admits motions that keep its appearance the same.

A symmetry is not visible.

If an object has symmetries, it is symmetric.

To examine this idea, you need to do the following exercises.

Reflections of the card are related to symmetry.

In order to simplify matters, I propose to ignore reflections for the moment, but to bring symmetry into the picture later.

A human face has leftright symmetry, but there is no actual motion of a face that is a symmetry.

All the symmetries of a rectangular card are cataloged.

Look at the card.
As you need it, mark its parts.
Write your conclusions and observations down.

For a square card, do the same.

Do the same for a brick with three edges.

It turns out that the rotation through 0 radians about any axis is necessary.

Nothing is one of the things that you could do to the card.
I could not tell if you had done anything or not.

There are infinitely many symmetries.

The center of mass is the intersection of the two diagonals, and the centroid is the center of a rectangle.

The issue of why there are exactly four will be taken up later.

Since the world doesn't have a solutions manual, we have to make our own decisions.

This is the choice that I will make, because distinguishing only the four symmetries leads to a rich and useful theory.

We have to consider rotation by 2 about one of the axes, and rotation by 3 about the other.
We can disregard the path of the motion and only take into account the final position of the parts of the card.
All of the parts of the card end up in the same place.

I propose to exclude reflection symmetries temporarily, for the sake of simplicity, as another issue is whether to include reflection symmetries as well as rotation symmetries.

If I leave the room and you perform two motions in a row, I will not be able to detect the result.

The result of two symmetries is the same.

The rectangle has symmetries.

The next step is to figure out all the products of the rectangular card and the square card.

A good way to record the results of your investigation is in a multiplication table with rows and columns labeled by the symmetries.
You have to fill in the rest of the table.

The beginning of the multiplication table.

Continue with the multiplication table for the square card after you've finished with the multiplication table for the rectangular card.

The eight symmetries of the square card have to be labeled in order for the multiplication table to be worked out.

If we agree on a labeling, it will be helpful to compare our results.

The square has symmetries.

The beginning of the multiplication table.

It's important to work things out for yourself when learning mathematics.

We have to keep track of the results of the multiplication tables for the symmetries of the square and rectangle.

It is possible to break the symmetry by labeling parts of the square.
I'm going to number both the locations of the four corners of the square and the corners themselves at the risk of overdoing it.
The numbers on the corners will travel with the card, and those on the locations will stay put.
The labeling for the square card is shown in the figure.

You can see that cd D r2 when you compare Figures 1.3.2 and 1.3.3.

The rule for computing all of the products is that the square of any element is nonmotion e.

The order of the elements is irrelevant in this multiplication table.

The product of two elements is the same.
Normally order does matter, but this is unusual.

The product of two powers of r (i.e., of two rotation around the axis through the centroid of the faces) is again a power of r. The nonmotion e is the square of the elements fa; b; c; d g.

The last property is obvious, without doing any close compu tation of the products.

The product of two symmetries that exchange the faces leaves the upper face above and the lower face below, so it has to be a power of r.

The order of the symmetries in this table does matter.

For example, ar D c.

The result of two symmetries followed by a third is the same as the result of the first and second symmetries.

The associative law is used for multiplication.
The law is expressed as s.t U/ D.st/u for any three symmetries.

The second symmetry is the nonmotion e composed with any other symmetry.

One can tell the inverse of a symmetry by looking at it.

The relation is satisfied by both of them.

We will pay a lot of attention to the consequences of these small observations.

The symmetries of an equilateral triangular plate are six.

The square card has symmetries.

Show that r k D r3k D rm, where m is the unique element of f0; 1; 2; and 3g.

Another way to list the symmetries of the square card makes it easy to compute the products of symmetries quickly.

We have been refining our idea of a symmetry of a geometric figure while looking at some examples.

A mathematical model is being developed for a physical phenomenon such as the symmetry of a brick or a card.
We have decided to ignore the path by which they arrived at the final position of the object.
A map from R to R is a symmetry.

The idea that a transformation is rigid or non-disturing can be formalised.

If for all points a, b, and R, we have d.

An affine isometry of R3 is always defined on a subset of R3.

If R is not contained in a two-dimensional plane, the affine extension is uniquely determined.
We will assume that these facts are established in Section 11.1.

If we assume that R is a square, we can say that it lies in the.x; y/-plane.

We can show how to map the set of R to itself.

The centroid of the figure is the intersection of the two diagonals of R. .C / D C is still equidistant from the four vertices since is an isometry.

We can assume that the figure is located with its centroid at 0, the origin of coordinates.
The results quoted in the previous paragraph relate to a linear isometry of R3.

The same argument and conclusion can be used for many other geometric figures.

There is at least one point that is mapped to itself by every symmetry of the figure.
If we place a point at the origin of coordinates, every symmetry of the figure extends to a linear isometry of R3.

A polyhedron is a three-dimensional object located with its centroid at the origin of coordinates.

The restriction to R of a linear isometry of R3 is every symmetry of R.

Our symmetries are implemented by 3 by 3 matrices.

There is a matrix A for all points in our figure,.x/ D Ax.5 The unique linear extension of the composed symmetry 12 is created by the composed linear transformation T1T2.
The matrix product A1A2 implements T1T2 if A1 and A2 are the matrices.
The product of the corresponding matrices can be used to compute the composition of symmetries.

We can use this observation to do the bookkeeping for symmetries.

We will find the matrix that implements the symmetry of the square or rectangle.

There is a brief review of elementary linear algebra in the appendix.

The linear transformation implementing each symmetry is unique if we insist on it.

The figure is in the.x; y/- plane with the sides parallel to the coordinate axes.

The coordinate axes will coincide with certain axes of symmetry.
We can orient the plane so that the axis of rotation for r1 coincides with the x- axis, the axis of rotation for r2 coincides with the y- axis, and the axis of rotation for r3 coincides with the z- axis.

The x-coordinate of a point in space is not changed by the rotation r1 The standard matrix of a linear transformation is determined by how the rotation r1 implements it.

If T is a linear transformation of R3 then the 3-by-3 matrix with columns T should be used.

Oe2/ D Oe2 and r1.

We can see what the rotation r2 and r3 do in terms of coordinates.

The square of any of the Ri's is E and the product of any two of the Ri's is the third.

The multiplication table for the matrices R1, R2, R3 and E is the same as for the symmetries r1, r2, r3 and e.

The axes of symmetry for the rotation a, b, and r coincide with the x-, y-, and Z-axes if you choose the orientation of the square in space.

The set of matrices fE; R; R2; R3; A; B; C; Dg have the same multiplication table as the corresponding set of symmetries.

We can conclude that CD D R2 if we compute it.

We can return to the question of whether we have found all the symmetries of the square.

We think that the figure lies in the.x; y/-plane with sides parallel to the coordinate axes and centroid at the origin of coordinates.
Since a symmetry is an isometry, it must take each edge to an edge of the same length.

The edges are at.

Oe1 and.w Oe2.
The symmetry is determined by.
Oe1 and.w Oe2 since they are the basis of the plane.

The only two edges that are longer than 2w are.

Oe1 and.
Oe2.
Oe2 must be.w Oe2 since these are the only two edges that are longer than 2.
There are at least four possible symmetries of the rectangle.
Four distinct symmetries have already been found.

For the square,.w Oe1 and.w Oe2 must be contained in the set f.w Oe1 and.w Oe2g.

If.w Oe1/ is.w Oe2, then.w Oe2/ is.w Oe1.

There are at least eight possible symmetries of the square.

There are eight distinct symmetries.

The multiplication table for the symmetries of the square should be reproduced by the products of the matrices E, R, R2, R3, A, B, C, D.

The equilateral triangle has six symmetries.

The multiplication table for the equilateral triangle is reproduced by the products of the matrices.

T.x/ D Ax C b is an affine transformation of R3.

The next four exercises outline an approach to showing that a sym metry of a square or rectangle sends a line to another line.

The approach is based on the idea that there is more than one truth.

There is a line segment called a2 in R3 and a2 D fsa1 C.1 s/a2 W 0 s 1g.

If a1 and a2 are elements of R and v 2, then a1 D a2.

V is not contained in any nontrivial line segment in R.

The configuration has symmetry because the objects are the same.

I could switch the objects around if you looked away, but you couldn't tell if I had moved them.
Symmetry is not a geometric concept.

There are three symmetries where two objects can be switched at the same time.

One object can be put in the place of a second, the second in the place of the third, and the third in the place of the first.
All the objects are left in place.
There are six symmetries.

The symmetries of a configuration of objects are called permutations.

We have to come up with a system for keeping track of things.

The objects occupy three positions.
We can describe each symmetry by recording the final position of the object that starts in position i.

The product of permutations is computed by following each object as it moves.

The element on the right is the first permutation and the one on the left is the second permutation.

The order in which permutations are ordered is important.

The permutations of n identical objects can be marked by two-line array of numbers, containing the numbers 1 through n in each row.

If a permutation moves an object from position i to position j, the corresponding two line array has j positioned below i.
The first row of numbers are usually arranged in increasing order.
The same rule is used for permutations of three objects.

The multiplication of permutations is an associative activity.

Each object is left in its original position by an identity permutation.

In either order, the product of e with any other permutation is.

There is an inverse permutation 1 for each permutation.

In either order, the product is e.

These properties are even more obvious because of a slightly different point of view.

If the range of f is all of Y, there is an x 2 X such that f.x/ D y.

The inverse of f and satisfying f is called X.
The unique element of X is f 1.y/.

Consider maps from a set.

The composition of maps is associative.
If f and g are on the same map, then the composition f i g is also on the same map.

The identity map idX is the identity element for this product; that is, for any f 2 Sym.X/, the composition of idX with f in either order.

The inverse of a map is the inverse for this product; that is, for any f 2 Sym.X/, the composition of f and f 1 in either order is idX.

Functions and sets are discussed in Appendix B.

A permutation of n objects can be identified if the function maps to j.

The permutation 1 2 3 4 5 6 7 D 4 3 1 2 6 5 7 in S7 is identified with a map of the stars.
The multiplication of permutations is the same as the composition of the maps.
The three properties listed for permutations are related to the corresponding properties of the maps.

We usually write Sn for the permutations of a set of n elements.

It's not hard to see that the size is nS D n.n 1/.2/.1/.
The image of 1 under an inverted map can be any of the n numbers 1, 2, and so forth.

S52 is not a different kind of object than S5 which has 120 elements, because we couldn't begin to write out the multiplication table for S52 or even list the elements of S52.

A permutation such as 1 4 2 3 that leaves all other numbers fixed is called a cycle.

The order of the entries is more important than the first entry.

If each leaves fixed the numbers moved by the other, there are two cycles.

The expression D 1 4 2 3 5 6 is a product of disjoint cycles.

Let's see how permutations are calculated.

The permutation on the right is taken first in calculating permutations.

The first of the permutations takes 1 to 4 and the second takes 4 to 7, so the product takes 1 to 7.
The product takes 7 to 6 because the first leaves 7 fixed and the second takes 7 to 6.
The product takes 6 to 4 and the first takes 6 to 5.
The product takes 4 to 2 after the first and second leaves.
The first takes 2 to 3 and the second takes 3 to 1, so the product takes 2 to 1.
This closes the cycle.
The product fixes 5 because the first permutation takes 5 to 6 and the second takes 6 to 5.
The product fixes 3 because the first takes 3 to 1 and the second takes 1 to 3.

The product of the cycles 1 4 2 3 and 5 6 is the permutation D. Their product is not related to the order in which they are multiplied.

If the kth power of the identity is greater than the lower power of the identity, a permutation has order k. A k-cycle has order k. The least common multiple of the lengths of the cycles is equal to the order of the disjoint cycles.

2 4 3 5 1 6 7 9 10 is the least common multiple of 4, 2, and 3.

There is a perfect shuffle of a deck of cards.

The first cut of the deck is perfect, with the half deck consisting of the first n cards placed on the left and the last n cards placed on the right.
The first card from the right goes on top, followed by the first card from the left, then the second card from the right, and so on.

The perfect shuffle of a deck of 10 cards is 10.

The order of the perfect shuffle is 6.

The permutation 2 Sn is written in cycle notation.

The first number (1 a1 n) is not fixed.
Write a2 D.a1/ a3 D.a2/ D.a1// a4 D.a3/ D...a1 Since each number is in f1, it cannot be all distinct.
There is a number k such that a1; a2; : : ; ak are all distinct, and D a1.
The permutation permutes the numbers fa1, a2, and 2.

Consider the first number b1 62 fa1; a2; : : ; akg that is not fixed.

Write b2 D.b1/ b3 D.b2/ D.b1// b4 D.b3/ D...b1 The numbers fa1; a2; : : ; akg; akg; b g among themselves, and the remaining numbers f1; 2; : : : ;ng n.FA1; are permuted.

"Shuffling cards and stopping times" was written by Aldous and P. Diaconis.

Continue this way until the product of disjoint cycles is written.

The phrase "continue in this way" is a signal that it is necessary to use mathematical induction to formalize the argument.

The factors are unique, and the order in which they are written is irrelevant, because disjoint cycles 1 and 2 commute.

There is no preference for the first entry in the cycle notation.
In order to not have to make an exception for the identity element, we regard e as the product of the empty collection of cycles.

Every permutation of a finite set can be written as a product of disjoint cycles.

If jXj D 1 is the only permutation of X, there is nothing to do since the only permutation of X is the identity e. Let be a non identity permutation of X.

Since jXj is finite, there is a number k such that x0; x1; and xk are all distinct.
The sets X1 D, X2 D, and X1 are each invariant and therefore the product of 1

2 D jX2 is a product of disjoint cycles.

It is also a product of disjoint cycles.

A variation on the same argu ment states that the cycle containing x0 is uniquely determined by the list.

Any expression of a disjoint cycle must contain it.

The decomposition of 2 as a product of disjoint cycles is unique and the product of the remaining cycles in the expression yields 2.
The cycle decomposition is unique.

Compare the multiplication table of S3 with the set of symmetries of an equilateral triangular card.

Matching elements of S3 with symmetries of the triangle makes the two multiplication tables agree.

There are a pair of 2-cycles that don't commute.

The perfect shuffle is a deck of 2n cards.

Suppose X is the union of two disjoint sets.

The permutation 2 Sym.X/ is invariant for X1 and X2.

Show that the assumption.xk/ D xl for some l, 1 l k leads to a contradiction.

We can do computations in these systems in order to find out the order of a perfect shuffle of a deck of cards.

We return to familiar mathematical territory in this section.

We study the set of numbers.
In elementary school, multiplication in the integers can be interpreted in terms of repeated addition, for example if the number is a and n and the number is a C.

The set of numbers f1;.2; : : g by Z and the set of natural numbers f1; 2; 3; : : g by N are given.

If n 2 N is positive, we write n 0 and say it's non- negative.

If a is non-negative, the absolute value is equal to a otherwise.

The prop erties of the integers are addition and multiplication.

A b is written for a C.

The sum and product of positive integers is positive.

We write a b if a b is 0 or a b if a b is 0.

The total order is on the integers.
It is never true that a; for distinct integers a; b; b; or b; or a; or a; or a; or c; or a; or a; or a; or a; or a; or a; or a; or a; or a;

There is a more precise version of the statement in the proposition.

Many of the most interesting properties of the integers have to do with divisibility, factorization, and prime numbers.

The emphasis in this section will be more on a systematic, logical development of the material, rather than exploration of unknown territory, since these concepts are already familiar to you from school mathematics.
The main goal is to show that every natural number has a unique factorization as a product of prime numbers; this is more difficult than one might think.

Divisibility is a definition.

If there is a quotient of aq D b, then b can be divided by remainder without the need for a remainder.

Let a, b, c, U, and v represent numbers.

U and v can't be zero.

Suppose first that both are positive.
We have a maxfu.
If it holds, then D v D 1.
If D 1 and D 2 are included, then also D 1 and D 2.

Both of them are.1.

Both have the same sign.

For (b), let the number b be the same as the number d and the number d be the same as the number vi.

0 D.uv 1/b.
The product of nonzero is either b D 0 or uv D 1.

The parts are left to the reader.

Some natural numbers like 2 and 13 are called prime because they can't be expressed as a product of smaller natural numbers.

Natural numbers can be written as a product of prime numbers, for example, 42 D 2 3 7.

If the number is greater than 1 and not divided by another number, it is prime.

To show that every natural number is a product of prime num bers, we have to use mathematical induction.

Natural numbers can be written as a product of prime numbers.

We need to show that all natural numbers can be written as a product of prime numbers.

The second form of mathematical induction is used to prove this statement.

For all natural numbers, the number r is a product of primes.

It is a product of primes if n happens to be a prime.
n can be written as a product if 1 a n and 1 b n are included.
According to the hypothesis, each of a and b can be written as a product of prime numbers.

In mathematics, 0 is the sum of an empty collection of numbers and 1 is the product of an empty collection of numbers.

A lot of arguments are saved by this convention.
The prime factorization is the product of an empty collection of primes.

There is a question of whether there are infinitely many prime numbers or finitely many.

Greek mathematicians solved the question before 300 B.C.
The solution is attributed to a book by the name of Elements: Theorem 1.6.6.
There are a lot of prime numbers.

There is at least one prime number for all natural numbers.

There is at least one prime number.
Suppose that there are at least k prime numbers.
We will show that there is at least one prime number.
Let p2 be a collection of k prime numbers.
Consider the number M D p1p2 pk C 1.

According to the previous proposition, M is a product of prime numbers.

If p is a prime dividing M, then fp1; p2; : : ; pk; is a collection of k C 1 prime numbers.

The fundamental fact about divisibility is that it is possible to divide an integer by any divisor and get a quotient q and a remainder r, with the remainder being smaller than the divisor.

The proof of this fact is related to what you learned in school.

We just put q D 0 and r D a.
If a d, first we guess an approximation q1 for the quotient such that 0 q1d a, and compute r1 D a q1d, which is less than a.

We put q D q1 and r D r1 if r1 d is finished.

We try to divide r1 by d, guessing a q2 such that 0 q2d r1, and compute r2 D r1 q2d.

We are done if q D q1 C q2 and r D r2 are put together.

We are trying to divide r2 by d.

It must be that 0 rn d because the rn are non negative.

We stopped and put q D q1 C q2 C qn and r D rn.

The procedure for a negative is the same.

A proof of the existence part of the following assertion can be found in the description of this algorithm.

The essential idea is the same in the following proof as it is in the previous proof.

There are unique integers q and r, D qd C r and 0 r d, given the numbers a and d.

The case should be considered a 0.

Take q D 0 and r D a.

Now suppose that's true.

Assume that the existence assertion holds all nonnegative integers that are smaller than a.

We are done after a D.q0 C 1/d C r.

The case is dealt with by 888-276-5932 888-276-5932 888-276-5932 888-276-5932.

Take q D 1 and r D aCd if d is 0 Assume that the existence assertion holds for all non positive integers whose absolute values are smaller than jaj.
It holds for a C d, so there are two integers q0 and r.

We have shown the existence of q and r. Suppose that a D qd C r and a D q0d C r0 have 0 r and d. It's the same as r r0 D.q0 q/d, so it's the same as r r0 The only possibility is r r0 D 0.

We have shown the existence of a prime factorization, but we have not shown that it is unique.

The issue is addressed in the following discussion.
The greatest common divisor of two integers can be computed without knowing their prime factorizations.

If V 2 N divides m and n, then V also divides, the natural number is the greatest common divisor of nonzero integers m and n.

If the greatest common divisor is unique, then notice.

Several important properties are recorded in the following proposition.

Let I.m; n/ D fam C bn W a; b 2 Zg; n and x 2 I.m; n and x 2 I.m.

A natural number, that is a common divisor of m and n and an element of I.m, is the greatest common divisor of m and n.

The greatest common divisor is m and n. Sequences jnj > n1 > n2 0 and q1 are defined.

After no more than n steps, this process must end.

We have a system of relations that include m D q1n C n1 n D q2n1 C n2.

A b is the first row of Q 1. nr is a common divisor of m and n.

The greatest common divisor is given by g.c.d.

q1 is the sequence of quotients, q2 is the sequence of quotients, q3 is the sequence of quotients, q4 is the sequence of quotients, q5 is the sequence of quotients, q6 is the sequence of quotients.

18 D 8233 is a result of Definition 1.6.12 If g:c:d:.m and n are nonzero, they are prime.

If 1 2 I.m; n/ is the case, m and n are relatively prime.

1 D 3 21 C 4 16 is a relatively prime number.

If p is a prime number and a is a nonzero number, then either p divides or p is a prime number.

The proof of uniqueness of prime factorizations can only be shown from here.

Let p be a prime number.

If p divides a, then p divides b.

If p does not divide a, then 1 D, a C Vp, and V are prime.

Multiplying by b shows that b is divisible by p.

Suppose that a prime number p divides a product a1a2 One of the factors is divided by p.

A natural number's prime factorization is unique.

The case n D 1 cannot be written as the product of any nonempty collection of prime numbers.

Assume that the assertion of unique factorization holds for all natural numbers less than n. q1 divides n D p1p2 and hence is equal to one of the pi.
It follows that p1 D q1.

It's based on the idea that r D s and qi D pi are for everyone.

If a factor is found, we can search for other factors.

We will continue this procedure until all of the factors are present.
This procedure is not very efficient.
No truly efficient methods are available for large natural numbers.

If V 2 N divides each ai, then V also divides, it's a natural number.

If the greatest common divisor is unique, then notice.

We prove existence by means of a generalization.

There is a natural number d and an n-by-n matrix Q if Q is not zero.

The proof of proposition 1.6.10 establishes the base case n D 2.
Fix n > 2 and assume the assertion holds for lists of less than zero nonzero numbers.

The most common divisor of nonzero integers a1; a2; : : : ; an exists and is a linear combination of a1; a2; : : : ; an.

The greatest common divisor is d.

The most common divisor of a1; : : ; an is denoted g:c:d.

Use the well-ordering principle to complete the sketch of the proof.

We have a D qd C r.

You have to show that if y 2 I.m; n/, then xy 2 I.m:n/.

You have to show that if y 2 I.m; n/, then b divides y.

For each of the following pairs of numbers, write g:c:d:.m; n:c:d:.m; and n:c:d:.m as a linear combination of m and n.

If you want your program to give the greatest common divisor, you need to explicitly give the given nonzero integers.

This method is taught in school.

In 7 hours it will be 4 o'clock if it is now 9 o'clock.

The clock number 12 is the identity for clock addition, which means that the clock will show the same time in 12 hours.
If it is now 12 o'clock, then after 5 periods of seven hours, it will be 11 o'clock.
All the usual laws of arithmetic are held except for the fact that division is not generally possible.

Imagine taking a number line with points marked and wrapping it around the clock face with 0 on it.

The numbers are put on the clock face and then on the number line.
The numbers are on the number line and on the clock face.

The numbers are on the number line.

When the distance between the two numbers on the number line is multiple of n, the interval between the two numbers on the number line wraps a number of times around the clock face.

If a is congruent to b modulo n, and if a is divisible by n, we write a b.mod n.

If a b and b c are both strued by n, then a D 0 is also strued by n.

When the number line is wrapped around the clock face, there is a set of all integer points that land at the same place.

The set OEa is called a congruence class.

The unique number of r is 0 r n and the r is divided by n. The interval f1; 1; : : ; n 1g is the unique element of OEa.

When the number line is wrapped around the clock face, the label on the circumference of the clock face is remn.a/.

The following are equivalent for b 2 Z.

If a.mod n/ is used, then a.mod n/ is used for any c 2 Z.

This shows that.
If c b.mod n/, then c a.mod n/.
This shows the meaning of (a) and (b).

The unique element of OEx f0; 1 is characterized by remn.x/.

This shows that is true.

There are distinct classes called OEn 1.

These classes are not related.

A0; b; b0 be in the same place as a0 and b.mod n/.

We now describe the natural structure of the set Zn.

Let A and B be elements of Zn, and let 2 A and B be elements of A and B.

The choice of representatives doesn't affect the classes.

If a0 is another representative of A and b0 another representative of B, then a0.mod n/ and b b0.mod n/.
It makes sense to define A C B D OEa C b andAB D OEab.

In order to see what the issue is in the preceding discussion, we need to look at another example in which we can't define operations on classes of numbers in the same way.

Let N and P be the set of negative and non-negative integers.
One of these two classes has every single digit in it.
The sum of a positive number and a negative number can be either positive or negative, depending on the choice of n and p. It doesn't make sense to define N C P D s.n C p/.

It is easy to check that sensible operations on Zn satisfy most of the usual rules of math once we have defined them.

The correspond ing property of Z is followed by the assertions in the proposition.

The commutativity of multiplication on Zn is shown.

The proof of multiplication proceeds the same way as the proof of associativity.

The reader is responsible for checking the details of the other assertions.

On the other hand, nonzero elements can have a zero product.

If there is a nonzero element, we call it a zero divisor.
Zero divisors are the case in Z6.

Many elements have multiplicative inverses and can be invertible if there is an element such as OEaOEb D.

There are multiplicative inverses in Z14, OE1, OE3 and OE9.

You can check to see if the remaining nonzero elements are zero divisors.
In Z14, every nonzero element is either a zero divisor or an inverted one, and there are just as many zero divisors as inverted elements.

In Z7, every nonzero element is invertible.

You should compare the behavior of multipli cation in Z to that of these phenomena.

In the exercises for this section, you are asked to investigate further the zero divisors and invertible elements in Zn, and solutions to the form ax b.mod n/.

Let's look at some examples of computations in Zn and congruences.

The main principle to remember is that a b.mod n/ is the same as a OEa D OEb in Zn.

This is easy because of 4 1.mod 5/.

Let's figure out a few powers of 4 modulo 9.

It follows that in Z9 and Z7 for all natural numbers.
We only have to find the conjugacy class of 237 modulo 3 if we want to compute 4237 modulo 9.

It is the same as the assertion for all OEa 2 Z7.

We only have to check this for the 7 elements in Z7, which is easy.
This is a special case of Fermat's little Theorem.

It is not necessarily true that OEb D OEc is in Zn.

If a and b are relatively prime, this is always possible.

The result can be extended to any number of simultaneous congruences.

If a and b are prime natural numbers, then, and V are also prime natural numbers.

There is an x that is.mod a/ and.mod b/.

They are used for studying the numbers.

Congruence modulo n is a fundamental relation in the integers and any statement about it is equivalent to a statement about Zn.
Sometimes it's easier to study a statement about congruence in the integers in terms of Zn.

The objects Zn are fundamental building blocks in a number of theories.

All finite fields are constructed using Zp.

The development of algebra is a fundamental part of modern digital engineering because the algebraic systems Zn were first studied in the nineteenth century without any view toward practical applications.

For digital communication, for error detection and correction, for cryptography, and for digital signal processing 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884

The exercises ask you to prove parts of the proposition.

The addition in Zn is both commutative and associative.

If you can prove that OE0 is an identity element for addition in Zn, you can prove it for all.

If you can prove that multiplication in Zn is commutative and associative, you can prove that OE1 is an identity element for multiplication.

Write out multiplication tables for Zn.

The zero divisors in Zn for n 10 can be catalog using your multiplication tables from the previous exercise.

Guess which elements in Zn are invertible and which are zero divisors based on your data.

A more analytical approach to invertibility and zero divisors is provided in the next three exercises.

Suppose a is relatively prime to n.

There is no such thing as C nt D 1 if a is not relatively prime to n.

It's possible that OEa isn't invertible in Zn.

The Zn is defined by LOEa.OEb.
LOEa is surjective because OE1 is not in the range of LOEa.

Consider Exercise 1.7.11.

The 8x 12.mod 125/ is a congruence.

This exercise will show you a proof of the Chinese remainder theorem.

You can find x such as that x 3.mod 4/ and x 5.mod 9/.

The set Q of rational numbers, the set R of real numbers, or the set C of complex numbers are Polynomials.

I am confident that most readers will not be disturbed by the informality of defining polynomials as an expression of the form.

It is possible to codify the concept by defining a polynomial to be an infinite sequence of elements of K, with all but one entry equal to zero.
The sequence.3; 2; 7; 0; 0; would be interpreted as 7x2 C 2x C 3.
The operations of addition and multiplication can be changed to other functions.

The operations of multiplication and addition satisfy the same properties as those listed for the integers.

The properties can be verified using the definitions of the operations and the corresponding properties of the operations in K.

The reader is responsible for verification of the remaining properties listed in the propo sition.

deg.p/ is the degree of p 2KOEx.

If p D Pj aj xj is a nonzero polynomial of degree k, the leading coefficients of p are ak and akxk.

If the leading coefficients of the polynomial are 1 it is said to be monic.

The leading coefficients are.2 and.7 and the degree of p D is 7.

The goal of this section is to show that the theory of divisibility is similar to the theory of divisibility for the integers, which was presented in Section 1.6.

The results of this section are similar to the results of Section 1.6, with the proof also following a nearly identical course.
The role that absolute value plays for integers is discussed in this discussion.

If both of U are nonzero, then part (a) is done.

If either of them had a positive degree, the other would also have a positive degree.

U and v are elements of K.

If g D vf and f D ug are included, then g D uvg or g.1 uv/ D 0.

k D 0 meets the requirement if g D 0 does the same.

The reader is responsible for the remaining parts.

The prime number's analogue should not admit any nontrivial factorization.

It is possible to "factor out" an arbitrary nonzero element of K, but this should not count as a genuine factorization.
A nontrivial factorization f.x/ D g.x/h.x/ is one in which both of the factors have a positive degree and the analogue of a prime number is a polynomial.
They are called irreducible rather than prime.

If the degree of the polynomial is positive and it cannot be written as a product of two different degrees, it is irreducible.

The following statement is an analogue of the existence of prime factorizations.

A positive degree polynomial can be written as a product of irreducible polynomials.

The degree is proof.

The definition of irreducibility states that every degree 1 is irreducible.
Let f be a polynomial of degree greater than 1, and make the hypothesis that every polynomial whose degree is less than f can be written as a product of irreducible polynomials.
If f is irreducible, then it can be written as a product.
Each gi is a product of irreducible polynomials, and so is f.

There are infinitely many irreducible polynomials in KOEx.

There are fields with finitely many elements.

The same proof can be applied for such fields.

It is not true that KOEx has irreducible polynomials.

Every irreducible polynomial has degree 1 in COEx and degree 2 in ROEx.
There are irreducible polynomials in QOEx.
If K is a finite field, KOEx has many irreducible polynomials.

Let's use an example to recall the process of long division.

Let p D 7x5 C 3x2 C x C 2 and d D 5x3 C 2x2 C 3x C 4.
We want to find polynomials such that p D qd C r and deg.r/ deg.d /.

Let p and d be elements of KOEx.

There is a monomial m D bxk 2KOEx and a polynomial p0 2KOEx.

Put m D.an and p0 D p md.

The leading term for both p and md is anxn.

Let the elements of KOEx be p and d.

There are two types of polynomials in KOEx: p D dq C r and deg.r.

We divide p by d and get a monomial quotient of less than the degree of p. We continue in this fashion until we get the rest of degree less than deg.d.

The formal proof is based on the degree of p.

If deg.p/ deg.d /, then put q D 0 and r D p.

We can write p D md Cp0 in the lemma.

We have p D qd C r if we put q D q0 C m.

If g 2 koEx divides p and q, then g also divides f, which is a common divisor of nonzero polynomials p and q.

We are going to show that two nonzero polynomials have the same common divisor.

The greatest common divisor is unique to multiplication by a nonzero element of K. The most common divisor is monic, which has a leading coefficient of 1.
The one that is monic is the greatest common divisor.
The monic greatest common divisor of p and q is q/.

The results are similar to the results for the integers, and the proof are almost identical to those for the integers.

The proof of the analogous result for the integers should be modeled on the proof of the other results.
You will be able to better understand the proofs for the integers and how they have to be modified to apply to polynomials.

The set I.m; n/ D famCbn W a; b 2 Zg was an important part of our discussion of the greatest common divisor.

The set is similar to the one for polynomials.

The greatest common divisor is an element of I.f; g/.

The proof of 1.6.10 should be mimicked with the proof of 1.8.13.

We do repeat division with remainder, each time getting a portion of smaller degree.

The process must end with a zero remainder.

The final nonzero remainder is a common divisor of f and g.

The division with remainder gives.

C 9 x 3 x2 6 x3 C 6 x4 3 x5 C x6/ D

The most common divisor of f.x/ and g.x/ is d1.

The definition is d1.x/ D.1=4/s.x/f.x/ C.1=4/t.x/g.x.

If g:c:d:.f; g/ D 1 are used, they are prime.

If 1 2 I.f; g/ is the prime, then the two polynomials f; g 2KOEx are relatively prime.

If p divides the product, then f or p divides g.

One of the factors is divided by p.

The factorization of a polynomial into irreducible factors is unique.

Up to multiplication by nonzero elements, the irreducible factors are unique.

This is the end of our treatment of unique factorization.

You haven't yet learned any general methods for recognizing irreducible polynomials or for carrying out the factorization of a polynomial by irreducible polynomials.

We finish this section with some important results.

There is a polynomial q that is called p.x/ D q.x/.x a/ C p.a/.

If, and only if, x splits p.

When the remainder r is a constant, write p.x/ D q.x/.x a/Cr.

Substituting a for x will give you p.a/ D r.

2 K is a root of a p 2 koEx if p. In the irreducible factorization of p, the multiplicity of the root is k if x.

A degree n degree p 2 koEx has most of its roots in K, counting with multiplicities.

The sum of all roots is at most n.

The exercises ask you to prove parts of the proposition.

If you want to prove that multiplication in KOEx is commutative and associative, you have to show that 1 is an identity element for multiplication.

Let p 2 be irreducible.

The following statements need to be proved.

For each of the following pairs, find the greatest common divisor.

A computer program can be used to compute the greatest common divisor.

Make your program look for polynomials that are similar to the one you're looking for.

Show that g:c:d.f1; f2; : ; fk/ and make a reasonable definition.

Counting Counting is a fundamental and pervasive technique.

The method of inclusion-exclusion and the binomial coefficients will be discussed in this section.
First, the binomial coefficients and the binomial theorem are used to prove Fermat's little theorem, then inclusion -exclusion is used to get a formula for it.
The formula for the'function is used to get the generalization of the little Fermat theorem.

There are some problems with counting subsets of a set.

There are 8 D 23 subsets.

There are two possibilities for the n elements: the element is in the subset or not.

2n subsets of a set with n elements are possible because the in/out choices for the n elements are independent.

Define a map from the set of subsets of f1; 2; : : ; ng to the set of sequence; sn/ If you put a 1 in the j th position if j 2 A and a 0 in the j th position otherwise, the corresponding sequence will be formed.

It's not hard to check that.
There are just as many subsets of f1; 2 as there are of n 0's and 1's.

There are 2n subsets in a set with n elements.

There are 10 subsets of f1; 2; : : ; 5g and you can list them all.

The number of two-element subsets should be marked by.

It's known and equal to n.n 1/2.

In the following way, build the nS permutations of f1; 2; : : : ; ng.

This can be accomplished in a number of ways.

The numbers have some properties.

The number k 2 Z is a natural number.

Part (a) can be seen from the definition.

Both sides of (c) are equal to 1.

Both sides of (c) are zero when k is negative or greater than n.

Consider paths in the.x; y/-plane from.0 to a point.

We only admit paths of a C b "step" in which each step goes one unit to the right or one unit up, and each step is either a horizontal or vertical segment.
A path can be specified by a sequence of R's and b U's if the steps to the right and b are exactly the same.

Let the numbers x and y be the variables.

The sum of 2n monomials, each obtained by choosing x from some of the factors, is called.x C y/n.

The total number of subsets of an n element set is the sum over k of the number of subsets with k elements.

The binomial theorem is followed by putting x D y D 1.
The binomial theorem is followed by putting x D 1; y D 1.
The two sums in part (c) are equal by part (b), and they add up to 2n by part (a); hence each is equal to 2n 1.

P is a prime number.

I hope you discovered the first part of the next lemma while doing the exercises.

If a element is relatively prime to n, it has a multiplicative inverse.

There are numbers such as C nt D 1 if a is relatively prime to n. A and n have a common divisor if a is not prime to n. Say k t D a and ks D n. Reducing modulo n means that OEa is a zero divisor in Zn, and therefore not invertible.

0 in Zp is prime to p, so OEa is in Zp by part.

P is a prime number.

Since the case a D 0 is trivial and the case a 0 follows from the case a > 0, it makes sense to show this for a natural number.

The proof can be seen by the way in which it is presented.
The assertion is obvious.
If that is the case, then assume a 1/p and a 1mod p/.

1/ C 1.mod p/ D a is where the first and second congruences are from.

The conclusion of part a is the same as the conclusion of part D. By the previous proposition, if a is not divisible by p, OEa has a multiplicative inverse.

Adding both sides of the equation gives OEap 1 D. This is the same as ap 1 1.mod p/.

If you have three subsets of a finite set, you want to count A, B and C. If you just add jAj C jBj C jC j, you might have too large a result, because any element that is in more than one of the sets has been counted multiple times.

If an element lies in A B C, then it has been counted three times in jAj C jB C j.

Adding back jA B C j would fix this.

Our next estimate is jA [ B [ C j D jAj C jBj C jC j jA Bj jA C jB C jA B C j: 1.9.

We would like to get a generalization of Formula for any number of subsets.

Let us be any set.

The characteristic function of the relative complement of X is 1U nX D 1 1X and the characteristic function of the intersection of two sets is 1XY D 1X 1Y.

X 0 is the relative complement of a subset X U.

Let A1 and A2 be subsets of U.

Evaluate both sides of the equation at some point.

Such permutations are called derangements.

Ai doesn't depend on the k subsets.

Dn is the closest to nS.

Diners leave coats in a restaurant.

The formula was obtained before.

There is a proof of this in the exercises.

Take D 7 and D 4.

Take D 16 and D 7.

It is possible to prove the binomial theorem.

That's 3n D PnkD0

There are three exercises that outline a proof.

The prime to the prime p is assumed to be relatively prime.

Fermat's little theorem says ap 1 D 1 C kp for some k.

Each i is divided by pki.

Function from G G to G is what an operation or product on a set G is.

There is a product that is associative.

There is an inverse.

It is easy to make a concept out of the common char acteristics.

The process of elevating recurring phenomena to a con cept has several advantages.

C is a group of numbers.

The group H is the same as the group R.

It's clear that OE1 2 is what it is.

This proof is intended to show the power of abstract thought.

R is an abelian group.

A multiplication is associative.

In a ring, multiplication doesn't need to be commutative.

The ring is called a commutative ring if it is multi plication.

The number systems R, Q, C are rings.

A noncommutative ring is a set of n-by-n matrices.

Many rings have an identity element for multiplication.

The only units in the ring are.

The units are not zero constant.

As with groups, rings may be related to one another, and under standing their relations helps us understand their structure.

Q is a subring of Z.

Let T be a matrix with real entries.

The idea of the direct sum of rings is needed first.

The map is clearly defined.

One to one is what the map says.

The second assertion is more subtle than the first.

Every nonzero element is a unit.

Three fields are R, Q, and C.

Z isn't a field.

ROEx isn't a field.

Zp is a field if p is a prime.

K can be any field.

x 1

OExb is a ring isomorphism.

It's the science of secret or secure communication.

R. Rivest, A. Shamir, and L. Adleman10 are well known.

Let D pq.

Write h D tm C.

It is easy to find large prime num bers and to do the computations needed to decode.

I choose a small prime r D 55589 and send you an email.

"ALGEBRA IS INTERESTING" is your secret text.

It is not possible to factor 300 digit numbers.

One doesn't do such computations by hand.

Chapter 2.

In the previous chapter, we saw many examples of groups, and finally arrived at a definition, or collection of axioms, for them.

We have e D ee0 since it is an identity element.

If hg D e, then h D g 1.

If gh D e, then h D g 1.

When gh D e is similar, the proof is there.

G is an element of a group D.

Since gg 1 D e, it follows from the proposition that g is inverse of g 1.

La stands for left multiplication.

G by Ra.x.

Ra stands for multiplication.

There is an assertion that the map La has an inverse map.

The computation shows that La i La 1 D idG.

The inverse maps of La and La 1 are shown.

Ra's proof is similar.

Consider Z12 with multiplication.

The order of a group is based on its size.

There are examples of groups of small order.

A group of order 4 is the set of symmetries of the rectangular card.

The isomorphism is the map.

If one of the two groups is abelian and the other is nonbelian, the two groups are not isomorphic.

S3 and Z6 are both abelians.

Z1 is the unique group of order 1.

Z2 is the unique group of order 2.

Z3 is the unique group of order 3.

Z5 is the unique group of order 5.

All groups of order are abelians.

This shows that.

A set M with an associative operation is referred to as juxt sitapoion.

The product of five elements at a time is well-defined and independent of the way that the elements are grouped for multiplication.

M is a juxtaposition.

The product of one element is D a.

The product is defined by (a) and (b).

Fix elements a1; : : an 2 M.

1 element at a time is fined.

Determine the symmetry group of the rhombus.

If H is abelian, G is too.

Group with a small num ber of elements are investigated by means of their multiplication tables.

The group also has four elements.

G is abelian.

For all a; b 2 G,.ab/ 1 D a 1b 1.

B 2 G, aba 1b 1 D e.

2.2. Subgroups and Cyclic Groups

M is a set with a not necessarily associative operation.

A1a2 an 2 N.

H is a subgroup of G.

The inverse h 1 is an element of H.

H can be a subgroup.

These observations can be used as a labor-saving device.

If At A D E is an n-by-n matrix, it is said to be orthogonal.

.AB/t D Bt At D B 1A 1 D.AB/ 1 is related.

O.n; R/, then.A 1/t D.At /t D A D.A 1/ 1, so A 1 2 O.n; R/.

G itself and feg are part of any group.

Appendix D can be found here.

There are eight 3-cycles in S4.

We will have a theory that will make the result transparent.

C is hA.

feg is the smallest subgroup.

There is a subgroup that appears in all groups.

The word lattice is used in many different ways in mathematics.

A k D is defined as a 1/k.

The generator is a part of the group.

There is a subgroup of Z called HD i D Z D fnd W n 2 Zg.

H1i D h 1i D Z is a cycle.

Since hOE1i D Zn, it is cyclic.

A group of order n with generator D e2i is called Cn.

There is a chance that two powers coincide.

The order of the by o.a/ is marked by us.

The order of OE5 is 6.

Any two groups of the same order are isomorphic.

There is a group G.

The isomorphic to Z is if a has infinite order.

The group Zn is isomorphic to a finite order.

In this section, we determine all of the groups.

H should be a subgroup of Z.

If d 2 N, then d Z S Z.

If b divides, then aZ bZ.

Let's check part (c) first.

If bja, then a 2 bZ, so aZ bZ.

We verify part a. H should be a subgroup of Z.

H D qd 2 Z.

We have shown that there is a d 2 N and a d Z D H.

Every subgroup of Z is related to Z.

Since d divides n, this is just n.

H is a subgroup of Zn.

Either H D fOE0g or there is a d > 0 such that H D hOEd i.

H is a subgroup of Zn.

The number is divided by d.

There is a subgroup of Zn.

The subgroup of Zn has a dividing n.

The parts are from the proposition.

Part (b) is an exercise for the reader.

The order of OEb in Zn is n/.

The generators are OE1, OE5 and OE11.

The subgroup of a group is called cyclic.

A group is an element of finite order.

A group is an element of finite order.

A group is an element of finite order.

The group has an order '.2n/ D 2n 1.

3 is not a pattern.

The subgroup of S4 is.12/.34/;.13/.

24/; and.14/.23/g.

They are either the same or inverse if they have three digits in common.

This is a 3-cycle.

H is a subgroup.

A group under matrix multiplication can be formed by the set of R that varies through real numbers.

Let S be a subset of the group G.

Let's be a part of the group.

It is a part of order 6.

The order of.1234 is 4.

The order of the two disjoint 2-cycles is 2.

The subgroup lattice of Z 24 should be determined.

The subgroup lattice of Z30 needs to be determined.

H 0 and jH j divide.

This is the conclusion of corollary 2.2.26(b).

There is an order of 2n 2 for n D 3; 4 and 5.

2.3. THE DIHEDRAL GROUPS

These figures are capable of rotation in three dimensions.

D is a subgroup of N D.

Write j D j0 for the flip over the x- axis.

These relations can be used to calculate all products in D.

A 1 D N for all a 2 D is achieved by the subgroup N of D.

The symmetries of the regular polygons are next.

The n-gon has a symmetry group.

The "flips" are symme tries of the n-gon.

There is a regular n-gon in the plane.

Writing N D, show that D D N.

Linear transformations of R3 are used to implement the symmetries of the disk.

JRt D R t J is related to JRt D R t J.

The group Dn is related to the n-gon.

2.4. HOMOMORPHISMS AND ISOMORPHISMS

There is a subgroup of D6 that is similar to D3.

In a regular 15-gon, every third vertex is painted red.

There is no requirement to be injective or surjective.

To make these observations more useful, we number the edges of S.

The square's edges are labeled.

To verify the homomorphism property from the symmetry group of the square into S4.

The diagonals of the square are labeled.

Let G be an abelian group with a fixed number.

Let's look at some general observations.

There are group morphisms.

H is a groupmorphism.

For each g 2 G, '.g 1/ D.

It's followed by the proposition 2.1.1(a) that '.eG/ D eH.

We recall a math ematical notation before we say the next proposition.

f 1.B/ is the conventional notation for the preimage of B.

H is a groupmorphism.

Each subgroup is a subgroup of H.

Under multiplication and inverses, '.A/ is closed.

The elements of '.A/ are h1 and h2.

A such that D '.ai for I D 1 and 2.

The proof of part is an exercise.

A normal subgroup is called G.

Here gNg 1 means 1 W and 2 N.

H is a groupmorphism.

According to the proposition, ker.

We now know that it is a normal subgroup.

H is a groupmorphism.

It's a normal subgroup of G.

It's enough to show that g ker.

The opposite containment needs to be shown.

If we replace g by g 1, we get that for all g 2 G, g 1 ker.

If it is injective, eG is the pre image of eH.

If that is the case, suppose that ker.

W k 2 Zg is the code for OEk.

There are more examples of homomorphisms explored in the exercises.

The sign is called the homomorphism.

The alternating group is also referred to as this subgroup.

The odd permutations contain.12/An.

If k is even, a k-cycle is even.

Do you know what the kernel is?

There is a group with the same name.

H be a combination of G and H.

Let x2 be variables.

It's a homomorphism from Sn to f1; 1g.

If a permutation is a product of k 2-cycles, then./ D.

If,./ D 1, the parity is even.

T./ is a form of Homomorphism.

Two elements a and b in a group G are said to be conjugate if there is an element g 2 G.

When two elements are conjugate is determined by this exercise.

Sm can be considered as a subgroup.

2.5. COSETS AND LAGRANGE'S THEOREM

It is an automorphism and preserves parity.

The set of matrices with positive determinant is a normal subgroup.

Rn is considered to be an abelian group under vector addition.

Let G be a group of abelians.

Consider the subgroup H D fe.

The right coset of H in G is called H g.

It takes a little labor to do this computation.

Similarly, (b) means that bH aH.

H is a subgroup of a group.

Each left coset is nonempty and the union of left coset is G.

If aH bH 6D, let c 2 aH.

The restriction to aH is a combination of aH and bH.

It makes sense for infinite groups.

There are co sets of nZ in Z.

G is the only subgroup other than feg.

There are two groups of prime order p that are isomorphic.

The order of the subgroup is the cardinality.

K H G are part of a subgroup.

We have to use another approach if the groups are infinite.

K H G are part of a subgroup.

The group S3 is considered.

N is not abnormal.

The left and right coset of N are the same.

Using the criterion of the previous exercise, N is normal.

2.6. EQUIVALENCE RELATIONS AND SET PARTITIONS

The center of a group is a normal subgroup.

Do the same for R/.

N is not abnormal.

An equivalence relation is an example of that.

If x and y are the same, then x and y are the same.

A set partition is an example.

In mathe matics equivalence relations are very common.

Equality is an equivalency relation on X.

It's a good idea to be a natural number.

The set of all lines in the plane is an equivalent relation to the parallelism of lines.

Is there an equivalent relation on X.

Let's say we have an equivalent relation on X.

x 2 OEx D OEy, so x y.

If it's 2 OEx, then it's x.

An equivalent relation on X and y 2 X is needed.

Consider an equivalence relation with a set.

The partition of the set X is the collection of equivalence classes.

This is an equivalency relation.

In x and y, the definition of x P y is clearly defined.

An equivalence relation on X is given by every partition of a set X.

We need to show that P D P 0.

That is, Oeda D A.

Let X be whatever you want it to be.

Group G and subgroup H are considered.

Be any map.

The fibers f 1.y/ for y 2 Y are the equivalent classes.

An equivalent relation on a set X is needed.

Such as that 0 D s I f.

Such as that 0 D s I f.

Such as that 0 D s I f.

Choose x 2 X such that y D is.x/.

This is the final proof that s is aective.

There is a set of left cosets of H in G.

We have 1.aH / D fb 2 G W bH D aH g D aH.

The classes for conjugacy are called equivalence classes.

2.7. Quotient Groups and Homomorphism Theorems

The conjugacy class of g is fgg.

Consider a map from a set X to another set Y.

Consider the permutation group with its normal subgroup.

The structure of a group can be found in the set G of left cosets of a N in G.

A D aN and B D bN are used.

We have to check that the result is independent of the choices of 2 A and 2 B.

We have a 1a0 2 N and b 1b0 2 N since aN D a0N and bN D b0N.

Since N is normal, b 1.a 1a0/b 2 N.

This multiplication is a group.

The quotient group of G is called G by N.

The quotient homomorphism is called G.

The normal subgroup is nZ D fn W 2 Zg.

It is surjective because it has the same range.

It is an isomorphism.

Z is a normal subgroup since R is abelian.

Z is precisely the kernels of '.

The group isomorphism is from the quotient group R to T.

The inverse of TA;b is A 1b.

There is a condition for two elements to be congruent.

A D A0 is equivalent to this.

It's an isomorphism of groups.

G be a surjective homomorphism.

In a group H, let's be an element of order.

A normal subgroup of GL.n; R/ is the setSL.n; R/ of matrices of determinant 1.

Z is a normal subgroup of G and is the center.

There are at least two natural models available.

It is the same as the invertible matrices Z.

For each subgroup B of G,'1.B/ is a subgroup of G.

A priori is contained in B.

For a subgroup of G containing N.

This is the same as a 1x 2 ker.

Let B D'1.B/.

B is normal in G.

B is normal in G.

Gxg 1 2 B, by normality of B, so Ng Nx Ng 1 2 '.B/ D B.

In G, B is normal.

Let K be a normal subgroup of G.

It is a composition of surjective homomorphisms.

That is, D K.

We can use the homomorphism theorem again to identify G with N.

The identification carries K to the image of K in G.

The quotient map is marked by G.

Is it K-N-G-N?

The conclusion follows from the first one.

If aN D bN, then b 1a 2 N K D ker.

Z, and in particular OEa has an order.

Z is a surjective homomorphism.

If k is divided by, then kZ Z Z.

A is a subgroup of G.

An is a subgroup of G.

The group of n-by-n invertible complex matrices is known as G D GL.n.

Z is a subgroup of the matrices.

A Z D A fE W is a root of unityg.

If n is odd, the same holds with C replaced by R.

If n is odd, then GL.n; R/ DSL.n; R/Z.

G be a surjective homomorphism.

'.aN / D '.a/ is well defined.

This gives an associative product on some subsets.

Take A D aN and B D bN.

N is a normal subgroup.

The inverse of TA;b is A 1b.

G is a finite group.

Aut.G/ is a group.

Show that the map is called Z.G/.

The map c isn't surjective.

Is every automorphism of S3 inner?

If G is abelian, you need to prove it.

Suppose G is abelian.

Chapter 3.

The multiplication is done coordinate by coordinate.

Za is also known as Zb.

Both of these groups are normal in A B.

It's evident that Assertion (c) is true.

It is similar to hOEbi hOEai S Za Zb.

The group of symmetries of the rectangle should be called G.

The difference is more psychological.

This is a follow up to example 3.1.3 and k.

Group operation C is for abelian groups.

Nr are subgroup with N1 C N2 C Nr D G.

Take G to be Z2 Z2.

The two subgroup have a trivial intersection.

The coordinate-by-coordinate multiplication on the product P D A1 An of groups makes it into a group.

Group P D A1 An has a direct product.

B are surjective.

A B is defined by D.

I am the same as in the previous exercise.

There are conditions for to be surjective.

Just as in a direct product, Dn D NA, N A D feg, and A S Dn is the same.

We have a commutation relation for r 2 N.

We have G D NA D AN, AN D feg, and A S G.

The construction is easy.

A is a part of the automorphism group of N.

N A is a group under the multiplication.n.

We need to check the associativity of the product.

Let.n; a/,.n0; a0/, and.n00 be elements of N I A.

It's easy to know if it's the identity of N I A.

We have to find the inverse of an element.

This is an exercise.

The direct product is a special case of the semidirect product, with the homomorphism, trivial,,.a/ D idN for all a 2 A.

Aut.Z7 is defined by,.OEk/ D 'k.

Do you know if Zn I Z2 is isomorphic to Dn?

Z4 has a subgroup that is similar to Z2.

The quotient Z4 is similar to Z2.

Z4 isn't a direct or semidirect product of Z2.

V 2 K, v 2 V are for all.

A K-vector space is a type of space over K.

The parts were proved above.

The range of T is T.V.

A linear transformation from V to C is called R 1 f.t/g.t/ dt.

One verification is needed for the sake of illustration.

The quotient map is a group Homomorophism.

V is a linear map of spaces.

V is the quotient map.

T respects multiplication by numbers.

The details are left to you.

There is a surjective linear map.

Let M be a subspace of V.

A linear isomorphism is defined by T.x/ C M.

T.x/CM is a group isomorphism.

V be a linear map.

The quotient map is marked by V.

M is V is N.

T.v/ defines a group homo morphism from V to M.

A and N should be subspaces of V.

A and N are contained in V.

S is dependent on other things.

A linear independent set can't contain the zero.

This shows that fen g is independent.

There are two things to notice: enem D enCm and e0 D 1.

Equation gives,1 D 0 as well.

A is a proper subset of B and v 2 B n A.

The relation is contrary to the independence of B.

Let A0 D A n fv1g.

A basis of V is B.

The set is called B.

B is a subset of V.

B is a basis.

It follows the same way as span.B/ D V.

V is said to be infinite-dimensional.

V has a finite basis if it is finitedimensional.

A subset of a set is a basis.

B is a basis of V.

If fv1 is linearly independent, then C is the zero matrix.

Let V have a finitedimensional space with a spanning set.

Let Y D be a linearly independent subset of V.

From Propostion 3.3.25, any basis of a finitedimensional space is finite.

Dim.V / D 1 is written if V is infinite-dimensional.

W is finite and dim.V is dim.

The basis of W is V.

Y is a subset of W.

The basis of V is contained.

A basis of V is the proposition 3.322, B.

The lemma is an axiom of set theory.

There is a zero-dimensional space with one element.

The basis of the empty set is unique.

Oen is the standard ordered basis of Kn.

The two n-dimensional spaces over K are linearly isomorphic.

The case is left to the reader.

Be a function.

There is an equivalency between (a), (c), and (d).

N is a subspace of V.

Let fw1 be a basis of W.

T I S.Pi,i wi / D T.Pi,i xi / D Pi,i wi.

Suppose v 2 ker.T / W 0.

This shows that ker.T / W 0 D f0g.

We showed that V D ker.T / W 0.

Let V be a finite-dimensional space and N be a sub space.

There is a subspace M of V such as V D N.

Complements of a subspace are not unique.

S is a subset of a space.

Rather than using the proposition.

V is not a straight line.

Every subset of V is finite.

V does not admit a sequence of linearly independent subsets.

It is infinite-dimensional.

V has a subset that is infinite.

Let fv1; v2; : : : ; vsg be a basis of N.

It's easy to check that S C T is also linear.

The map defined by is the inverse of a linear map.

V should be over a field K.

HomK.V; W is a vector space.

V D HomK.V; K/ is a vector space.

B spans V.

The coefficients are zero, which shows that B is linearly independent.

The basis of V dual to B is called B.

The dimensions of V are 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609- 888-609-

The number f.v/ is given by the pair of v 2 V and f 2 V.

There is a map called K.

V is the same.

V is a finitedimensional space over a field.

fvg is linearly independent if v is not a zero.

There is a basis B of V with v 2 B.

V is a linear isomorphism.

It's essential that finite dimensionality is present.

.V is not surjective.

S and T are subsets of V and W is a subspace of V.

I'm a subspace of V.

The parts are left to the reader as exercises.

We have W W II and W W III for part (d) and part (c) respectively.

All these subspaces are equal.

The dual space of sub spaces and quotients can be described with the aid of annihilators.

We can check the surjectivity of this map.

Let T 2 HomK.V; W /.

HomK.V has dim.V and dim.W.

The reader is invited to look at the map.

Kn is a linear isomorphism, meaning that it takes a vector in W to its coor dinate vector with respect to C.

HomK.V; W is the first column of OET C;B.

OET C;B is injected.

T.vj / D PniD1 ai;j wiWD.

It is surjective.

It's immediate from (a).

Let B D.v1 and C D.w1 be used.

OES T D;B and OES D;C OET C;B agree.

There is a ring isomorphism between EndK.V and Matn.K.

The inverse OEidC;B is invertible.

There is a problem of determining the matrix of a linear trans formation with respect to two different bases.

Two bases of V are B and B0

The part was proved above.

Let A be an n-by-n matrix, Q an inverted n-by-n ma trix, and set A0 D Q.

Let E D.

B0 is a basis of Kn.

It's just Q 1.

It's enough to be able to compute OEidB;E and OEidE;B.

Matn.K is an equivalent relation to EndK.V.

The determinant and trace are important similarity invariants.

The sum of the diagonal entries is the trace of a square matrix.

Let C 1 D.di;j be an inverted matrix.

The trace is similar to the one before it.

Section 8.3 shows how the determinant is dicussed.

There are parts to Lemma 3.4.6.

Let W be a subspace and let V be a finitedimensional space.

Let W be a subspace and let V be a finitedimensional space.

The entry is equal to the one on the wii.

Both T 0 and T 0 are linear transformations.

As D fa1 C C as W ai 2 Ai for all ig.

A finite group is finitely generated.

There is a Pi ai xi.

This is the standard basis of Zn.

Abelian groups don't need to be free.

xng is a basis of G.

The map in (b) is a group homomorphism.

This shows that there is no difference between (a) and (b).

S is a subset of Zn.

It makes sense to consider linear independence of S over Z or over Q.

A minimal generating set is the basis of a free abelian group.

Let b 2 B n B0

B0 is contrary to the linear independence of B.

B0 doesn't generate G.

Any basis of a free abelian group is finite.

Since S is finite, it is contained in the subgroup B0 of B.

Let G be a free abelian group.

The base case n D 1 is taken care of by Z.

Let N be a subgoup.

The subgroup of Z is composed of Zn to Z and,n.N.

Let A be a subgroup of G.

A is a generating set with no more than n elements.

N is free.

If ai;j D 0 is diagonal, then D j is.

A should be an m-by-n matrix.

A version of Gauss ian elimination is used to Diagonalize the matrix A.

In Matm.Z/, E CVEi;j is inverse.

Two rows are interchanged in the second type of elementary row operation.

Pi;j is its own inverse.

There is a Smith normal form on my page.

Elementary row oper ations are similar to elementary column operations.

A has a nonzero entry in the.1; 1/ position.

Minimum size is nonzero.

Take note of the.1; 1/ entry of the matrix.

A is the row matrix A D.

A1; : : ; an.

Let.v1 be a sequence of elements of Zn.

The j th column of the n-by-n matrix is vj.

In Matn.Z/, P is invertible.

trivially means condition

Let A D.ai;j.

If N is not a trivial subgroup, there is nothing to do.

Let xsg be a generating set for minimum cardinality.

All the dj are nonzero.

Vng is a basis of zinc.

The dj's are all nonzero.

It's free, of rank no more than n.

N should be used to indicate the kernels of '.

There are be abelian groups and Bi Ai subgroup.

If it is a finitely generated tor sion group, an abelian group is finite.

The least positive element of ann.G/ is the period of G.

The period and order of G coincide if G is cyclic.

Z2 Z2 has periods 2 and 4.

G is a finitely generated abelian group.

If G is free, it is abelian.

Part a is an exercise.

A is a free submodule and B is a free submodule.

The rank of B is determined.

This proves something.

Any module that is free is free.

If p splits, then Zx is Zx S Zp.

If p does not divide, then pZx D Zx.

PZx is Zx.

Zx D pZx follows.

p is a prime number if G is an abelian group.

G is a space over Zp.

G is an abelian group.

Check the space axioms.

G is a surjective group.

The proof of uniqueness is in Theorem 3.6.2.

The period of G is marked as D bt D a.

It is followed by the unique of dimensions.

P is a prime number.

Fix a problem.

We have ai D bi D p.

The in variant factor decomposition is called the direct product decomposition.

Where p is a prime, let G be an abelian group of order.

There is a unique sequence of exponents.

A partition is a sequence of natural numbers such as Pi ni D n.

Let G be part of a group of order.

The type of G is the partition.

P should be a prime.

Gi D 2 G W, i x D 0g.

The most common divisor is rsg.

Suppose that xj 2 Gj for 1 j s and Pj xj D 0.

G is a finite abelian group.

If p does not divide the order of G, GOEp D f0g.

p is a prime number.

GOEp is a subgroup of G.

The result is given by applying the previous proposition with D pki.

The primary decompo sition of G is the decomposition of Theorem 3.6.14

The Zn D G1 Gs can be computed.

GOE5 D hOE12i is a subgroup of Z60 of size 5.

We can get a procedure for resolving any number of simultaneous congruences using the same considerations.

There is an x such that is x xi.mod.

Up to congruence mod n D, x is unique.

Take ri D n as a first approximation.

5 D 60, r1 D n=4 D 15, r2 D n=3 D 20, and r3 D n=5 D 12.

Put y1 D 45, y2 D 40, and y3 D 36.

mod 60 can be reduced to find a unique solution.

A subgroup of a finite abelian group G is GOEp.

The sum of those Ci whose order is a power of p is noted.

Consider G D Z30 Z50 Z28.

For all other primes, GOEp D 0.

The abelian group is a direct product of the prime power order.

The elementary divisors of G are the orders of the factors.

Look at the example G D Z30 Z50 Z28 again.

The elementary divisors are 4; 2; 3; 25; and 5.

To get the largest in variant factor, multiply the largest entries of each list.

The largest entry should be removed.

Continue until all the lists are exhausted.

The biggest invariant factor is 4 25 3 7 D.

Classify the abelian groups.

The order '.N' is where the function is.

There is a multiplicative group of nonzero elements.

They are both cyclic.

Part (a) is a follow up to the first part.

Let p be an odd prime.

It is a cycle.

Let G be a group of abelians.

Show that ZB is free and that G is a group.

Find the elementary divisor and invariant factor decompositions for each group.

Find the elementary divisor and invariant factor decompositions for each group.

Consider Z36.

Consider Z180.

Let G be a finite abelian group.

Chapter 4.

The regular polyhedra are the cube, the dodecahedron, and the icosahedron.

If you have models of the regular polyhedra, we can get their rotation groups.

The first thing we do is the tetrahedron.

Including the identity, we have 12.

The crossproduct of Ov1, Ov2, and Ov3 is referred to as Ovi.

One of the symmetry of the tetrahedron is 5.

The cube is now being considered.

We need to find an injective homomorphism into some permuta tion group that is caused by the action of the rotation group on some geometric objects associated with the cube.

The rotation group of the cube is changed into S4.

S4 is similar to the group of 3-by-3 signed permutation matrices.

The f Ovi g form a right-handed orthonormal basis with the help of two further unit vectors.

R should be a polyhedron.

The dodecahedron has six 5-fold axes through the centroids of the oppo site faces.

The dodecahedron has 15 2-fold axis through the center of the oppo site edges.

The dodecahedron has 60 tries.

Consider any edge of the dodecahedron.

One of our families has six edges.

There is a cube in the dodecahedron.

By explicit computation, we could do this.

The rotation group has 60 elements and must be A5.

The rotation groups of the dodecahedron and icosahedron are similar to the group of even permutations A5.

The data of the previous exercise and the method of Exercises 4.1.2 and 4.1.3 can be used to calculate the matrices for the icosahedron rotation.

To be a unit, we need to be a unit with a P.

I mean the inner product in R3.

Symmetry of the figure is a type of reflection that sends a geometric figure onto itself.

It is easy to check the multiplication table under multiplication and inverse.

We have 14 matrices that allow us to compute a lot of products.

These are new matrices.

There are sixteen symmetries of the square tile, eight rotation, five reflections, and two reflection rotation.

A group is the set of sixteen symmetries.

The formula for the reflection J needs to be verified.

J is linear.

The standard basis of R3 is found with its matrix.

There is a plane that doesn't pass through the origin.

There are eight non rotation matrices in the set of eight rotation matrices.

The sixteen symmetries form a group.

It preserves distance, d.a/, b/, for all.

Let's use an isometry of Rn such as.0/ D 0.

Another characterization of matrices is given by exercise 4.4.3

F D ffi g is an orthonormal basis.

There is a linear map on Rn.

There is a matrix with respect to some orthonormal basis.

Every orthonormal basis has a matrix.

U is an isometry.

The determinant is.1.

Let A be a matrix.

A change of basis matrix is called V.

A linear isometry's determinant is.1.

The SO.n; R/ is a set of n-by-n matrices with a determinant equal to 1.

A matrix is a reflection matrix.

The role of reflections and rotation in SO.2; R/ should be sorted out next.

Consider the situation in three dimensions.

C1 is an eigenvalue for any element of SO.3.

jP must be a reflection.

A 2 SO.3 would be R/.

A has an eigenvector v with eigenvalue 1.

jP is a rotation of P.

Suppose A 2 O.3; R/ n SO.3; R/.

So is a reflection.

It is a reflection.

Show that j is not a straight line.

Show that j, 1 D j., if it's a linear isometry.

If A is a matrix with a single digit number, show it.

The columns of A are normal.

An orthonormal basis is the rows of A.

There are three reflection matrices that make up a matrix of rotation-reflection.

Every symmetry is implemented by a linear isometry of R3 for "reasonable" figures.

I don't distinguish between linear isome tries and their standard matrices.

R is an index 2 subgroup of G and G D.

The full symmetry group of the cube has 48 elements.

The figure must come in two versions that are mirror images of each other.

A band with symmetry.

Observe that i2 D 1 and that for any rotation.

Chapter 5.

Let's do an action of G on X.

x0 is sent to x00, then x1 is sent to x00.

O.x/ is the orbit of x.

Group G acts on its own by left multiplication.

Let G be any group and H any subgroup.

The action is not straight forward.

If they have the same cycle structure, they are conjugate.

The set of subgroup of G should be any group and X.

G acts on X by conjugate, cg.H / D gHg 1.

Let G do something.

Stab.x/ D fg 2 G W gx D xg.

Stab.x/ is a subgroup of G.

Let G act on X and x 2 X.

This calculation shows that is defined.

G is finite.

Consider the actions of a group G.

The partition of n determines the parameters of cycle structures.

The identity is the only element of cycle structure 14.

There are three elements.

There are three-cycles.

There are four-cycles.

Let X be the family of k element subsets.

The proof is the same as before.

Let D r1 C r2 C rk.

Group G can act on a set.

Gx D fgx W g 2 Gg.

On the set f1; 2; : : : ;ng, the symmetric group Sn acts naturally.

H D fe; 2/g S3.

H is normal.

There are two blue and two white beads.

There are four arrangements and two other arrangements.

Consider the set F D f.g; x/ 2 G X W gx D xg.

Every arrangement is fixed.

The rotation of the nonagon is 2.

3 and hrki have three different paths.

A white bead is required for any fixed arrangement.

There are 12 fixed points in X.

There are many ways to color the faces of a dodecahedron.

A mathematical object is a set.

Aut.G/ is a subgroup of Int.G/.

W. R. Scott wrote Group Theory.

Z, and let r D.

The previous result and corollary 3.6.26 have been followed.

Several abelian groups are deter mined in the exercises.

Aut.Z2 Z2/ S S3 can be shown in two different ways.

Consider the actions of a group G.

24 D 1 C 6 C 3 C 8 C 6 is given.

There are nonidentity elements in G.

G, of order p2, is not a constant.

It is possible to choose h 2 G n hgi since o.g/ D p.

They commute when g and h are both of order p.

I say that hgi hhi D feg.

G is abelian.

Let G be a group of order.

Every subgroup is normal if G is abelian.

The power of a prime number is called jGj D pn.

Suppose the result holds for all groups of order pn0, where n0 is.

There is a quotient map.

The existence of sub groups of order a power of prime is being investigated.

If b 2 G and b D e, then also ba D e.

If ap/ 2 X, then ap 1/ 2 X as well.

The fixed point of X is e.

OEG W H is divided by p.

H should act by left multiplication.

H is fixed.

Let's look at the condition for a coset xH to be fixed.

This is possible for each h 2 H, hxH D xH.

That is the number of h 2 H x 1hx 2 H.

The set of g 2 G is called gHg 1 D H.

The normalizer is NG.H / H.

We can considerNG.H, which has a size that is divisible by p.

P is a p-Sylow subgroup.

The set of conjugates of P is found in the family X of p-Sylow subgroups.
P should act on X.

There is a subgroup of order.

Let p and q be primes.

Any group of order pq is cyclic.

Group of order pq.

The subgroup of G of order pq is called PQ D QP.

G S Zp Zq S Zpq must be trivial.

The number n5 of conjugates of R is con gruent to 1 and divides 30. n5 2 f1; 6g.

R has 6 conjugates if it is not normal.

We have Aut.Z15/S Aut.Z5/Aut.Z3/S S Z4 Z2.

The value of Aut.Z3/ is 'i'.

We can determine all groups of order 28, up to isomorphism.

Let G be a group.

G is congruent to 1 and divides 28, so n7 D 1.

A subgroup of order 4 is called a 2-Sylow subgroup.

The abelian groups of order 28 are Z7 Z4 and Z2 Z2.

A7 D b4 D 1 and bab 1 D a 1 are elements that make up this group.

There are two exercises called 5.4.13 and 5.4.14.

The subgroup is similar to Z2 Z2.

Z2 is a surjective group.

The show is determined by the kernel.

The non-abelian group of order 20 should be categorized.

The non-abelian group of order 18 should be categorized.

The non-abelian group of order 12 should be categorized.

Without loss of continuity, this section can be omitted.

The techniques of the previous Sec tions will be used to classify the transitive subgroup of S5.

A5 is the only normal subgroup of S5 other than S5 and feg.

The subgroup of S5 that is not equal to S5 or A5 is called G.

Then OES5 W G 5; that is, jGj 24.

I claim that this action is faithful because of the morphism of S5 into Sym.X and S Sd.

The kernels is a subgroup of S5.

2 G is an element of order 5.

A 5-cycle is the only possible structure.

D.1 2 3 4 5/.

That is normal in G.

We know that for any 2 S5.

The group's cardinality is no more than 20.

We have G h; I.

Let H be a subgroup and let G be a finite group.

Group G acts on sets X and Y.

On a set X, let G act transitively.

The subgroup generated by.1234/ and.14/.23/ is D4 S4.

There is a set of conjugates of D4 in S4.

The resulting homomorphism of S4 to S3 should be explicitly computed.

P should be a prime.

The power of p is equal to G.

N is a normal subgroup of a group.

Let G be a finite group with a prime and a subgroup.

P should be a 7-Sylow subgroup.

Chapter 6.

The definitions of rings and fields were encountered by us.

R is an abelian group.

A multiplication is associative.

That is, Ra.nb/ D nLa.b.

The number systems Z, Q, R, and C are rings.

In Section 1.8, we have talked about polynomials in one variable over a field.

QOEx is an element of 7xyz C 3x2yz2 C 2yz3.

The set ROEx1; is a commutative ring with multiplicative identity.

Let R be a field and let X be any set.

The element of S is the opposite of x.

If no proper subring of R contains S, R is generated as a ring.

Let S be a subset of EndK.V.

The following construction is inspired by the previous example.

Let G be a finite group.

You are asked to confirm that ZG is in the exercises.

Rn is the direct sum of R1 and R2.

There is a field.

The group Z2 is written as fe; g, where 2 D e.

It is not automatic.

R can be any ring with multiplicative identity.

First, for any 2 R and 2 Z,.n 1/a D and a.

It is not always injective.

Consider the ring C.R/ of continuous real-valued functions on R.

There are more examples of ring homomorphisms in the exercises.

We define 'a' by this formula.

R0 is a ring.

xn to R0 extending'and sending each xj to aj is what we would like.

If the elements aj are mutually commuted, it makes sense.

R0 is a ring.

The proof is the same as the one variable substitution principle.

We usually use p.a/.

Set a D x.

Let T 2 EndK.V be a vector space over K.

What does this mean, and how does it follow from the proposition EndK.V?

Moreover, 'T.Pi i xi / D Pi.i I D Pi i T i.

We usually write 'T.p/'.

R is a ring with a multiplicative identity element.

The map is in the other direction.

S is a ring ker.

Since it is a subgroup, it is a Homomorphism of abelian groups.

If R and x are the same.

R x 2 ker.

Similarly, xr 2 ker.

nZ is the name of the person.

R can be any ring with a multiplicative identity element.

The latter does not.

Consider the situation of Corollary.

K can be a field.

If K is finite, it is never injected.

Any field is a simple ring.

Part (a) is an exercise.

There are 2 N such as x 2 Ik and y 2 I.

x C y 2 in I

If x 2 I and r 2 R are combined, then there is 2 N such as x 2 In.

I'm an ideal.

R is a left ideal of RS.

R has an identity element.

It's easy to check thatRS is a left ideal.

hSi CRS is a left ideal.

J is a left ideal.

The ideal is two-sided.

If R has an identity element, then hSi C RSR D RSR.

Every ideal is a principal.

The ideal in the field is the principal.

S is a principal ideal of Z.

The proof of (c) is similar to the proof of 2.2.21

g D qf 2 koExf.

Consider a direct sum of rings.

Ri is an ideal person.

The map is a ring isomorphism.

Show that the ring is a unital one.

S should be a part of a set X.

Since y is a matrix, it is possible to write it down.

A nonzero homomorphism of a field is injective.

M is an ideal in a ring R with identity and a 2 R n M.

R should be a ring without identity.

The multiplication in R is well defined.

It is easy to check the ring axioms once we have checked that the multiplication is defined.

If 1 is the multiplicative identity in R, then 1 C I is the multiplicative identity in R.

The ring Z is the quotient of the ring Zn.

G.x/ C.f / D r.x/ C.f.

The example is K D R and f.x/ D x2 C 1.

The following is the basic homomorphism theorem.

It also respects multiplication.

All of the polynomials are in the kernel.

x2 C 1 is a multiple of g.

Since C is a field, ROEx is a field.

The rule of multiplication in C is the same as the rule of ROEx.

Let B D i Bi go.

There is a problem with showing that is surjective.

The subgroup of R and the subgroup of R containing J are both represented by 1.B/.

Each of the next three results is an analogue for the rings of a homomor phism theorem.

Let me be an ideal of R and D.

R is a ring isomorphism onto R.

'.x/ C I.

The quotient map is marked by R and J.

Is it R or J?

The group homo morphism from R to R is defined by '.x/.

R be a surjective homomorphism of rings.

I D fa C r W a 2 A and I g.

It's simple.

If x 2 R is a nonzero element, then R is simple.

Since R is simple, R D Rx is nonzero.

This follows from the previous ones.

Rather than appealing to the result of the proposition.

If.n is a prime, show that nZ is maximal ideal in Z.

The Chinese remainder theorem is given a version by this exercise.

State and prove a generalization of the result for the ring of polynomials.

The ring's nonzero elements can be zero.

You are asked to verify the examples.

An integral domain is the ring of Z.

Any field is an inseparable part of it.

If R is an integral domain, then ROEx is an integral domain.

There are two common constructions of fields.

One construction is treated in two places.

The field of fractions1 is a construction.

R should be an integral domain.

The quotient of S should be indicated by the relation between Q.R/ and S.

The field of fractions and quotient ring should not be confused with the quotient field.

R can be seen as a subring of Q.R/.

Q.R/ is a field.

R is injected into Q.R/.

There are two exercises 6.4.11 and 6.4.12.

Q.R/ is a field containing R as a subring if R is an integral domain.

The field of rational functions is called Q.KOEx.

The field is called K.x/.

This field is called K.x1.

It is said that R has a characteristic p.

J in a ring R has no nontrivial zero divi sors that can be easily characterized.

Every maximal ideal is prime in a commutative ring.

If M is a maximal ideal, then R is an integral domain.

M is a prime ideal.

Every ideal in Z has a form called d Z.

R should be a commutative ring.

The set N of nilpotent elements of R is ideal.

Show that there are no nilpotent elements.

S is a synonym for N ker.

An integral domain is the ROEx with coefficients in R.

The results of the previous problem should be generalized.

Q.R/ is a field.

Q.R/ is a field containing R.

The function for the numbers is d.n.

The function is d.f for the poly nomials.

If it admits a Euclidean function, call it an integral domain R.

Take d.z/ D jzj2 for the "degree" map.

We have d.w/.

Let the nonzero elements be in ZOEi.

A nonzero, nonunit element is in an integral domain.

Proper factors of a are said to be elements b and c.

The irreducible elements are not always prime.

Two elements are said to be associates if they divide the other.

If d divides each ai, then d divides c.

There is a gcd of fai g.

R should be a Euclidean domain.

The ideal in R is the main one.

Every Euclidean domain is a prin cipal ideal domain and a unique factorization domain.

It's natural to ask if the implications can be reversed.

ZOEx is a unique factorization domain.

This means that ZOEx is a UFD.

There are only a few possibilities.

5 W x 2 Zg.

These are two different factorizations.

5 is not a unique factorization domain because some elements are irreducible.

A proper factorization is 1 x/.

x does not have factorization by irreducibles.

A principal ideal domain is a unique factorization domain.

We show that an element can't have two different factorizations.

Suppose that R is an ideal domain.

Let me D. Since R is a PID, there exists an element b 2 I such that I D bR.

Let R be the ideal domain.

Let R be the ideal domain.

A proper factorization of a D bc is admitted.

There is a sequence that is contrary to Lemma 6.5.15.

R should be an integral domain.

R is an integral domain and p 2 R is a nonzero nonunit element.

The "pR prime" is tautological.

Sup pose p is prime and p has a factorization.

P divides a D pr.

b is a unit.

Suppose that R is an ideal domain.

If r is a unit, then J D aR D pR.

There is at least one factorization by irreducibles.

If r D 0 reads 1 D q1 qs.

It's not invertible since irreducibles are not.

p1 is prime according to Lemma.

The ideal in R is the main one.

Try N.z/ D jzj2 for the function.

You can try N.z/ D jzj2 for the function.

There is a unit called a D ub.

A principal ideal domainR is irreducible.

Exercise 6.3.11 is a generalization of the Chinese remainder theorem.

The elements x and y are in good shape.

Show that R is not a ideal domain.

R is a field if it is irreducible.

Show that KOEOEx is an ideal domain.

Divisors in a unique factor ization domain are discussed in the first part of this section.

Write a letter B.

If c is not a unit, consider irreducible factorizations of b and c.

Let a1; : : ; as be nonzero elements in a unique factorization domain.

Let m.j / D mini fNJ be used for each j.

D is a common divisor of fa.

For each j, k.j, and m.j. E divides d.

If b is a unit, then p divides it.

P divides a or b.

R is a factorization domain.

If p is irreducible, there is no need for maximal pR.

ZOEx is a UFD.

If the coefficients are relatively prime, call an element of ROEx primitive.

This discussion can be extended to elements of F OEx.

Such as d1b2 D ud2b1.

Take R D Z.

The product of two primitive elements is primitive.

F.x/ is primitive.

The existence of irreducible factorizations is implied by one property.

The number of irreducible factors appearing in any irreducible factorization of a.

If b is a proper factor, then m.b/ m.a/.

The sequence is finite if it follows.

This is what is shown in the proof.

This is what was shown in the proof.

The ascending chain condition for principal ideals is satisfied by R.

Every irreducible in R is prime.

Lemma 6.6.4 and Lemmas 6.6.13 are the previous ones.

There are elements in 5 that are not prime.

In R, irreducibles are prime.

R is a unique factorization domain.

Where f1 is primitive, let f.x/ 2 ROEx.

The previous exercise should be generalized over a unique factorization domain.

Let R be a UFD.

Show that.R is anintegrative domain for all irreducible p.

The ascending chain condition requires 5 to have elements that are not prime.

Without loss of continuity, this section can be skipped.

The rings ZOEx and KOEx are not ideal.

The ascend ing chain condition for ideals is an equivalent condition to the finite generation property.

The ideal is generated by a subset of a ring R.

Every ideal of R is finite.

The ascending chain condition is satisfied by R.

Every ideal of R is finitely generated.

This shows that R is good for ideals.

R has an ideal that is not finitely generated.

R doesn't meet the ideals of the ACC.

Noetherian rings are named after someone.

The minimum degrees of nonzero elements of J are given.

Noetherian is the name of ROEx.

J D J0 is what we claim.

F 2 J has a degree and leading term.

The Hilbert basis theorem is followed by this.

R is a commutative ring with an identity element.

5 is not a UFD.

In this section, we will look at some elementary techniques for determining whether a polynomial is irreducible.

The problem of determining whether a polynomial is irreducible is limited.

Reducing the polyno Mial modulo a prime is a basic technique for testing for irreducibility.

F is irreducible over Q.

F is irreducible over Q.

Over Q, f.x/ is irreducible.

The last equation shows that,0 is a contradiction.

The Eisenstein criterion says that x3 C 14x C 7 is irreducible.

This is irreducible by Eisenstein's criterion.

First, then f is irreducible.

The details are provided for example 6.8.6.

Chapter 7.

The solution of polynomial equations is the most traditional concern of algebra.

During the Middle Ages, Arabic scholars preserved and augmented this knowledge.

For another 250 years, the matter stood.

Neither of them had time to celebrate his success.

The structure of the field and the symmetry of the roots are what we will be concerned with.

The element i2 is called a discriminant.

The method described in this section can be used to find the roots of f.

The fields in this section are general.

The exercises ask you to check this.

L has something over K.

The basis of M over K is fi j g.

L is related to K.

The set of numbers is a countable union of sets.

On dimK.L/, the second statement is proved.

There is nothing left to prove.

There is an element a1 2 L n K.

We have I, D f.x/KOEx, where f.x/ is an element of minimum degree in I.

This is part a.

This shows both parts.

To find the in verse of an element.

It follows that.

The desired inverse is D 1, so r.

There is a converse to the proposition.

L is a finite extension of K.

We have KiC1 D Ki.aiC1, where aiC1 is a combination of Ki and K.

Ki D KOEa1; : : ; ai for all I and, in par ticular, L DKOEa1; : : ; an.

L is over K.

We have the following propo sition.

K L should be a field extension.

It suffices that K.a; b/ A.

The set of numbers is count able, so it is a countable field.

Show that the finiteness of dimK.L/ is not true.

Let D e2i be the number.

The dimensions are 6 over Q.

Refer to Exercise 7.2.8.

Over Q, 2x C 2 is irreducible.

There are elements of the form a0 C a1 C a22 C a33 C a44.

K L should be a field extension.

I have 62 K and dimK.E/ D 6.

If I have 2 K, then dimK.E/ D 3.

Otherwise, dimK.E/ D 6.

K.,1, and E have dimensions over K.

So dimK.E/ D 3 would be 2 K.

Mapwise and pointwise, 1 to 2.

Aut.L/ is a group of automorphisms.

Aut.L/ is a subgroup of F automorphisms of L.

An automorphism that fixes,j and interchanges the other two roots can be found in AutK.E/.

Fix.H / D fa 2 E W.a/ D for all 2 H g.

K Fix.H is a field for each subgroup.

Part a is an exercise.

Let K D fix.AutK.E.

There are no proper subgroup of AutK.E/ S Z3 because of multiplicativity of dimensions.

The case dimK.E/ D 6 needs to be considered.

K M E is an intermediate field.

H is one of the groups just listed.

h.x/ splits into linear factors in EOEx.

The splitting field for h.x/ is E.

There is an automorphism of E that takes a to b.

K M E is an intermediate field.

Let M D Fix.H.

If H is one of the Hi, then M D M D K.

K.,i

Let K be a subfield of C, let f.x/ 2KOEx be an irreducible cubic polynomial, and let E be the splitting field of f.x/ in C.

The result is remarkable.

There is an infinite set with the split ting field E.

It is irreducible over Q.

K C should be a field.

If f.x/ 2KOEx is an irreducible polynomial, what would it be?

Suppose f.x/ 2KOEx is a irreducible square root.

Why is the denominator not zero?

K.,i

Show that if M is any of the fields.

If M is K.i, then AutM.E/ D A3.

K.0/ that fixes K pointwise.

There is a priori that K Fix.AutK.E.

Since dimK.E/ is finite, E is related to K.

Let V 2 E and p.x/ show the minimal polynomial of V over K.

Let V D V1 be the distinct elements of f.V/ W 2 AutK.E/g.

It follows that p.x/ divides g.x/ since V is a root of g.x/.

The extension is called Galois over K.

Section will give the details of the proof of the equality.

The range of AutM.E/ is all of the subgroups of AutK.E/.

Their roots are 3/.

The fixed field of V is Q.

This result is also proved in a general setting.

The roots are in C.

Degree 4 is over Q and E D Q.

The automorphism is restricted to E.

It is of order 4 and 2.

D4 has 10 groups.

There are ten intermediate fields, including the ground field and the splitting field, in the Galois correspondence.

It's possible to determine each field explicitly.

The field extensions of Q are called Galois.

Chapter 8.

X was fined for g 2 G and x 2 X.

From R to End.M is a ring Homomorophism.

R has a multiplicative identity and M is unital.

V is made into a unital koEx-module.

The corresponding homomorphism is End.V.

There is a structure arising from the linear map T.

An R-module is also known as a left R-module.

A right R-module can be defined similarly.

A ring R is a right R module.

R is the ring of n-by-n matrices.

M should be an R-module and R should be a ring.

By left multiplication, let R act on itself.

M is an R-module.

Any collection of submodules of M should be called fM.

Let M be an increasing sequence of submodules.

Two submodules of M should be A and B.

M and S M are modules.

M is a submodule of RS.

M is finite.

R-module isomorphism is a form of R-module homomorphism.

An R-module homomorphism from M to M is called an R-module endomorphism.

M stands for EndR.M.

R is a commutative ring.

Rn is the set of n-by-1 matrices over R.

R is non-commutative.

R is the ring of n-by-n matrices.

In the next section, we will discuss module homomorphisms.

The direct sum of M1 and M2 is called M1 and M2 respectively.

There is an equivalency between (a), (c), and (d).

The basis for M is a linearly independent set.

The field K is free as a K-module.

Modules over other rings are not free.

xng is a basis of M.

The map in (b) is an R-module homomorphism.

If the map is surjective, B will generate M. This shows that there is no difference between (a) and (b).

A minimal generatiing set is any basis of M.

The elements of B0 are R- linear.

B0 does not generate M.

R can be any ring with multiplicative identity.

R always means a ring and M is an R-module.

The submodule of M is calledIM.

M is a submodule of N.

V is an n-dimensional space over a field K.

V n is a free module over EndK.V.

Let V be a finitedimensional space over a field.

M is a submodule of N.

HomR.M; N is an abelian group.

EndR.M is a ring with addition and multiplication.

M is an R-module and N is an R-submodule.

The quotient map is an R-module homomorphism.

M is an R-module and N is an R-submodule.

R-modules is a homomorphism.

There are ana logues for modules in all of the homomorphism theorems.

The analogous theorem for abelian groups is used to prove each of the theorems.

There is a surjective homomorphism of R-modules.

M is the quotient homomorphism.

Also respects the R actions.

Let R be any ring, M any R-module, and x 2 R.

Ann.x/ is a submodule of R and is a left ideal.

R-module homomorphism of M onto M is allowed.

The quotient map is marked by M.

K is M and N is N.

M is an R-module.

The Factorization Theorem can be proved.

The Diamond Isomorphism Theorem needs to be proved.

R should be a ring with an identity element.

R is a ring with a multiplicative identity element.

There is a special case that all the Mi are equal.

We have to show that to make sure.

Then also y.j and D x.j.

R-bi linear and alternating:.x; x/ D 0 for all x 2 M.

A permutation can be written as a product of transposi tions.

A multi linear function is a multi linear function.

S.'/ D P2S is a multi linear functional.

It is an alternating multi linear functional.

A is shown in a similar argument.

Is alternating.

Let x1, x2, and xn 2 M be used.

I claim that the sum is zero.

This shows that A. is correct.

A sequence of elements of Rn.

Lemma 8.3.5 states that it is an alternating multi linear function.

: ; an/ D

: ; an/ D.a1;

The properties are given in the proposition.

An alternating map.

Let A and B be matrices over R.

Adding a multiple of one column will get B from A.

The formula used to compute the determinants is very inefficient.

When R is a field, the previous proposition provides an efficient method of computing determinants.

The sign of the determinant can be changed by operations of the first and second types.

The number of row interchanges performed in the reduction is 1/k det.B.

The proof of the other is the same as the proof of the other.

The determinant is called the.i; j; minor of A.

The cofactor matrix of A is called 1/iCj det.Ai;j.

The expansion of the determinant is called the cofactor expansion.

Moreover, D ak;j.

1/iCj det.

The cofactor matrix of At is C t.

If the determinant is a unit in R, an element of Matn.R/ is non-invertible.

Det.A/ is nonzero and R is an integral domain.

The following trick can eliminate the additional hypotheses.

Consider the matrix X D.xi;j /1i;j n in Matn.R0/.

It's easy to check that.

Look at that '.

Mi;j / D '.M /i;j.
It follows that '.,.M // D,' using these two observations.

The trick is worth remembering.

If k > n, show that.Rn/k has no nonzero alternating multi linear functions with values in R.

Replacing the j -th column of A with a matrix called Aj is how it is obtained.

We will determine the struc ture of finitely generated modules over a principal ideal domain in the following section.

Represent elements of the R-module M n by 1-by-n matrices.

If fv1 is linearly independent over R and OEv1, then C is the zero matrix.

R is a commutative ring with an identity element.

We will show that.

A D.ai;j is the n-by-m matrix.

The matrix B D.bi;j is a m-by-n matrix.

The vj's linear independence makes it necessary for us to haveAB D En.

We have A0B0 D.

Any two basis have the same qualities.

R is a commutative ring with an identity element.

The zero module is free of rank zero.

This is more than just a convention, it follows from the definitions.

A principal ideal domain is what R means for the rest of this section.

F 0 D span.ff1; : : ; fn 1g/

Let N 0 D N F 0 be a submodule of F.

Every element has a unique expansion.

D y C rh 2 N 0 C Rh.

N D Rh C N 0.

Suppose that M has a finite set.

N is free.

If ai;j D 0 is diagonal, then D j is.

There is a Smith normal form on my page.

A version of Gauss ian elimination is used to Diagonalize the matrix A.

We will only be interested in the case of R in applications.

Two rows are interchanged in the third type of elementary row operation.

The inverse of Pi;j is Matm.R/.

I j is inverse U.

I; j.

Left multiplication by U.

Elementary row oper ations are similar to elementary column operations.

We need another measure of size if R is not Euclidean.

Each nonzero element can be factored as a D up1p2 p since R is a unique factorization domain.

The number of irreducibles appearing in a factorization is unique.

The length of a to be is defined.

A has a nonzero entry in the.1; 1/ position.

Proceed as before if, divides V.

We are done.

Lemma 3.5.10 was replaced by 8.4.8.

It's a field.

A means of computing the greatest common divisor d.x/ is provided by the diagonalization procedure.

The row matrix A D.a1.x/ should be marked with the letters A.

The first column of Q should be marked with.t1.x/.

The matrix is C D.ci;j.

The matrix D.di;j should be marked with D.

We conclude that CD D En is the n-by-n identity matrix.

R and C are in Matn.R/.

In Matn.R/, C is inverse C 1.

N is a free module.

All the dj are nonzero.

R should be a PID.

R is a principal ideal domain in the remaining exercises.

The proof of Theorem 8.4.12 shows that N is free.

A module is a module.

The previous exercise should be retained.

dswsg is a basis of range.

The kernels can be found by finding the one in Zn.

To find a description of ker.A/, use the diagonalization A0 D PAQ.

Let x1 be a set of generators of minimal cardinality.

Dsvg is a basis of N and di divides dj for i j.

The be R-modules and Bi Ai submodules are included.

M is a finitely generated module.

For all I, then s D t, D k and.ai.

The idea of torsion is introduced before addressing the unique statement in the theorem.

0 and r1r2.sx Cty/ D 0.

M is a module if M D Mtor is correct.

There is a chance that M is free of torsion.

G is a finite abelian group.

G is a finitely generated module.

Since it is finite, G is finitely generated.

V is a finitedimensional space over a field.

Let T 2 end.

0 and.Pi,i xi /v D Pi,i T i.v/ D 0.

If a 2 R is a period of 2 M, then Rx S R is a period of 2 M.

Any submodule of M is finitely generated.

If M is free, it is a free module.

Part a is an exercise.

A is a free submodule and B is a free submodule.

The rank of B is determined.

This proves something.

Any module that is free is free.

Let x 2 M, let ann.x/ D.a/, and let p 2 R be irreducible.

If p divides a, then Rx is S R.

If p does not divide, then p Rx.

.p/ is the prescription drug name.

It's that Rx D.

Suppose p 2 R and pM D are irreducible.

R is the quotient map.

A surjective R-module homomorphism is called M.

We are ready for the proof of being unique.

The unique rank is Lemma rank.A0/ D rank.B0/.

Periods of M are as and bt.

We can assume it's D bt D m.

Fix a problem.

We have.ai / D.bi / D.p/.

The invariant factors of M are called the Structure Theorem.

Up to multiplication by units is how they are determined.

There is a unique sequence of exponents.

It is easy to check that M is a submodule.

The period m of M is divided by pr.

If p does not divide, M OEp D f0g.

There is an irreducible factorization of a period of M.

The most common divisor is rsg.

Suppose that xj 2 M OEpj for 1 j s and Pj xj D 0.

It follows that xi D 0.

Let R be the ideal domain.

The period of OEri is pmi.

R should be a principal idea domain.

Set,1.x/ D x2 C 1,,2.x/ D.x 1/, and,3.x/ D.x 3/3.

We can get a procedure for resolving any number of simultaneous congruences using the same considerations.

Take ri D a=,i as a first approximation.

Let a.x/ D,1.x/,2.x/,3.x/.

We have to find ui.x/ such that ui.x/ri.x/ 1mod for each.

We can take ui.x/ to be ri.x/.

There is a list of irreducibles appearing in an irreducible factorization of a period m of M.

AOEpj D M OEpj is the internal direct product of the submodules.

M is an R-module and S is a subset of R.

R is an integral domain with a non-principal ideal J.

Let R be the ideal domain.

P is an irreducible of R.

V is finitely generated as a K-module.

T v 2 V1 for all.

The invariant factors of T are called the polynomials ai.x/.

The companion matrix is by Ca.

The matrix of T is the companion matrix of a.x/.

Both conditions are equivalent.

J / D xd C J D.a0 C ad 1xd 1/ C J.

The generator v0 and a.x/ 2 ann.v0/ are both cyclic with V.

This shows that condition is implied.

The invariant factors of T are V.

The rational canonical form is an invariant for similarity of linear transformations.

T2 D U T1U 1 follows.

The rational canonical form of A is a unique matrix that is similar to a matrix A.

The invariant factors of A are what they are called.

The rational canonical form in Matn.F is the same as the one in Matn.K.

If A and B are the same in Matn.K/, they are the same in Matn.F/.

If they are similar in Matn.F /, then they have the same form in Matn.F.

K F should be fields.

K F should be fields.

Let A 2 Matn.K/ Matn.F.

We need to find the source of the code.

The matrix A can be used to "lift" the transformation T.

D Pi ai;j fi.

We have all f 2 F.

Finally, if Pi, i 2 ker.

Pi, I fi / D Pi, Iei.

D 0 is for all.

T is a part of the module.

As shown in Equation (8.6.1).

V1 has period aj.x/.

Let's say the degree is aj.x/.

We are going to look at how to find a basis for the transformation T determined by multiplication by A is.

The matrix has columns in v1, v2, v3 and v4.

Let A 2 Matn.K.

Let T 2 EndK.V /.

The characteristic polynomial of T is called T.x/.

Let A be the matrix of T with some basis of V.

The product of A is a 2 Matn.K.

The result is related to the proposition.

T has a degree at most dim.V.

There is a relation between the mini mal polynomial and the characteristic polynomial.

The roots of T 2 EndK.V have an important characterization.

Let T 2 end.

If, and only if, T has an eigenvector in V with eigenvalue.

The computation of the rational canonical form of T was made in the text.

All of the coefficients are related to A.x/.

T 2 EndK.V is where V is n-dimensional.

Let A 2 Matn.K be any field.

It is followed by A.A/ D 0.

The analysis continues from the previous section.

There are m blocks and d basis elements in the ordered basis B.

Write p.x/ D xd C ad 1xd 1 C a0

Cp is the companion matrix of p.x/.

Consider the special case that p.x/ is linear.

Up to permutation of the Jordan blocks, the matrix of T in Jordan is unique.

The elementary divisors of T are V.

A 2 Matn.K/ Matn.F is allowed.

Let A 2 Matn.K.

We can give a typical application of these ideas to matrix theory.

The matrix in Matn.K/ is similar to the transpose.

It is similar to Jm.0/.

It's similar to its transpose.

If Ai is similar to Bi, then A is similar to B.

The Jordan matrix is similar to the transpose.

The proof of the proposition is let a 2 Matn.K/.

It is possible to show that A and At are similar.

Similarity in Matn.F is defined in the following way.

J is similar to Jt.

The V is a module with a period power of.x.

x 1 D x 0, x 2 D x 1, etc.

There are no solutions for /x D x r.

The number of Jordan blocks is the number of independent eigenvectors of A.

1 doesn't have a solution.

There are two possible Jordan forms.

Consider one of the submodules.

Call the generator w0, the sub module W, and the period a.x/.

A has a Jordan form in Mat5OEQ.

We can see that the Jordan form of A is J2.1/.

The second method from the text is used to repeat the previous exercise.

There is a basis consisting of eigenvectors.

The Jordan form of A is diagonal.

The divisors of A are linear.

A0 C N, where A0 is diagonalizable, N is nilpotent, and A0N D NA0

N 2 Matn.K should be a nilpotent matrix.

There is a characteristic N.x/ D xn.

The trace of N is zero.

S 1AS is a form of Jordan.

Chapter 9.

The fields in this chapter are not subfields of the complex numbers.

It is a finite field extension.

In particular, b is algebraic over K.

Part (b) is related to part (a).

The smallest subfield of L containing E and F is E F.

If E is over K and F, then E is over F.

Let a 2 E F.

The dimensions of Q.

We have a different point of view regarding the extensions.

K pointwise is fixed by isomorphisms.

There is a technical variation of part (c) of the proposition.

The splitting field is unique to isomor phism.

The idea is to use both of them.

The required isomorphism is L.

L are the roots of p1.x/.

Q, and extending.

Over Q, 3 is irreducible.

A finite-dimensional extension of a finite field is a finite field.

In this section, we look at multiple roots of polynomials and their relation to the formal derivative.

A multiple root is a root of more than one.

The rule from calcu lus is that we define D.xn/ D nxn 1 and extend linearly.

If, and only if, f.x/ is a constant polynomial, Df.x/ D 0 will be.

If K is a field, then f.x/ 2KOEx is irreducible.

If the characteristic of a field K is zero, any irreducible polynomial in KOEx has only simple roots in any field extension.

The map is a combination of two things.

K is a finite field.

Any irreducible polynomial has simple roots in any field extension.

AutK.L/ is a subgroup.

It fixes K pointwise and sends it to V.

L and K.V/ L give the result.

The left coset of AutM.L/ has 1 and 2 in it.

If 2 AutK.L/, then permutes the roots of f.x/.

.,/ is a root of f.x/.

There are only two possibilities for the Galois group.

We want to get similar results in general.

H is a subgroup of Aut.L/.

H is Fix.H and D is D for all 2 H g.

Fix.H is a subfield of L.

If K1 K2 L are fields, then K01.

An element in a field extension of K is said to be separable over K if it has a minimal polynomial.

There is a claim that a 2 M0 D K.

There is nothing to show if Mj 1 D Mj is the case.

L is over K because dimK.L/ is finite.

Both are monic.

The extension is called Galois.

The extension is separable and can be divided into linear factors over L.

M L is an extension of K M L.

L is the splitting field of a separable polynomial over 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884

L is finite by the hypothesis applied to K.,/, jAutK.,/,L/j D dimK.

We have jAutK.M, jIsoK.M, and L/j D dimK M.

There is an element 2 L that is called L D K.

If the char acteristic is zero, then K is infinite.

V/ are different.

For the moment, put D k, C V.

D jIsoK.K./; E/j n D dimK L.

If the characteristic is zero, Separability is automatic.

Put F D Fix.H.

L is Galois over F.

First note that jH j jAutK.L/j dimK.L/ is finite.

Then F D Fix.AutF.L is by a proposition so L is Galois over F.

There is a V such that L D F.V/.

K L should be a Galois field extension.

AutK.L/, H D AutFix.H/.L/.

M is an intermediate field.

If AutM.L/ is normal, M is invariant.

This is the final part of the proof.

We need the following variant of the proposition.

L is a Galois field extension.

H D AutF.L/ and dimF.L/ D jH j.

The minimal polynomial of V over F is marked by p.x/.

Both are monic.

D deg.p/ jH j is dimF.F.V.

With K D F, we can reach the con clusions.

L was generated by A.

An important technical result is used in the sequel.

Without loss of continuity, the remainder of this section can be omitted.

L should be a field.

Any collection of L automorphisms is linearly independent.

There is a collection of distinct automorphisms of L.

We have n D 0.

Let K L be a field extension.

nC1g is a subset of AutK.L/.

The collection of j is linearly dependent.

The j can't be all distinct by the previous proposition.

L and M are finite-dimensionalg.

The proof of the proposition is given by this exercise.

Show that A is Galois over A.

Let xn be variables, and let K be any field.

Kd OEx1; xn KS OEx1; xn for d 0.

xn/ is a subfield of K.x1.

The Galois group AutKS.x1; is called xn/.K.x1.

By Exercise 9.6.1, xn is the fixed field.

The extension is called Galois and the group is called the Galois group.

If j > n, we put j.x1; : : xn/ D 0

The i is the same as the monic monomials.

A partition is a finite decreasing sequence of nonnegative numbers.

The partition's size is jj D P i.

We need to produce a more obvious basis in order to do this.

The person is positive.

For any permutation, note that.

This proves the parts.

Take n D 4 and D 3.

For any ring A, the ring of symmetric polynomials in AOEx1, xn equals AOE1, and so on.

Go ahead and do this procedure.

P holds.

The least partition is in P.d.

This shows that P holds.

An example is used to illustrate the algorithm.

Consider the numbers in three variables.
Take p D x3 C y3 C.

It is possible for such computations to be automated.

For each d, Kd OEx1; : : : ; xn is invariant.

Let's use variables.

Un denote the roots of this polynomial in a splitting field E.

The Galois group AutK.t1; is a symmetric group.

Fix K and take it to fi.

We know that I differentiate odd and even permutations.

Even if, and only if,.i/ D i, every permutation is satisfying.

Suppose that f is irreducible and separable.

The irreducible separable polynomial is f 2KOEx.

The discriminant has a square root in K.

We don't need to know the roots of f to compute the dis criminant.

F.x/ is irreducible over Q.

The matrix has ai and n columns.

If the matrix R.f has a determinant zero, f and g have a nonconstant common divisor.

We want to show that.

We need to change our point of view at this point.

So has a summand.

1/n.n 1//2 R.f; f 0/.

Check to see if x4 C x2 C x C 1 is irreducible over Z3.

One such is that f.x/.x/ D g.x/'.x/.

The Galois groups of quartic polynomials are determined in this section.

The Galois group of g is over K.

We distinguish cases based on whether the resolvent is irreducible.

It is not possible to distinguish between these cases.

If G is D4 or Z4 then 2 is either 4 or 2.

If H.x/ splits over K.i/, the Galois group of g is Z4.

Let L D K.i/ D K.1; 1 and 3.

The preceding poly nomial is H.x/.

Suppose that H.x splits over L.

OEE W L D 2 and G D Z4 were discussed before the lemma.

The Galois group G is Z4.

We have.1 D.

Soon, some examples of the use of this lemma will be given.

There are two square roots of 3.

It is possible to choose the square roots so that their product is q.

It's not a square in Q.

The group is called Galois.

Take K D Q and f.x/ D 21 C 12 x C 6 x2 C 4 x3 C x4.

If g is not irreducible, it must factor into two irreducible quadratics.

G is either Z4 or D4.

The nonzero roots are 8.

The Galois group is D4.

The coefficients of f are invariant because of the automorphism of C.

i2 is assumed to be negative.

The Galois group G of f is in order 4.

The extension of K is not fixed pointwise.

This is not the same as the other one.

The ground field should be Q and f.x/ D x4 C 5x C 5.

The Eisenstein criterion shows the irreducibility of f.

The alternating group contains G of f.

The resolving field is divided into Q.i/ D Q.

The root of the resolvent is 5.

H.x splits over Q.

The Galois group is Z4.

x4 C px C p is irreducible with the Galois group S4.

The Galois group should be used for the following polynomials.

The computation of Galois groups of polynomials in QOEx of degree 5 or more will be discussed in this section.

The degree partition is m1, m2 and m3.

F is an irreducible polynomial with coefficients in Z.

We won't prove Theorem 9.9.1 here.

The alternating group A5 contains the Galois group G of f.

3 is irreducible and the reduction modulo 5 has a degree partition.

We have to show that some cycle types don't show up.

The group is called D5 rather than A5.

The situation is interesting.

It works and is based on mathematics that you know.

To find the probable Galois groups for each of the quintic polynomials, look at the reductions of the modulo primes.

The article was written by Soicher and McKay.

Chapter 10.

The section deals with a decomposition of a finite group into simple pieces.

Simple is what N is called.

A composition series is a finite group.

A composition series and a group of order 1 are both simple.

A finite group can be solved if it has a composition series in which the successive quotients are in order.

The section develops by means of exercises a description of solvability of groups.

The commutator subgroup OEG; G is normal.

There is a solution to G.

The trivial subgroup feg is a natural number.

G is easy to solve.

We show that G.k/ Hk for all k, in particular, G.r/ D feg.

Exercise 10.2.4 that G.1/ H1 follows since it is abelian.

This is the end of the argument.

You can use whatever criterion you want in the following exercises.

The alternating groups An for n 5 are not solvable in this section.

An is a simple group for n 5.

Conjugacy classes in the symmetric group are mined by cycle structure.

This fact is frequently used.

Two-cycles can be written as a product of one or two 3-cycles.

The elements are a result of two disjoint 2-cycles.

There is a 3-cycle.

All 3-cycles are conjugate in An.

Any 3-cycle is conjugate in An to.123/.

0 D.45/ is even, and 0.123/0 1 D is not.

There is a 3-cycle that does not commute.

N has a 3-cycle in the last two cases.

Since n 5, An has an element.a2a4a5.

N has a 3-cycle.

These computations may look disorganized.

It is necessary to prove that A5 and A6 are easy.

If the centralizer of N is An, then X has 2 elements.

N is the other conjugate of N.

N M D feg is assumed to be minimality.

M N is a subgroup of An.

M 1N 1 D NM D is normal.

A4 is normal.

In G, Z.A/ of A is normal.

The center of An is trivial for all 4 people.

The center of A4 is feg.

An is not abelian.

A normal subgroup of An is 10.3.4.

If you observe that a simple nonabelian group is not solvable, then you can conclude that An and Sn are not.

By inspection, the assertion is valid.

Q is a field extension of degree '.n/ over Q.

n.x/ is irreducible over Q.

Suppose p is not a root of f.

The injective group homomorphism from AutK.u/.E/ into the cyclic group was created by U.

AutK.u/.E/ is a group with an order that divides.

The length of the tower is called r.

L is over K.

The statement is proved by the act of inducting.

The root is f.x/.

There is a splitting field of f.x/ over F.

L is a radical extension of F.

The length of the radical tower is used to prove this.

The Galois group is feg if r D 0 is true.

AutK.K1 is a Galois over K group.

L is Galois over K./ and fur thermore, AutK./.L., S AutK./L.L/.

S AutK./L.L/ is solvable.

L is abelian, so solvable.

Both N and G are solvable.

Without loss of continuity, this section can be omitted.

There is a converse to the proposition.

There is a subtle proof of this.

K E should be a Galois extension.

A multiplicative 1-cocycle is called E.

This nonzero element is called a 1.

D f 1f

This gives me something.

K E should be a Galois extension.

The Galois extension is called E.

We can be a generator of AutK.E.

The order of the group is AutK.E/.

K E is a Galois field extension.

AutK./.E.

will be solved.

E./ is an extension of K./.

Chapter 11.

The norms are preserved.

Affine hyperplanes are not necessarily passing through the origin.

P is always of the form 2 Rn W hx.

There are two distinct points in Rn.

This is not a positive of the proposition.

Every extension in part a is linear if 0 2 R and.0/ D 0 are included.

Any isometry of Rn is affine.

The special case that R contains 0 and.0/ D 0 should be considered first.

Let V D span.R and let fa; akg be a basis of V contained in R.

I claim that T1 is not a straight line.

Rn is a measurement.

This is the final proof of part a.

Rn that ends 0.

There is only one linear extension of 0.

This proves it.

Suppose that 0 2 R and.0/ D 0.

L.x/ C b0 is an affine extension.

Part (d) follows by taking R D Rn.

Any element of O.n; R/ is a reflection matrix.

p D has the desired properties if the orthogonal reflection is in this hyperplane.

The structure of the isometries of Rn is being worked on.

Define the translation for b 2 Rn.

Let be an isometry of Rn and D.

A translation and a linear isometry are what D b is.

A linear isometry is called B.

The isometry is a translation.

It's affine reflection.

Otherwise h j, D tV.s,j is a glide-reflection.

A, B, C, and D are noncoplanar points.

The identity map is called D.

Show that for any two-cycles, b/, T, and b/ is the reflection in the hyperplane.

If, and only if, det.T is even if.

Show that any element of Sn is a product of 2-cycles.

That 2p C 2q > qp follows.

There are five regular polyhedra.

The figure shows a graph.

A tree is a graph that admits no cycles.

The next page shows a tree.

The number of connected components of S n G is G.

G 0 is still connected.

Hence,.G / D.G 0/ D 2.

Hence,.G / D.G 0/ D 2.

The proof has been completed by this.

The sphere has a dodecahedron projected on it.

On the facing page, Figure 11.2.3 shows the projection of a dodecahedron onto the sphere.

For two dimensions, it's easy to get the corresponding result: a fi nite subgroup of SO.

The rotation group SO.3 has a finite subgroup called G.

The number of G's acting on P should be marked by M.

Let xM be a representative of the M.

The exercises give a guide.

The analysis of the last exercise should be extended to the case n D 2.

The choice is a v 2 O nfu.

Consider the size of the O P.

O must contain with any of its elements.

Let the sun set at 2 o'clock.

The choice is a v 2 O nfu.

Consider the size of the O P.

Let v 2 O satisfy all y 2 Onfug.

The subgroup of O.3 is called iH S H Z2.

Define G D H.

The groups are classified by their mode of action, which is more than a classification up to group isomorphism.

V D R3 is a set of linear combinations.

Physical crystals are what we have defined as a crystal in three dimensions.

Solid table salt has a huge array of sodium and chlorine atoms in a ratio of one to one.

The unit cube is a fundamental domain of this lattice.

The translations of the crystal are called L.

The point group is defined by this example.

There is a point group in a crystal.

Take x 2 L and A 2 G0 into account.

Under G0, L is invariant.

The rotation order in G0 is 2, 3, 4, or 6.

This is followed by Exercise 11.3.

L is a two-dimensional lattice.

The basis for L is the linear span of fa.

For x 2 L1, there is a @ 2 GL.R2/ such that '.x/ D@.x/ for x 2 L1.

This proves something.

This proves something.

The statements will be verified in the exercises.

The entire symmetry groups are found in the lattice L.

A line through the origin.

L B is where.v/ is located.

L0 is the lattice generated by L .A.

The semidirect product of L and D1 is 2 G. This isomor phism class is called D1m.

Then G is generated by L.

Put v D.u/.

G is a product of L and D1.

The New York Times has a type D1g crystal.

There is no restriction on the lattice of the group G.

The point group D1 has to consider two cases.

L0 is the lattice generated by L .A.

D2mg is a isomorphism class.

D2c is a isomorphism class.

The other two are not.

The hexagonal lattice is generated by two equal lengths at an angle of 3.

G0 contains reflections in two lines.

There are reflections in the lines spanned by and b.

The case is called D4m.

The case is called D4g.

L and Z6 are semidirect products of G.

The group G has a rotation of order 6 and the lattice is hexagonal.

This is the final proof of the Theorem.

It is possible to classify three-dimensional crystals.

L is a two-dimensional lattice.

By using jja.

L0 is the linear span of fa.

Show that this means v D 0.

Let G be a subgroup.

This is a definite, bi linear function.

If g 2 G, then hhgxjgyii D.

T is an element of GL.n; R/.

T g D '.g/T for g 2 G1 should be verified.

G1 and G2 are conjugate in O.n.

These are considered in turn.

I'm satisfied with your explanation.

I'm not happy with your explanation.

The decision was supported by all of the committee members.

Some committee members did not agree with the decision.

One of the committee members did not support the deci sion.

The decision was opposed by one of the committee members.

The following is not a correct expression.

The decision was not supported by all of the committee members.

We might try to combine the negation with the con junction or the disjunction.

"not(not(A))" is equivalent to A.

Write out the truth table for A and B.

To verify this, write out the truth tables.

"if" is used in the sentences.

The picnic will be ruined if it rains tomorrow.

We notice something amiss when we form the negation.

Tomorrow's rain will not ruin our picnic.

Uncertainty is mitigated by the negation.

It's sometimes used instead of for emphasis.

We are going to discuss quantifiers in a more systematic way.

The quantity x2 is positive for every nonzero x.

If f is differentiable, then it is continu ous.

A positive square is a real number.

Here are a few examples.

How can we form the negation of sentences with tifiers?

Statements and deductions are not the only concerns of logic.

Every subgroup of an abelian group is normal.

3Z is a subgroup of Z.

Every car is going to end up in a pile of rust.

Miata is a vehicle.

Most statements require proof.

We have to assume A and prove B under this assumption.

The language of sets is the subject of this essay.

A set is a collection of objects.

If each element of A is also an element of B, we write A B.

Two sets are the same if they have the same elements.

W x 2 A and x 2 Bg.

There is a relation between intersection and logical conjunction.

Q.x/g, then B D 2 C W P.x/ or Q.x/g.

Every set has a subset of the empty set.

The properties of set operations are discussed here.

A B D B A.

It's just a matter of checking definitions.

Show that A n B D A C.B/.

The Cartesian product is an important construction.

The set of such pairs is called A B.

The surjective function is shown in Figure B.1.

If it is both injective and surjective, f is said to be a synonym.

The image of A under f is referred to as f.A/.

If B is a subset of Y, we write f 1.B/ for 2 X W and 2 Bg.

It is obvious to you that Tn2N Xn D X1 is the set of all nonneg ative real numbers.

A set is not finite.

It is clear that every set is finite or infinite, but it is not possible to determine which is which.

A subset of a set is finite.

A union of finitely many sets is not.

A countable set is 0g.

You need to climb a ladder.

There is a unique composition for all n and for all permutations.

You can get from one rung to the next.

P.k/ means P.k C 1/ for all k 2 N.

This is the final proof of the identity.

I prefer to leave out the proof of the equivalence because it is more abstract than the actual subject matter.

The properties just listed are satisfied by an defined procedure.

A group, a ring, and a field are elements of an associative multiplication.

1 gets us onto the grid.

As intuitive as this picture may be, it seems to require a new principle of induction on two variables.

The following situation is useful to consider.

This suffices to show that for all 2 N and 2 Tn.

Since each t 2 T belongs to some Tn, P.t holds for all t 2 T.

P.1; n/ holds for all 2 N; that is, P.m; n/ holds for all 2 T1.

P.m; n/ holds for all.m; n/ in Tk [ f.k C 1; n/ W n 2 Ng D TkC1.

R is used to solve the equation.

To show that C is a field, you have to check that nonzero Ele ments have multiplicative inverses.

In the form reit, where r > 0 and t 2 R.

The polar form is the form for a complex number.

It's easy to multiplication two numbers written in polar form: r1eit r2eis D r1r2ei.sCt/.

For a complex number, we have Zn D rneint.

A quick review of linear algebra is provided in this appendix.

S is dependent on other things.

A linear independent set can't contain the zero.

V is its own span.

F0g and Kn are subspaces.

V is a subspace of Kn.

R3 is the basis of 5g.

The equation is equivalent to 1 D 0, 2 D 0, and 3 D 0.

M x is a combination of the columns.

Matm;n.K/ and HomK.Kn; Km/ are related to the same thing.

The reader is left to check it.

The identity transformation on Kn is based on En.

M and M 0 are two matrices and M x D M 0x is for all x 2 Kn.

M j D M Oej D M 0 Oej D.M 0/j for all j, so M D M 0.

The person is injective.

The person is surjective.

This proves it.

We need to compute T to do that.

The kernel of T is 2 V W T.x/ D 0g.

Km is a subspace of Kn.

There is a nonzero vectors in the kernels.

S is dependent on something.

Aa D 0 doesn't have zero solutions.

Let fv1 be a set of s vectors.

The span should be all of Kn.

The identity matrix has columns.

Every basis of Kn has elements.

V satisfy.PsiD1,ivi/ C Vv D 0.

Then vj D Pi.

If the span is equal to V, call a set of vectors.

Every set contains a finite set.

Let V be a subspace of Kn.

Any basis of V has no more than n elements.

The basis of V is the linearly independent set.

The basis of V is a subset of any spanning set.

T 1 is the inverse map from V to Ks.

The dimensions of V are called the cardinality of a basis.

The empty set is the basis of V.

The dimensions of f0g are 0.

If V is not the zero subspace, then any basis of V has car dinality greater than 0.

Such that I R D idKm.

The identity matrix is marked by E.

The columns of M are independent.

The M columns have the same span as Kn.

T is a linear isomorphism.

There is a left inverse.

T has an inverse.

M is not spherical.

There is a left inverse.

M has an inverse.

M t is not spherical.

The rows of M are independent.

The rows of M have the same span.

The equivalency of (a) and (c) is given.

T is injective if it has a left inverse.

We have (f) H, (c) H, (e) H.

W be a linear map.

Let S be a set of numbers.

It is also linear.

T.S is independent.

If T.S is a basis of W, then show that S is a basis of V.

The basis of f0g is the empty set.

The map should be a linear one.

Show that if T is injective.

Show the dimensions of the range of T.

If T is surjective, show that.

The standard inner product on Rn is hx.

The norm of x 2 Rn is defined by jjxjj D hx.

If their inner product is zero, the two vectors are orthogonal.

Since fvi, vng is a basis of Rn.

There are many normal bases of Rn.

The Gram-Schmidt procedure goes as follows.

We can assume that the ai are not zero.

Put B1 D fa1

Then wj is related to the other.

If that's the case, put BjC1 D Bj.

A person named Bj C1 is orthonormal.

There are a few suggestions.

One of the most useful topics in algebra is linear and multi linear.

Both group theory and field theory are related to num ber theory.

The books are challenging, but they are accessible with a knowledge of this course.

Document Outline

Preface

Chapter 1. Algebraic Themes 1.1. What Is Symmetry? 1.2. Symmetries of the Rectangle and the Square 1.3. Multiplication Tables 1.4. Symmetries and Matrices 1.5. Permutations 1.6. Divisibility in the Integers 1.7. Modular Arithmetic 1.8. Polynomials 1.9. Counting 1.10. Groups 1.11. Rings and Fields 1.12. An Application to Cryptography

Chapter 3. Products of Groups 3.1. Direct Products 3.2. Semidirect Products 3.3. Vector Spaces 3.4. The dual of a vector space and matrices 3.5. Linear algebra over Z 3.6. Finitely generated abelian groups

Chapter 5. Actions of Groups 5.1. Group Actions on Sets 5.2. Group Actions---Counting Orbits 5.3. Symmetries of Groups 5.4. Group Actions and Group Structure 5.5. Application: Transitive Subgroups of S_5 5.6. Additional Exercises for Chapter 5

Chapter 7. Field Extensions -- First Look 7.1. A Brief History 7.2. Solving the Cubic Equation 7.3. Adjoining Algebraic Elements to a Field 7.4. Splitting Field of a Cubic Polynomial 7.5. Splitting Fields of Polynomials in C[x]

Chapter 9. Field Extensions -- Second Look 9.1. Finite and Algebraic Extensions 9.2. Splitting Fields 9.3. The Derivative and Multiple Roots 9.4. Splitting Fields and Automorphisms 9.5. The Galois Correspondence 9.6. Symmetric Functions 9.7. The General Equation of Degree n 9.8. Quartic Polynomials 9.9. Galois Groups of Higher Degree Polynomials

Chapter 11. Isometry Groups 11.1. More on Isometries of Euclidean Space 11.2. Euler's Theorem 11.3. Finite Rotation Groups 11.4. Crystals

Home

Chapter 1.

Understanding the "found" objects in the mathematical landscape is the first task of mathematics.

We will examine objects that are familiar and concrete, but we will sometimes pose questions about them that are not easy to answer.
The purpose is to introduce the themes that will be studied in the rest of the text, but also to begin looking at concrete situations.

We look into the idea of symmetry.

We may find it hard to give an adequate answer if we ask ourselves what we mean by symmetry.
When we see that symmetry can be given an operational definition, we will be able to associate a structure with each object.

The goal of the book is to encourage you to think for yourself.

Take some time to think about the questions.
Start with a list of objects that are symmetrical: a sphere, a circle, a cube, a square, a rectangular box, etc.

Take some time to think about the questions you're reading about.

A rectangular card is an example of a symmetric object.

The card admits motions that keep its appearance the same.

A symmetry is not visible.

If an object has symmetries, it is symmetric.

To examine this idea, you need to do the following exercises.

Reflections of the card are related to symmetry.

In order to simplify matters, I propose to ignore reflections for the moment, but to bring symmetry into the picture later.

A human face has leftright symmetry, but there is no actual motion of a face that is a symmetry.

All the symmetries of a rectangular card are cataloged.

Look at the card.
As you need it, mark its parts.
Write your conclusions and observations down.

For a square card, do the same.

Do the same for a brick with three edges.

It turns out that the rotation through 0 radians about any axis is necessary.

Nothing is one of the things that you could do to the card.
I could not tell if you had done anything or not.

There are infinitely many symmetries.

The center of mass is the intersection of the two diagonals, and the centroid is the center of a rectangle.

The issue of why there are exactly four will be taken up later.

Since the world doesn't have a solutions manual, we have to make our own decisions.

This is the choice that I will make, because distinguishing only the four symmetries leads to a rich and useful theory.

We have to consider rotation by 2 about one of the axes, and rotation by 3 about the other.
We can disregard the path of the motion and only take into account the final position of the parts of the card.
All of the parts of the card end up in the same place.

I propose to exclude reflection symmetries temporarily, for the sake of simplicity, as another issue is whether to include reflection symmetries as well as rotation symmetries.

If I leave the room and you perform two motions in a row, I will not be able to detect the result.

The result of two symmetries is the same.

The rectangle has symmetries.

The next step is to figure out all the products of the rectangular card and the square card.

A good way to record the results of your investigation is in a multiplication table with rows and columns labeled by the symmetries.
You have to fill in the rest of the table.

The beginning of the multiplication table.

Continue with the multiplication table for the square card after you've finished with the multiplication table for the rectangular card.

The eight symmetries of the square card have to be labeled in order for the multiplication table to be worked out.

If we agree on a labeling, it will be helpful to compare our results.

The square has symmetries.

The beginning of the multiplication table.

It's important to work things out for yourself when learning mathematics.

We have to keep track of the results of the multiplication tables for the symmetries of the square and rectangle.

It is possible to break the symmetry by labeling parts of the square.
I'm going to number both the locations of the four corners of the square and the corners themselves at the risk of overdoing it.
The numbers on the corners will travel with the card, and those on the locations will stay put.
The labeling for the square card is shown in the figure.

You can see that cd D r2 when you compare Figures 1.3.2 and 1.3.3.

The rule for computing all of the products is that the square of any element is nonmotion e.

The order of the elements is irrelevant in this multiplication table.

The product of two elements is the same.
Normally order does matter, but this is unusual.

The product of two powers of r (i.e., of two rotation around the axis through the centroid of the faces) is again a power of r. The nonmotion e is the square of the elements fa; b; c; d g.

The last property is obvious, without doing any close compu tation of the products.

The product of two symmetries that exchange the faces leaves the upper face above and the lower face below, so it has to be a power of r.

The order of the symmetries in this table does matter.

For example, ar D c.

The result of two symmetries followed by a third is the same as the result of the first and second symmetries.

The associative law is used for multiplication.
The law is expressed as s.t U/ D.st/u for any three symmetries.

The second symmetry is the nonmotion e composed with any other symmetry.

One can tell the inverse of a symmetry by looking at it.

The relation is satisfied by both of them.

We will pay a lot of attention to the consequences of these small observations.

The symmetries of an equilateral triangular plate are six.

The square card has symmetries.

Show that r k D r3k D rm, where m is the unique element of f0; 1; 2; and 3g.

Another way to list the symmetries of the square card makes it easy to compute the products of symmetries quickly.

We have been refining our idea of a symmetry of a geometric figure while looking at some examples.

A mathematical model is being developed for a physical phenomenon such as the symmetry of a brick or a card.
We have decided to ignore the path by which they arrived at the final position of the object.
A map from R to R is a symmetry.

The idea that a transformation is rigid or non-disturing can be formalised.

If for all points a, b, and R, we have d.

An affine isometry of R3 is always defined on a subset of R3.

If R is not contained in a two-dimensional plane, the affine extension is uniquely determined.
We will assume that these facts are established in Section 11.1.

If we assume that R is a square, we can say that it lies in the.x; y/-plane.

We can show how to map the set of R to itself.

The centroid of the figure is the intersection of the two diagonals of R. .C / D C is still equidistant from the four vertices since is an isometry.

We can assume that the figure is located with its centroid at 0, the origin of coordinates.
The results quoted in the previous paragraph relate to a linear isometry of R3.

The same argument and conclusion can be used for many other geometric figures.

There is at least one point that is mapped to itself by every symmetry of the figure.
If we place a point at the origin of coordinates, every symmetry of the figure extends to a linear isometry of R3.

A polyhedron is a three-dimensional object located with its centroid at the origin of coordinates.

The restriction to R of a linear isometry of R3 is every symmetry of R.

Our symmetries are implemented by 3 by 3 matrices.

There is a matrix A for all points in our figure,.x/ D Ax.5 The unique linear extension of the composed symmetry 12 is created by the composed linear transformation T1T2.
The matrix product A1A2 implements T1T2 if A1 and A2 are the matrices.
The product of the corresponding matrices can be used to compute the composition of symmetries.

We can use this observation to do the bookkeeping for symmetries.

We will find the matrix that implements the symmetry of the square or rectangle.

There is a brief review of elementary linear algebra in the appendix.

The linear transformation implementing each symmetry is unique if we insist on it.

The figure is in the.x; y/- plane with the sides parallel to the coordinate axes.

The coordinate axes will coincide with certain axes of symmetry.
We can orient the plane so that the axis of rotation for r1 coincides with the x- axis, the axis of rotation for r2 coincides with the y- axis, and the axis of rotation for r3 coincides with the z- axis.

The x-coordinate of a point in space is not changed by the rotation r1 The standard matrix of a linear transformation is determined by how the rotation r1 implements it.

If T is a linear transformation of R3 then the 3-by-3 matrix with columns T should be used.

Oe2/ D Oe2 and r1.

We can see what the rotation r2 and r3 do in terms of coordinates.

The square of any of the Ri's is E and the product of any two of the Ri's is the third.

The multiplication table for the matrices R1, R2, R3 and E is the same as for the symmetries r1, r2, r3 and e.

The axes of symmetry for the rotation a, b, and r coincide with the x-, y-, and Z-axes if you choose the orientation of the square in space.

The set of matrices fE; R; R2; R3; A; B; C; Dg have the same multiplication table as the corresponding set of symmetries.

We can conclude that CD D R2 if we compute it.

We can return to the question of whether we have found all the symmetries of the square.

We think that the figure lies in the.x; y/-plane with sides parallel to the coordinate axes and centroid at the origin of coordinates.
Since a symmetry is an isometry, it must take each edge to an edge of the same length.

The edges are at.

Oe1 and.w Oe2.
The symmetry is determined by.
Oe1 and.w Oe2 since they are the basis of the plane.

The only two edges that are longer than 2w are.

Oe1 and.
Oe2.
Oe2 must be.w Oe2 since these are the only two edges that are longer than 2.
There are at least four possible symmetries of the rectangle.
Four distinct symmetries have already been found.

For the square,.w Oe1 and.w Oe2 must be contained in the set f.w Oe1 and.w Oe2g.

If.w Oe1/ is.w Oe2, then.w Oe2/ is.w Oe1.

There are at least eight possible symmetries of the square.

There are eight distinct symmetries.

The multiplication table for the symmetries of the square should be reproduced by the products of the matrices E, R, R2, R3, A, B, C, D.

The equilateral triangle has six symmetries.

The multiplication table for the equilateral triangle is reproduced by the products of the matrices.

T.x/ D Ax C b is an affine transformation of R3.

The next four exercises outline an approach to showing that a sym metry of a square or rectangle sends a line to another line.

The approach is based on the idea that there is more than one truth.

There is a line segment called a2 in R3 and a2 D fsa1 C.1 s/a2 W 0 s 1g.

If a1 and a2 are elements of R and v 2, then a1 D a2.

V is not contained in any nontrivial line segment in R.

The configuration has symmetry because the objects are the same.

I could switch the objects around if you looked away, but you couldn't tell if I had moved them.
Symmetry is not a geometric concept.

There are three symmetries where two objects can be switched at the same time.

One object can be put in the place of a second, the second in the place of the third, and the third in the place of the first.
All the objects are left in place.
There are six symmetries.

The symmetries of a configuration of objects are called permutations.

We have to come up with a system for keeping track of things.

The objects occupy three positions.
We can describe each symmetry by recording the final position of the object that starts in position i.

The product of permutations is computed by following each object as it moves.

The element on the right is the first permutation and the one on the left is the second permutation.

The order in which permutations are ordered is important.

The permutations of n identical objects can be marked by two-line array of numbers, containing the numbers 1 through n in each row.

If a permutation moves an object from position i to position j, the corresponding two line array has j positioned below i.
The first row of numbers are usually arranged in increasing order.
The same rule is used for permutations of three objects.

The multiplication of permutations is an associative activity.

Each object is left in its original position by an identity permutation.

In either order, the product of e with any other permutation is.

There is an inverse permutation 1 for each permutation.

In either order, the product is e.

These properties are even more obvious because of a slightly different point of view.

If the range of f is all of Y, there is an x 2 X such that f.x/ D y.

The inverse of f and satisfying f is called X.
The unique element of X is f 1.y/.

Consider maps from a set.

The composition of maps is associative.
If f and g are on the same map, then the composition f i g is also on the same map.

The identity map idX is the identity element for this product; that is, for any f 2 Sym.X/, the composition of idX with f in either order.

The inverse of a map is the inverse for this product; that is, for any f 2 Sym.X/, the composition of f and f 1 in either order is idX.

Functions and sets are discussed in Appendix B.

A permutation of n objects can be identified if the function maps to j.

The permutation 1 2 3 4 5 6 7 D 4 3 1 2 6 5 7 in S7 is identified with a map of the stars.
The multiplication of permutations is the same as the composition of the maps.
The three properties listed for permutations are related to the corresponding properties of the maps.

We usually write Sn for the permutations of a set of n elements.

It's not hard to see that the size is nS D n.n 1/.2/.1/.
The image of 1 under an inverted map can be any of the n numbers 1, 2, and so forth.

S52 is not a different kind of object than S5 which has 120 elements, because we couldn't begin to write out the multiplication table for S52 or even list the elements of S52.

A permutation such as 1 4 2 3 that leaves all other numbers fixed is called a cycle.

The order of the entries is more important than the first entry.

If each leaves fixed the numbers moved by the other, there are two cycles.

The expression D 1 4 2 3 5 6 is a product of disjoint cycles.

Let's see how permutations are calculated.

The permutation on the right is taken first in calculating permutations.

The first of the permutations takes 1 to 4 and the second takes 4 to 7, so the product takes 1 to 7.
The product takes 7 to 6 because the first leaves 7 fixed and the second takes 7 to 6.
The product takes 6 to 4 and the first takes 6 to 5.
The product takes 4 to 2 after the first and second leaves.
The first takes 2 to 3 and the second takes 3 to 1, so the product takes 2 to 1.
This closes the cycle.
The product fixes 5 because the first permutation takes 5 to 6 and the second takes 6 to 5.
The product fixes 3 because the first takes 3 to 1 and the second takes 1 to 3.

The product of the cycles 1 4 2 3 and 5 6 is the permutation D. Their product is not related to the order in which they are multiplied.

If the kth power of the identity is greater than the lower power of the identity, a permutation has order k. A k-cycle has order k. The least common multiple of the lengths of the cycles is equal to the order of the disjoint cycles.

2 4 3 5 1 6 7 9 10 is the least common multiple of 4, 2, and 3.

There is a perfect shuffle of a deck of cards.

The first cut of the deck is perfect, with the half deck consisting of the first n cards placed on the left and the last n cards placed on the right.
The first card from the right goes on top, followed by the first card from the left, then the second card from the right, and so on.

The perfect shuffle of a deck of 10 cards is 10.

The order of the perfect shuffle is 6.

The permutation 2 Sn is written in cycle notation.

The first number (1 a1 n) is not fixed.
Write a2 D.a1/ a3 D.a2/ D.a1// a4 D.a3/ D...a1 Since each number is in f1, it cannot be all distinct.
There is a number k such that a1; a2; : : ; ak are all distinct, and D a1.
The permutation permutes the numbers fa1, a2, and 2.

Consider the first number b1 62 fa1; a2; : : ; akg that is not fixed.

Write b2 D.b1/ b3 D.b2/ D.b1// b4 D.b3/ D...b1 The numbers fa1; a2; : : ; akg; akg; b g among themselves, and the remaining numbers f1; 2; : : : ;ng n.FA1; are permuted.

"Shuffling cards and stopping times" was written by Aldous and P. Diaconis.

Continue this way until the product of disjoint cycles is written.

The phrase "continue in this way" is a signal that it is necessary to use mathematical induction to formalize the argument.

The factors are unique, and the order in which they are written is irrelevant, because disjoint cycles 1 and 2 commute.

There is no preference for the first entry in the cycle notation.
In order to not have to make an exception for the identity element, we regard e as the product of the empty collection of cycles.

Every permutation of a finite set can be written as a product of disjoint cycles.

If jXj D 1 is the only permutation of X, there is nothing to do since the only permutation of X is the identity e. Let be a non identity permutation of X.

Since jXj is finite, there is a number k such that x0; x1; and xk are all distinct.
The sets X1 D, X2 D, and X1 are each invariant and therefore the product of 1

2 D jX2 is a product of disjoint cycles.

It is also a product of disjoint cycles.

A variation on the same argu ment states that the cycle containing x0 is uniquely determined by the list.

Any expression of a disjoint cycle must contain it.

The decomposition of 2 as a product of disjoint cycles is unique and the product of the remaining cycles in the expression yields 2.
The cycle decomposition is unique.

Compare the multiplication table of S3 with the set of symmetries of an equilateral triangular card.

Matching elements of S3 with symmetries of the triangle makes the two multiplication tables agree.

There are a pair of 2-cycles that don't commute.

The perfect shuffle is a deck of 2n cards.

Suppose X is the union of two disjoint sets.

The permutation 2 Sym.X/ is invariant for X1 and X2.

Show that the assumption.xk/ D xl for some l, 1 l k leads to a contradiction.

We can do computations in these systems in order to find out the order of a perfect shuffle of a deck of cards.

We return to familiar mathematical territory in this section.

We study the set of numbers.
In elementary school, multiplication in the integers can be interpreted in terms of repeated addition, for example if the number is a and n and the number is a C.

The set of numbers f1;.2; : : g by Z and the set of natural numbers f1; 2; 3; : : g by N are given.

If n 2 N is positive, we write n 0 and say it's non- negative.

If a is non-negative, the absolute value is equal to a otherwise.

The prop erties of the integers are addition and multiplication.

A b is written for a C.

The sum and product of positive integers is positive.

We write a b if a b is 0 or a b if a b is 0.

The total order is on the integers.
It is never true that a; for distinct integers a; b; b; or b; or a; or a; or a; or c; or a; or a; or a; or a; or a; or a; or a; or a; or a;

There is a more precise version of the statement in the proposition.

Many of the most interesting properties of the integers have to do with divisibility, factorization, and prime numbers.

The emphasis in this section will be more on a systematic, logical development of the material, rather than exploration of unknown territory, since these concepts are already familiar to you from school mathematics.
The main goal is to show that every natural number has a unique factorization as a product of prime numbers; this is more difficult than one might think.

Divisibility is a definition.

If there is a quotient of aq D b, then b can be divided by remainder without the need for a remainder.

Let a, b, c, U, and v represent numbers.

U and v can't be zero.

Suppose first that both are positive.
We have a maxfu.
If it holds, then D v D 1.
If D 1 and D 2 are included, then also D 1 and D 2.

Both of them are.1.

Both have the same sign.

For (b), let the number b be the same as the number d and the number d be the same as the number vi.

0 D.uv 1/b.
The product of nonzero is either b D 0 or uv D 1.

The parts are left to the reader.

Some natural numbers like 2 and 13 are called prime because they can't be expressed as a product of smaller natural numbers.

Natural numbers can be written as a product of prime numbers, for example, 42 D 2 3 7.

If the number is greater than 1 and not divided by another number, it is prime.

To show that every natural number is a product of prime num bers, we have to use mathematical induction.

Natural numbers can be written as a product of prime numbers.

We need to show that all natural numbers can be written as a product of prime numbers.

The second form of mathematical induction is used to prove this statement.

For all natural numbers, the number r is a product of primes.

It is a product of primes if n happens to be a prime.
n can be written as a product if 1 a n and 1 b n are included.
According to the hypothesis, each of a and b can be written as a product of prime numbers.

In mathematics, 0 is the sum of an empty collection of numbers and 1 is the product of an empty collection of numbers.

A lot of arguments are saved by this convention.
The prime factorization is the product of an empty collection of primes.

There is a question of whether there are infinitely many prime numbers or finitely many.

Greek mathematicians solved the question before 300 B.C.
The solution is attributed to a book by the name of Elements: Theorem 1.6.6.
There are a lot of prime numbers.

There is at least one prime number for all natural numbers.

There is at least one prime number.
Suppose that there are at least k prime numbers.
We will show that there is at least one prime number.
Let p2 be a collection of k prime numbers.
Consider the number M D p1p2 pk C 1.

According to the previous proposition, M is a product of prime numbers.

If p is a prime dividing M, then fp1; p2; : : ; pk; is a collection of k C 1 prime numbers.

The fundamental fact about divisibility is that it is possible to divide an integer by any divisor and get a quotient q and a remainder r, with the remainder being smaller than the divisor.

The proof of this fact is related to what you learned in school.

We just put q D 0 and r D a.
If a d, first we guess an approximation q1 for the quotient such that 0 q1d a, and compute r1 D a q1d, which is less than a.

We put q D q1 and r D r1 if r1 d is finished.

We try to divide r1 by d, guessing a q2 such that 0 q2d r1, and compute r2 D r1 q2d.

We are done if q D q1 C q2 and r D r2 are put together.

We are trying to divide r2 by d.

It must be that 0 rn d because the rn are non negative.

We stopped and put q D q1 C q2 C qn and r D rn.

The procedure for a negative is the same.

A proof of the existence part of the following assertion can be found in the description of this algorithm.

The essential idea is the same in the following proof as it is in the previous proof.

There are unique integers q and r, D qd C r and 0 r d, given the numbers a and d.

The case should be considered a 0.

Take q D 0 and r D a.

Now suppose that's true.

Assume that the existence assertion holds all nonnegative integers that are smaller than a.

We are done after a D.q0 C 1/d C r.

The case is dealt with by 888-276-5932 888-276-5932 888-276-5932 888-276-5932.

Take q D 1 and r D aCd if d is 0 Assume that the existence assertion holds for all non positive integers whose absolute values are smaller than jaj.
It holds for a C d, so there are two integers q0 and r.

We have shown the existence of q and r. Suppose that a D qd C r and a D q0d C r0 have 0 r and d. It's the same as r r0 D.q0 q/d, so it's the same as r r0 The only possibility is r r0 D 0.

We have shown the existence of a prime factorization, but we have not shown that it is unique.

The issue is addressed in the following discussion.
The greatest common divisor of two integers can be computed without knowing their prime factorizations.

If V 2 N divides m and n, then V also divides, the natural number is the greatest common divisor of nonzero integers m and n.

If the greatest common divisor is unique, then notice.

Several important properties are recorded in the following proposition.

Let I.m; n/ D fam C bn W a; b 2 Zg; n and x 2 I.m; n and x 2 I.m.

A natural number, that is a common divisor of m and n and an element of I.m, is the greatest common divisor of m and n.

The greatest common divisor is m and n. Sequences jnj > n1 > n2 0 and q1 are defined.

After no more than n steps, this process must end.

We have a system of relations that include m D q1n C n1 n D q2n1 C n2.

A b is the first row of Q 1. nr is a common divisor of m and n.

The greatest common divisor is given by g.c.d.

q1 is the sequence of quotients, q2 is the sequence of quotients, q3 is the sequence of quotients, q4 is the sequence of quotients, q5 is the sequence of quotients, q6 is the sequence of quotients.

18 D 8233 is a result of Definition 1.6.12 If g:c:d:.m and n are nonzero, they are prime.

If 1 2 I.m; n/ is the case, m and n are relatively prime.

1 D 3 21 C 4 16 is a relatively prime number.

If p is a prime number and a is a nonzero number, then either p divides or p is a prime number.

The proof of uniqueness of prime factorizations can only be shown from here.

Let p be a prime number.

If p divides a, then p divides b.

If p does not divide a, then 1 D, a C Vp, and V are prime.

Multiplying by b shows that b is divisible by p.

Suppose that a prime number p divides a product a1a2 One of the factors is divided by p.

A natural number's prime factorization is unique.

The case n D 1 cannot be written as the product of any nonempty collection of prime numbers.

Assume that the assertion of unique factorization holds for all natural numbers less than n. q1 divides n D p1p2 and hence is equal to one of the pi.
It follows that p1 D q1.

It's based on the idea that r D s and qi D pi are for everyone.

If a factor is found, we can search for other factors.

We will continue this procedure until all of the factors are present.
This procedure is not very efficient.
No truly efficient methods are available for large natural numbers.

If V 2 N divides each ai, then V also divides, it's a natural number.

If the greatest common divisor is unique, then notice.

We prove existence by means of a generalization.

There is a natural number d and an n-by-n matrix Q if Q is not zero.

The proof of proposition 1.6.10 establishes the base case n D 2.
Fix n > 2 and assume the assertion holds for lists of less than zero nonzero numbers.

The most common divisor of nonzero integers a1; a2; : : : ; an exists and is a linear combination of a1; a2; : : : ; an.

The greatest common divisor is d.

The most common divisor of a1; : : ; an is denoted g:c:d.

Use the well-ordering principle to complete the sketch of the proof.

We have a D qd C r.

You have to show that if y 2 I.m; n/, then xy 2 I.m:n/.

You have to show that if y 2 I.m; n/, then b divides y.

For each of the following pairs of numbers, write g:c:d:.m; n:c:d:.m; and n:c:d:.m as a linear combination of m and n.

If you want your program to give the greatest common divisor, you need to explicitly give the given nonzero integers.

This method is taught in school.

In 7 hours it will be 4 o'clock if it is now 9 o'clock.

The clock number 12 is the identity for clock addition, which means that the clock will show the same time in 12 hours.
If it is now 12 o'clock, then after 5 periods of seven hours, it will be 11 o'clock.
All the usual laws of arithmetic are held except for the fact that division is not generally possible.

Imagine taking a number line with points marked and wrapping it around the clock face with 0 on it.

The numbers are put on the clock face and then on the number line.
The numbers are on the number line and on the clock face.

The numbers are on the number line.

When the distance between the two numbers on the number line is multiple of n, the interval between the two numbers on the number line wraps a number of times around the clock face.

If a is congruent to b modulo n, and if a is divisible by n, we write a b.mod n.

If a b and b c are both strued by n, then a D 0 is also strued by n.

When the number line is wrapped around the clock face, there is a set of all integer points that land at the same place.

The set OEa is called a congruence class.

The unique number of r is 0 r n and the r is divided by n. The interval f1; 1; : : ; n 1g is the unique element of OEa.

When the number line is wrapped around the clock face, the label on the circumference of the clock face is remn.a/.

The following are equivalent for b 2 Z.

If a.mod n/ is used, then a.mod n/ is used for any c 2 Z.

This shows that.
If c b.mod n/, then c a.mod n/.
This shows the meaning of (a) and (b).

The unique element of OEx f0; 1 is characterized by remn.x/.

This shows that is true.

There are distinct classes called OEn 1.

These classes are not related.

A0; b; b0 be in the same place as a0 and b.mod n/.

We now describe the natural structure of the set Zn.

Let A and B be elements of Zn, and let 2 A and B be elements of A and B.

The choice of representatives doesn't affect the classes.

If a0 is another representative of A and b0 another representative of B, then a0.mod n/ and b b0.mod n/.
It makes sense to define A C B D OEa C b andAB D OEab.

In order to see what the issue is in the preceding discussion, we need to look at another example in which we can't define operations on classes of numbers in the same way.

Let N and P be the set of negative and non-negative integers.
One of these two classes has every single digit in it.
The sum of a positive number and a negative number can be either positive or negative, depending on the choice of n and p. It doesn't make sense to define N C P D s.n C p/.

It is easy to check that sensible operations on Zn satisfy most of the usual rules of math once we have defined them.

The correspond ing property of Z is followed by the assertions in the proposition.

The commutativity of multiplication on Zn is shown.

The proof of multiplication proceeds the same way as the proof of associativity.

The reader is responsible for checking the details of the other assertions.

On the other hand, nonzero elements can have a zero product.

If there is a nonzero element, we call it a zero divisor.
Zero divisors are the case in Z6.

Many elements have multiplicative inverses and can be invertible if there is an element such as OEaOEb D.

There are multiplicative inverses in Z14, OE1, OE3 and OE9.

You can check to see if the remaining nonzero elements are zero divisors.
In Z14, every nonzero element is either a zero divisor or an inverted one, and there are just as many zero divisors as inverted elements.

In Z7, every nonzero element is invertible.

You should compare the behavior of multipli cation in Z to that of these phenomena.

In the exercises for this section, you are asked to investigate further the zero divisors and invertible elements in Zn, and solutions to the form ax b.mod n/.

Let's look at some examples of computations in Zn and congruences.

The main principle to remember is that a b.mod n/ is the same as a OEa D OEb in Zn.

This is easy because of 4 1.mod 5/.

Let's figure out a few powers of 4 modulo 9.

It follows that in Z9 and Z7 for all natural numbers.
We only have to find the conjugacy class of 237 modulo 3 if we want to compute 4237 modulo 9.

It is the same as the assertion for all OEa 2 Z7.

We only have to check this for the 7 elements in Z7, which is easy.
This is a special case of Fermat's little Theorem.

It is not necessarily true that OEb D OEc is in Zn.

If a and b are relatively prime, this is always possible.

The result can be extended to any number of simultaneous congruences.

If a and b are prime natural numbers, then, and V are also prime natural numbers.

There is an x that is.mod a/ and.mod b/.

They are used for studying the numbers.

Congruence modulo n is a fundamental relation in the integers and any statement about it is equivalent to a statement about Zn.
Sometimes it's easier to study a statement about congruence in the integers in terms of Zn.

The objects Zn are fundamental building blocks in a number of theories.

All finite fields are constructed using Zp.

The development of algebra is a fundamental part of modern digital engineering because the algebraic systems Zn were first studied in the nineteenth century without any view toward practical applications.

For digital communication, for error detection and correction, for cryptography, and for digital signal processing 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884 888-349-8884

The exercises ask you to prove parts of the proposition.

The addition in Zn is both commutative and associative.

If you can prove that OE0 is an identity element for addition in Zn, you can prove it for all.

If you can prove that multiplication in Zn is commutative and associative, you can prove that OE1 is an identity element for multiplication.

Write out multiplication tables for Zn.

The zero divisors in Zn for n 10 can be catalog using your multiplication tables from the previous exercise.

Guess which elements in Zn are invertible and which are zero divisors based on your data.

A more analytical approach to invertibility and zero divisors is provided in the next three exercises.

Suppose a is relatively prime to n.

There is no such thing as C nt D 1 if a is not relatively prime to n.

It's possible that OEa isn't invertible in Zn.

The Zn is defined by LOEa.OEb.
LOEa is surjective because OE1 is not in the range of LOEa.

Consider Exercise 1.7.11.

The 8x 12.mod 125/ is a congruence.

This exercise will show you a proof of the Chinese remainder theorem.

You can find x such as that x 3.mod 4/ and x 5.mod 9/.

The set Q of rational numbers, the set R of real numbers, or the set C of complex numbers are Polynomials.

I am confident that most readers will not be disturbed by the informality of defining polynomials as an expression of the form.

It is possible to codify the concept by defining a polynomial to be an infinite sequence of elements of K, with all but one entry equal to zero.
The sequence.3; 2; 7; 0; 0; would be interpreted as 7x2 C 2x C 3.
The operations of addition and multiplication can be changed to other functions.

The operations of multiplication and addition satisfy the same properties as those listed for the integers.

The properties can be verified using the definitions of the operations and the corresponding properties of the operations in K.

The reader is responsible for verification of the remaining properties listed in the propo sition.

deg.p/ is the degree of p 2KOEx.

If p D Pj aj xj is a nonzero polynomial of degree k, the leading coefficients of p are ak and akxk.

If the leading coefficients of the polynomial are 1 it is said to be monic.

The leading coefficients are.2 and.7 and the degree of p D is 7.

The goal of this section is to show that the theory of divisibility is similar to the theory of divisibility for the integers, which was presented in Section 1.6.

The results of this section are similar to the results of Section 1.6, with the proof also following a nearly identical course.
The role that absolute value plays for integers is discussed in this discussion.

If both of U are nonzero, then part (a) is done.

If either of them had a positive degree, the other would also have a positive degree.

U and v are elements of K.

If g D vf and f D ug are included, then g D uvg or g.1 uv/ D 0.

k D 0 meets the requirement if g D 0 does the same.

The reader is responsible for the remaining parts.

The prime number's analogue should not admit any nontrivial factorization.

It is possible to "factor out" an arbitrary nonzero element of K, but this should not count as a genuine factorization.
A nontrivial factorization f.x/ D g.x/h.x/ is one in which both of the factors have a positive degree and the analogue of a prime number is a polynomial.
They are called irreducible rather than prime.

If the degree of the polynomial is positive and it cannot be written as a product of two different degrees, it is irreducible.

The following statement is an analogue of the existence of prime factorizations.

A positive degree polynomial can be written as a product of irreducible polynomials.

The degree is proof.

The definition of irreducibility states that every degree 1 is irreducible.
Let f be a polynomial of degree greater than 1, and make the hypothesis that every polynomial whose degree is less than f can be written as a product of irreducible polynomials.
If f is irreducible, then it can be written as a product.
Each gi is a product of irreducible polynomials, and so is f.

There are infinitely many irreducible polynomials in KOEx.

There are fields with finitely many elements.

The same proof can be applied for such fields.

It is not true that KOEx has irreducible polynomials.

Every irreducible polynomial has degree 1 in COEx and degree 2 in ROEx.
There are irreducible polynomials in QOEx.
If K is a finite field, KOEx has many irreducible polynomials.

Let's use an example to recall the process of long division.

Let p D 7x5 C 3x2 C x C 2 and d D 5x3 C 2x2 C 3x C 4.
We want to find polynomials such that p D qd C r and deg.r/ deg.d /.

Let p and d be elements of KOEx.

There is a monomial m D bxk 2KOEx and a polynomial p0 2KOEx.

Put m D.an and p0 D p md.

The leading term for both p and md is anxn.

Let the elements of KOEx be p and d.

There are two types of polynomials in KOEx: p D dq C r and deg.r.

We divide p by d and get a monomial quotient of less than the degree of p. We continue in this fashion until we get the rest of degree less than deg.d.

The formal proof is based on the degree of p.

If deg.p/ deg.d /, then put q D 0 and r D p.

We can write p D md Cp0 in the lemma.

We have p D qd C r if we put q D q0 C m.

If g 2 koEx divides p and q, then g also divides f, which is a common divisor of nonzero polynomials p and q.

We are going to show that two nonzero polynomials have the same common divisor.

The greatest common divisor is unique to multiplication by a nonzero element of K. The most common divisor is monic, which has a leading coefficient of 1.
The one that is monic is the greatest common divisor.
The monic greatest common divisor of p and q is q/.

The results are similar to the results for the integers, and the proof are almost identical to those for the integers.

The proof of the analogous result for the integers should be modeled on the proof of the other results.
You will be able to better understand the proofs for the integers and how they have to be modified to apply to polynomials.

The set I.m; n/ D famCbn W a; b 2 Zg was an important part of our discussion of the greatest common divisor.

The set is similar to the one for polynomials.

The greatest common divisor is an element of I.f; g/.

The proof of 1.6.10 should be mimicked with the proof of 1.8.13.

We do repeat division with remainder, each time getting a portion of smaller degree.

The process must end with a zero remainder.

The final nonzero remainder is a common divisor of f and g.

The division with remainder gives.

C 9 x 3 x2 6 x3 C 6 x4 3 x5 C x6/ D

The most common divisor of f.x/ and g.x/ is d1.

The definition is d1.x/ D.1=4/s.x/f.x/ C.1=4/t.x/g.x.

If g:c:d:.f; g/ D 1 are used, they are prime.

If 1 2 I.f; g/ is the prime, then the two polynomials f; g 2KOEx are relatively prime.

If p divides the product, then f or p divides g.

One of the factors is divided by p.

The factorization of a polynomial into irreducible factors is unique.

Up to multiplication by nonzero elements, the irreducible factors are unique.

This is the end of our treatment of unique factorization.

You haven't yet learned any general methods for recognizing irreducible polynomials or for carrying out the factorization of a polynomial by irreducible polynomials.

We finish this section with some important results.

There is a polynomial q that is called p.x/ D q.x/.x a/ C p.a/.

If, and only if, x splits p.

When the remainder r is a constant, write p.x/ D q.x/.x a/Cr.

Substituting a for x will give you p.a/ D r.

2 K is a root of a p 2 koEx if p. In the irreducible factorization of p, the multiplicity of the root is k if x.

A degree n degree p 2 koEx has most of its roots in K, counting with multiplicities.

The sum of all roots is at most n.

The exercises ask you to prove parts of the proposition.

If you want to prove that multiplication in KOEx is commutative and associative, you have to show that 1 is an identity element for multiplication.

Let p 2 be irreducible.

The following statements need to be proved.

For each of the following pairs, find the greatest common divisor.

A computer program can be used to compute the greatest common divisor.

Make your program look for polynomials that are similar to the one you're looking for.

Show that g:c:d.f1; f2; : ; fk/ and make a reasonable definition.

Counting Counting is a fundamental and pervasive technique.

The method of inclusion-exclusion and the binomial coefficients will be discussed in this section.
First, the binomial coefficients and the binomial theorem are used to prove Fermat's little theorem, then inclusion -exclusion is used to get a formula for it.
The formula for the'function is used to get the generalization of the little Fermat theorem.

There are some problems with counting subsets of a set.

There are 8 D 23 subsets.

There are two possibilities for the n elements: the element is in the subset or not.

2n subsets of a set with n elements are possible because the in/out choices for the n elements are independent.

Define a map from the set of subsets of f1; 2; : : ; ng to the set of sequence; sn/ If you put a 1 in the j th position if j 2 A and a 0 in the j th position otherwise, the corresponding sequence will be formed.

It's not hard to check that.
There are just as many subsets of f1; 2 as there are of n 0's and 1's.

There are 2n subsets in a set with n elements.

There are 10 subsets of f1; 2; : : ; 5g and you can list them all.

The number of two-element subsets should be marked by.

It's known and equal to n.n 1/2.

In the following way, build the nS permutations of f1; 2; : : : ; ng.

This can be accomplished in a number of ways.

The numbers have some properties.

The number k 2 Z is a natural number.

Part (a) can be seen from the definition.

Both sides of (c) are equal to 1.

Both sides of (c) are zero when k is negative or greater than n.

Consider paths in the.x; y/-plane from.0 to a point.

We only admit paths of a C b "step" in which each step goes one unit to the right or one unit up, and each step is either a horizontal or vertical segment.
A path can be specified by a sequence of R's and b U's if the steps to the right and b are exactly the same.

Let the numbers x and y be the variables.

The sum of 2n monomials, each obtained by choosing x from some of the factors, is called.x C y/n.

The total number of subsets of an n element set is the sum over k of the number of subsets with k elements.

The binomial theorem is followed by putting x D y D 1.
The binomial theorem is followed by putting x D 1; y D 1.
The two sums in part (c) are equal by part (b), and they add up to 2n by part (a); hence each is equal to 2n 1.

P is a prime number.

I hope you discovered the first part of the next lemma while doing the exercises.

If a element is relatively prime to n, it has a multiplicative inverse.

There are numbers such as C nt D 1 if a is relatively prime to n. A and n have a common divisor if a is not prime to n. Say k t D a and ks D n. Reducing modulo n means that OEa is a zero divisor in Zn, and therefore not invertible.

0 in Zp is prime to p, so OEa is in Zp by part.

P is a prime number.

Since the case a D 0 is trivial and the case a 0 follows from the case a > 0, it makes sense to show this for a natural number.

The proof can be seen by the way in which it is presented.
The assertion is obvious.
If that is the case, then assume a 1/p and a 1mod p/.

1/ C 1.mod p/ D a is where the first and second congruences are from.

The conclusion of part a is the same as the conclusion of part D. By the previous proposition, if a is not divisible by p, OEa has a multiplicative inverse.

Adding both sides of the equation gives OEap 1 D. This is the same as ap 1 1.mod p/.

If you have three subsets of a finite set, you want to count A, B and C. If you just add jAj C jBj C jC j, you might have too large a result, because any element that is in more than one of the sets has been counted multiple times.

If an element lies in A B C, then it has been counted three times in jAj C jB C j.

Adding back jA B C j would fix this.

Our next estimate is jA [ B [ C j D jAj C jBj C jC j jA Bj jA C jB C jA B C j: 1.9.

We would like to get a generalization of Formula for any number of subsets.

Let us be any set.

The characteristic function of the relative complement of X is 1U nX D 1 1X and the characteristic function of the intersection of two sets is 1XY D 1X 1Y.

X 0 is the relative complement of a subset X U.

Let A1 and A2 be subsets of U.

Evaluate both sides of the equation at some point.

Such permutations are called derangements.

Ai doesn't depend on the k subsets.

Dn is the closest to nS.

Diners leave coats in a restaurant.

The formula was obtained before.

There is a proof of this in the exercises.

Take D 7 and D 4.

Take D 16 and D 7.

It is possible to prove the binomial theorem.

That's 3n D PnkD0

There are three exercises that outline a proof.

The prime to the prime p is assumed to be relatively prime.

Fermat's little theorem says ap 1 D 1 C kp for some k.

Each i is divided by pki.

Function from G G to G is what an operation or product on a set G is.

There is a product that is associative.

There is an inverse.

It is easy to make a concept out of the common char acteristics.

The process of elevating recurring phenomena to a con cept has several advantages.

C is a group of numbers.

The group H is the same as the group R.

It's clear that OE1 2 is what it is.

This proof is intended to show the power of abstract thought.

R is an abelian group.

A multiplication is associative.

In a ring, multiplication doesn't need to be commutative.

The ring is called a commutative ring if it is multi plication.

The number systems R, Q, C are rings.

A noncommutative ring is a set of n-by-n matrices.

Many rings have an identity element for multiplication.

The only units in the ring are.

The units are not zero constant.

As with groups, rings may be related to one another, and under standing their relations helps us understand their structure.

Q is a subring of Z.

Let T be a matrix with real entries.

The idea of the direct sum of rings is needed first.

The map is clearly defined.

One to one is what the map says.

The second assertion is more subtle than the first.

Every nonzero element is a unit.

Three fields are R, Q, and C.

Z isn't a field.

ROEx isn't a field.

Zp is a field if p is a prime.

K can be any field.

x 1

OExb is a ring isomorphism.

It's the science of secret or secure communication.

R. Rivest, A. Shamir, and L. Adleman10 are well known.

Let D pq.

Write h D tm C.

It is easy to find large prime num bers and to do the computations needed to decode.

I choose a small prime r D 55589 and send you an email.

"ALGEBRA IS INTERESTING" is your secret text.

It is not possible to factor 300 digit numbers.

One doesn't do such computations by hand.

Chapter 2.

In the previous chapter, we saw many examples of groups, and finally arrived at a definition, or collection of axioms, for them.

We have e D ee0 since it is an identity element.

If hg D e, then h D g 1.

If gh D e, then h D g 1.

When gh D e is similar, the proof is there.

G is an element of a group D.

Since gg 1 D e, it follows from the proposition that g is inverse of g 1.

La stands for left multiplication.

G by Ra.x.

Ra stands for multiplication.

There is an assertion that the map La has an inverse map.

The computation shows that La i La 1 D idG.

The inverse maps of La and La 1 are shown.

Ra's proof is similar.

Consider Z12 with multiplication.

The order of a group is based on its size.

There are examples of groups of small order.

A group of order 4 is the set of symmetries of the rectangular card.

The isomorphism is the map.

If one of the two groups is abelian and the other is nonbelian, the two groups are not isomorphic.

S3 and Z6 are both abelians.

Z1 is the unique group of order 1.

Z2 is the unique group of order 2.

Z3 is the unique group of order 3.

Z5 is the unique group of order 5.

All groups of order are abelians.

This shows that.

A set M with an associative operation is referred to as juxt sitapoion.

The product of five elements at a time is well-defined and independent of the way that the elements are grouped for multiplication.

M is a juxtaposition.

The product of one element is D a.

The product is defined by (a) and (b).

Fix elements a1; : : an 2 M.

1 element at a time is fined.

Determine the symmetry group of the rhombus.

If H is abelian, G is too.

Group with a small num ber of elements are investigated by means of their multiplication tables.

The group also has four elements.

G is abelian.

For all a; b 2 G,.ab/ 1 D a 1b 1.

B 2 G, aba 1b 1 D e.

2.2. Subgroups and Cyclic Groups

M is a set with a not necessarily associative operation.

A1a2 an 2 N.

H is a subgroup of G.

The inverse h 1 is an element of H.

H can be a subgroup.

These observations can be used as a labor-saving device.

If At A D E is an n-by-n matrix, it is said to be orthogonal.

.AB/t D Bt At D B 1A 1 D.AB/ 1 is related.

O.n; R/, then.A 1/t D.At /t D A D.A 1/ 1, so A 1 2 O.n; R/.

G itself and feg are part of any group.

Appendix D can be found here.

There are eight 3-cycles in S4.

We will have a theory that will make the result transparent.

C is hA.

feg is the smallest subgroup.

There is a subgroup that appears in all groups.

The word lattice is used in many different ways in mathematics.

A k D is defined as a 1/k.

The generator is a part of the group.

There is a subgroup of Z called HD i D Z D fnd W n 2 Zg.

H1i D h 1i D Z is a cycle.

Since hOE1i D Zn, it is cyclic.

A group of order n with generator D e2i is called Cn.

There is a chance that two powers coincide.

The order of the by o.a/ is marked by us.

The order of OE5 is 6.

Any two groups of the same order are isomorphic.

There is a group G.

The isomorphic to Z is if a has infinite order.

The group Zn is isomorphic to a finite order.

In this section, we determine all of the groups.

H should be a subgroup of Z.

If d 2 N, then d Z S Z.

If b divides, then aZ bZ.

Let's check part (c) first.

If bja, then a 2 bZ, so aZ bZ.

We verify part a. H should be a subgroup of Z.

H D qd 2 Z.

We have shown that there is a d 2 N and a d Z D H.

Every subgroup of Z is related to Z.

Since d divides n, this is just n.

H is a subgroup of Zn.

Either H D fOE0g or there is a d > 0 such that H D hOEd i.

H is a subgroup of Zn.

The number is divided by d.

There is a subgroup of Zn.

The subgroup of Zn has a dividing n.

The parts are from the proposition.

Part (b) is an exercise for the reader.

The order of OEb in Zn is n/.

The generators are OE1, OE5 and OE11.

The subgroup of a group is called cyclic.

A group is an element of finite order.

A group is an element of finite order.

A group is an element of finite order.

The group has an order '.2n/ D 2n 1.

3 is not a pattern.

The subgroup of S4 is.12/.34/;.13/.

24/; and.14/.23/g.

They are either the same or inverse if they have three digits in common.

This is a 3-cycle.

H is a subgroup.

A group under matrix multiplication can be formed by the set of R that varies through real numbers.

Let S be a subset of the group G.

Let's be a part of the group.

It is a part of order 6.

The order of.1234 is 4.

The order of the two disjoint 2-cycles is 2.

The subgroup lattice of Z 24 should be determined.

The subgroup lattice of Z30 needs to be determined.

H 0 and jH j divide.

This is the conclusion of corollary 2.2.26(b).

There is an order of 2n 2 for n D 3; 4 and 5.

2.3. THE DIHEDRAL GROUPS

These figures are capable of rotation in three dimensions.

D is a subgroup of N D.

Write j D j0 for the flip over the x- axis.

These relations can be used to calculate all products in D.

A 1 D N for all a 2 D is achieved by the subgroup N of D.

The symmetries of the regular polygons are next.

The n-gon has a symmetry group.

The "flips" are symme tries of the n-gon.

There is a regular n-gon in the plane.

Writing N D, show that D D N.

Linear transformations of R3 are used to implement the symmetries of the disk.

JRt D R t J is related to JRt D R t J.

The group Dn is related to the n-gon.

2.4. HOMOMORPHISMS AND ISOMORPHISMS

There is a subgroup of D6 that is similar to D3.

In a regular 15-gon, every third vertex is painted red.

There is no requirement to be injective or surjective.

To make these observations more useful, we number the edges of S.

The square's edges are labeled.

To verify the homomorphism property from the symmetry group of the square into S4.

The diagonals of the square are labeled.

Let G be an abelian group with a fixed number.

Let's look at some general observations.

There are group morphisms.

H is a groupmorphism.

For each g 2 G, '.g 1/ D.

It's followed by the proposition 2.1.1(a) that '.eG/ D eH.

We recall a math ematical notation before we say the next proposition.

f 1.B/ is the conventional notation for the preimage of B.

H is a groupmorphism.

Each subgroup is a subgroup of H.

Under multiplication and inverses, '.A/ is closed.

The elements of '.A/ are h1 and h2.

A such that D '.ai for I D 1 and 2.

The proof of part is an exercise.

A normal subgroup is called G.

Here gNg 1 means 1 W and 2 N.

H is a groupmorphism.

According to the proposition, ker.

We now know that it is a normal subgroup.

H is a groupmorphism.

It's a normal subgroup of G.

It's enough to show that g ker.

The opposite containment needs to be shown.

If we replace g by g 1, we get that for all g 2 G, g 1 ker.

If it is injective, eG is the pre image of eH.

If that is the case, suppose that ker.

W k 2 Zg is the code for OEk.

There are more examples of homomorphisms explored in the exercises.

The sign is called the homomorphism.

The alternating group is also referred to as this subgroup.

The odd permutations contain.12/An.

If k is even, a k-cycle is even.

Do you know what the kernel is?

There is a group with the same name.

H be a combination of G and H.

Let x2 be variables.

It's a homomorphism from Sn to f1; 1g.

If a permutation is a product of k 2-cycles, then./ D.

If,./ D 1, the parity is even.

T./ is a form of Homomorphism.

Two elements a and b in a group G are said to be conjugate if there is an element g 2 G.

When two elements are conjugate is determined by this exercise.

Sm can be considered as a subgroup.

2.5. COSETS AND LAGRANGE'S THEOREM

It is an automorphism and preserves parity.

The set of matrices with positive determinant is a normal subgroup.

Rn is considered to be an abelian group under vector addition.

Let G be a group of abelians.

Consider the subgroup H D fe.

The right coset of H in G is called H g.

It takes a little labor to do this computation.

Similarly, (b) means that bH aH.

H is a subgroup of a group.

Each left coset is nonempty and the union of left coset is G.

If aH bH 6D, let c 2 aH.

The restriction to aH is a combination of aH and bH.

It makes sense for infinite groups.

There are co sets of nZ in Z.

G is the only subgroup other than feg.

There are two groups of prime order p that are isomorphic.

The order of the subgroup is the cardinality.

K H G are part of a subgroup.

We have to use another approach if the groups are infinite.

K H G are part of a subgroup.

The group S3 is considered.

N is not abnormal.

The left and right coset of N are the same.

Using the criterion of the previous exercise, N is normal.

2.6. EQUIVALENCE RELATIONS AND SET PARTITIONS

The center of a group is a normal subgroup.

Do the same for R/.

N is not abnormal.

An equivalence relation is an example of that.

If x and y are the same, then x and y are the same.

A set partition is an example.

In mathe matics equivalence relations are very common.

Equality is an equivalency relation on X.

It's a good idea to be a natural number.

The set of all lines in the plane is an equivalent relation to the parallelism of lines.

Is there an equivalent relation on X.

Let's say we have an equivalent relation on X.

x 2 OEx D OEy, so x y.

If it's 2 OEx, then it's x.

An equivalent relation on X and y 2 X is needed.

Consider an equivalence relation with a set.

The partition of the set X is the collection of equivalence classes.

This is an equivalency relation.

In x and y, the definition of x P y is clearly defined.

An equivalence relation on X is given by every partition of a set X.

We need to show that P D P 0.

That is, Oeda D A.

Let X be whatever you want it to be.

Group G and subgroup H are considered.

Be any map.

The fibers f 1.y/ for y 2 Y are the equivalent classes.

An equivalent relation on a set X is needed.

Such as that 0 D s I f.

Such as that 0 D s I f.

Such as that 0 D s I f.

Choose x 2 X such that y D is.x/.

This is the final proof that s is aective.

There is a set of left cosets of H in G.

We have 1.aH / D fb 2 G W bH D aH g D aH.

The classes for conjugacy are called equivalence classes.

2.7. Quotient Groups and Homomorphism Theorems

The conjugacy class of g is fgg.

Consider a map from a set X to another set Y.

Consider the permutation group with its normal subgroup.

The structure of a group can be found in the set G of left cosets of a N in G.

A D aN and B D bN are used.

We have to check that the result is independent of the choices of 2 A and 2 B.

We have a 1a0 2 N and b 1b0 2 N since aN D a0N and bN D b0N.

Since N is normal, b 1.a 1a0/b 2 N.

This multiplication is a group.

The quotient group of G is called G by N.

The quotient homomorphism is called G.

The normal subgroup is nZ D fn W 2 Zg.

It is surjective because it has the same range.

It is an isomorphism.

Z is a normal subgroup since R is abelian.

Z is precisely the kernels of '.

The group isomorphism is from the quotient group R to T.

The inverse of TA;b is A 1b.

There is a condition for two elements to be congruent.

A D A0 is equivalent to this.

It's an isomorphism of groups.

G be a surjective homomorphism.

In a group H, let's be an element of order.

A normal subgroup of GL.n; R/ is the setSL.n; R/ of matrices of determinant 1.

Z is a normal subgroup of G and is the center.

There are at least two natural models available.

It is the same as the invertible matrices Z.

For each subgroup B of G,'1.B/ is a subgroup of G.

A priori is contained in B.

For a subgroup of G containing N.

This is the same as a 1x 2 ker.

Let B D'1.B/.

B is normal in G.

B is normal in G.

Gxg 1 2 B, by normality of B, so Ng Nx Ng 1 2 '.B/ D B.

In G, B is normal.

Let K be a normal subgroup of G.

It is a composition of surjective homomorphisms.

That is, D K.

We can use the homomorphism theorem again to identify G with N.

The identification carries K to the image of K in G.

The quotient map is marked by G.

Is it K-N-G-N?

The conclusion follows from the first one.

If aN D bN, then b 1a 2 N K D ker.

Z, and in particular OEa has an order.

Z is a surjective homomorphism.

If k is divided by, then kZ Z Z.

A is a subgroup of G.

An is a subgroup of G.

The group of n-by-n invertible complex matrices is known as G D GL.n.

Z is a subgroup of the matrices.

A Z D A fE W is a root of unityg.

If n is odd, the same holds with C replaced by R.

If n is odd, then GL.n; R/ DSL.n; R/Z.

G be a surjective homomorphism.

'.aN / D '.a/ is well defined.

This gives an associative product on some subsets.

Take A D aN and B D bN.

N is a normal subgroup.

The inverse of TA;b is A 1b.

G is a finite group.

Aut.G/ is a group.

Show that the map is called Z.G/.

The map c isn't surjective.

Is every automorphism of S3 inner?

If G is abelian, you need to prove it.

Suppose G is abelian.

Chapter 3.

The multiplication is done coordinate by coordinate.

Za is also known as Zb.

Both of these groups are normal in A B.

It's evident that Assertion (c) is true.

It is similar to hOEbi hOEai S Za Zb.

The group of symmetries of the rectangle should be called G.

The difference is more psychological.

This is a follow up to example 3.1.3 and k.

Group operation C is for abelian groups.

Nr are subgroup with N1 C N2 C Nr D G.

Take G to be Z2 Z2.

The two subgroup have a trivial intersection.

The coordinate-by-coordinate multiplication on the product P D A1 An of groups makes it into a group.

Group P D A1 An has a direct product.

B are surjective.

A B is defined by D.

I am the same as in the previous exercise.

There are conditions for to be surjective.

Just as in a direct product, Dn D NA, N A D feg, and A S Dn is the same.

We have a commutation relation for r 2 N.

We have G D NA D AN, AN D feg, and A S G.

The construction is easy.

A is a part of the automorphism group of N.

N A is a group under the multiplication.n.

We need to check the associativity of the product.

Let.n; a/,.n0; a0/, and.n00 be elements of N I A.

It's easy to know if it's the identity of N I A.

We have to find the inverse of an element.

This is an exercise.

The direct product is a special case of the semidirect product, with the homomorphism, trivial,,.a/ D idN for all a 2 A.

Aut.Z7 is defined by,.OEk/ D 'k.

Do you know if Zn I Z2 is isomorphic to Dn?

Z4 has a subgroup that is similar to Z2.

The quotient Z4 is similar to Z2.

Z4 isn't a direct or semidirect product of Z2.

V 2 K, v 2 V are for all.

A K-vector space is a type of space over K.

The parts were proved above.

The range of T is T.V.

A linear transformation from V to C is called R 1 f.t/g.t/ dt.

One verification is needed for the sake of illustration.

The quotient map is a group Homomorophism.

V is a linear map of spaces.

V is the quotient map.

T respects multiplication by numbers.

The details are left to you.

There is a surjective linear map.

Let M be a subspace of V.

A linear isomorphism is defined by T.x/ C M.

T.x/CM is a group isomorphism.

V be a linear map.

The quotient map is marked by V.

M is V is N.

T.v/ defines a group homo morphism from V to M.

A and N should be subspaces of V.

A and N are contained in V.

S is dependent on other things.

A linear independent set can't contain the zero.

This shows that fen g is independent.

There are two things to notice: enem D enCm and e0 D 1.

Equation gives,1 D 0 as well.

A is a proper subset of B and v 2 B n A.

The relation is contrary to the independence of B.

Let A0 D A n fv1g.

A basis of V is B.

The set is called B.

B is a subset of V.

B is a basis.

It follows the same way as span.B/ D V.

V is said to be infinite-dimensional.

V has a finite basis if it is finitedimensional.

A subset of a set is a basis.

B is a basis of V.

If fv1 is linearly independent, then C is the zero matrix.

Let V have a finitedimensional space with a spanning set.

Let Y D be a linearly independent subset of V.

From Propostion 3.3.25, any basis of a finitedimensional space is finite.

Dim.V / D 1 is written if V is infinite-dimensional.

W is finite and dim.V is dim.

The basis of W is V.

Y is a subset of W.

The basis of V is contained.

A basis of V is the proposition 3.322, B.

The lemma is an axiom of set theory.

There is a zero-dimensional space with one element.

The basis of the empty set is unique.

Oen is the standard ordered basis of Kn.

The two n-dimensional spaces over K are linearly isomorphic.

The case is left to the reader.

Be a function.

There is an equivalency between (a), (c), and (d).

N is a subspace of V.

Let fw1 be a basis of W.

T I S.Pi,i wi / D T.Pi,i xi / D Pi,i wi.

Suppose v 2 ker.T / W 0.

This shows that ker.T / W 0 D f0g.

We showed that V D ker.T / W 0.

Let V be a finite-dimensional space and N be a sub space.

There is a subspace M of V such as V D N.

Complements of a subspace are not unique.

S is a subset of a space.

Rather than using the proposition.

V is not a straight line.

Every subset of V is finite.

V does not admit a sequence of linearly independent subsets.

It is infinite-dimensional.

V has a subset that is infinite.

Let fv1; v2; : : : ; vsg be a basis of N.

It's easy to check that S C T is also linear.

The map defined by is the inverse of a linear map.

V should be over a field K.

HomK.V; W is a vector space.

V D HomK.V; K/ is a vector space.

B spans V.

The coefficients are zero, which shows that B is linearly independent.

The basis of V dual to B is called B.

The number f.v/ is given by the pair of v 2 V and f 2 V.

There is a map called K.

V is the same.

V is a finitedimensional space over a field.

fvg is linearly independent if v is not a zero.

There is a basis B of V with v 2 B.

V is a linear isomorphism.

It's essential that finite dimensionality is present.

.V is not surjective.

S and T are subsets of V and W is a subspace of V.

I'm a subspace of V.

The parts are left to the reader as exercises.

We have W W II and W W III for part (d) and part (c) respectively.

All these subspaces are equal.

The dual space of sub spaces and quotients can be described with the aid of annihilators.

We can check the surjectivity of this map.

Let T 2 HomK.V; W /.

HomK.V has dim.V and dim.W.

The reader is invited to look at the map.

Kn is a linear isomorphism, meaning that it takes a vector in W to its coor dinate vector with respect to C.

HomK.V; W is the first column of OET C;B.

OET C;B is injected.

T.vj / D PniD1 ai;j wiWD.

It is surjective.

It's immediate from (a).

Let B D.v1 and C D.w1 be used.

OES T D;B and OES D;C OET C;B agree.

There is a ring isomorphism between EndK.V and Matn.K.

The inverse OEidC;B is invertible.

There is a problem of determining the matrix of a linear trans formation with respect to two different bases.

Two bases of V are B and B0

The part was proved above.

Let A be an n-by-n matrix, Q an inverted n-by-n ma trix, and set A0 D Q.

Let E D.

B0 is a basis of Kn.

It's just Q 1.

It's enough to be able to compute OEidB;E and OEidE;B.

Matn.K is an equivalent relation to EndK.V.

The determinant and trace are important similarity invariants.

The sum of the diagonal entries is the trace of a square matrix.

Let C 1 D.di;j be an inverted matrix.

The trace is similar to the one before it.

Section 8.3 shows how the determinant is dicussed.

There are parts to Lemma 3.4.6.

Let W be a subspace and let V be a finitedimensional space.

Let W be a subspace and let V be a finitedimensional space.

The entry is equal to the one on the wii.

Both T 0 and T 0 are linear transformations.

As D fa1 C C as W ai 2 Ai for all ig.

A finite group is finitely generated.

There is a Pi ai xi.

This is the standard basis of Zn.

Abelian groups don't need to be free.

xng is a basis of G.

The map in (b) is a group homomorphism.

This shows that there is no difference between (a) and (b).

S is a subset of Zn.

It makes sense to consider linear independence of S over Z or over Q.

A minimal generating set is the basis of a free abelian group.

Let b 2 B n B0

B0 is contrary to the linear independence of B.

B0 doesn't generate G.

Any basis of a free abelian group is finite.

Since S is finite, it is contained in the subgroup B0 of B.

Let G be a free abelian group.

The base case n D 1 is taken care of by Z.

Let N be a subgoup.

The subgroup of Z is composed of Zn to Z and,n.N.

Let A be a subgroup of G.

A is a generating set with no more than n elements.

N is free.

If ai;j D 0 is diagonal, then D j is.

A should be an m-by-n matrix.

A version of Gauss ian elimination is used to Diagonalize the matrix A.

In Matm.Z/, E CVEi;j is inverse.

Two rows are interchanged in the second type of elementary row operation.

Pi;j is its own inverse.

There is a Smith normal form on my page.

Elementary row oper ations are similar to elementary column operations.

A has a nonzero entry in the.1; 1/ position.

Minimum size is nonzero.

Take note of the.1; 1/ entry of the matrix.

A is the row matrix A D.

A1; : : ; an.

Let.v1 be a sequence of elements of Zn.

The j th column of the n-by-n matrix is vj.

In Matn.Z/, P is invertible.

trivially means condition

Let A D.ai;j.

If N is not a trivial subgroup, there is nothing to do.

Let xsg be a generating set for minimum cardinality.

All the dj are nonzero.

Vng is a basis of zinc.

The dj's are all nonzero.

It's free, of rank no more than n.

N should be used to indicate the kernels of '.

There are be abelian groups and Bi Ai subgroup.

If it is a finitely generated tor sion group, an abelian group is finite.

The least positive element of ann.G/ is the period of G.

The period and order of G coincide if G is cyclic.

Z2 Z2 has periods 2 and 4.

G is a finitely generated abelian group.

If G is free, it is abelian.

Part a is an exercise.

A is a free submodule and B is a free submodule.

The rank of B is determined.

This proves something.

Any module that is free is free.

If p splits, then Zx is Zx S Zp.

If p does not divide, then pZx D Zx.

PZx is Zx.

Zx D pZx follows.

p is a prime number if G is an abelian group.

G is a space over Zp.

G is an abelian group.

Check the space axioms.

G is a surjective group.

The proof of uniqueness is in Theorem 3.6.2.

The period of G is marked as D bt D a.

It is followed by the unique of dimensions.

P is a prime number.

Fix a problem.

We have ai D bi D p.

The in variant factor decomposition is called the direct product decomposition.

Where p is a prime, let G be an abelian group of order.

There is a unique sequence of exponents.

A partition is a sequence of natural numbers such as Pi ni D n.

Let G be part of a group of order.

The type of G is the partition.

P should be a prime.

Gi D 2 G W, i x D 0g.

The most common divisor is rsg.

Suppose that xj 2 Gj for 1 j s and Pj xj D 0.

G is a finite abelian group.

If p does not divide the order of G, GOEp D f0g.

p is a prime number.

GOEp is a subgroup of G.

The result is given by applying the previous proposition with D pki.

The primary decompo sition of G is the decomposition of Theorem 3.6.14

The Zn D G1 Gs can be computed.

GOE5 D hOE12i is a subgroup of Z60 of size 5.

We can get a procedure for resolving any number of simultaneous congruences using the same considerations.

There is an x such that is x xi.mod.

Up to congruence mod n D, x is unique.

Take ri D n as a first approximation.

5 D 60, r1 D n=4 D 15, r2 D n=3 D 20, and r3 D n=5 D 12.

Put y1 D 45, y2 D 40, and y3 D 36.

mod 60 can be reduced to find a unique solution.

A subgroup of a finite abelian group G is GOEp.

The sum of those Ci whose order is a power of p is noted.

Consider G D Z30 Z50 Z28.

For all other primes, GOEp D 0.

The abelian group is a direct product of the prime power order.

The elementary divisors of G are the orders of the factors.

Look at the example G D Z30 Z50 Z28 again.

The elementary divisors are 4; 2; 3; 25; and 5.

To get the largest in variant factor, multiply the largest entries of each list.

The largest entry should be removed.

Continue until all the lists are exhausted.

The biggest invariant factor is 4 25 3 7 D.

Classify the abelian groups.

The order '.N' is where the function is.

There is a multiplicative group of nonzero elements.

They are both cyclic.

Part (a) is a follow up to the first part.

Let p be an odd prime.

It is a cycle.

Let G be a group of abelians.

Show that ZB is free and that G is a group.

Find the elementary divisor and invariant factor decompositions for each group.

Find the elementary divisor and invariant factor decompositions for each group.

Consider Z36.

Consider Z180.

Let G be a finite abelian group.

Chapter 4.

The regular polyhedra are the cube, the dodecahedron, and the icosahedron.

If you have models of the regular polyhedra, we can get their rotation groups.

The first thing we do is the tetrahedron.

Including the identity, we have 12.

The crossproduct of Ov1, Ov2, and Ov3 is referred to as Ovi.

One of the symmetry of the tetrahedron is 5.

The cube is now being considered.

We need to find an injective homomorphism into some permuta tion group that is caused by the action of the rotation group on some geometric objects associated with the cube.

The rotation group of the cube is changed into S4.

S4 is similar to the group of 3-by-3 signed permutation matrices.

The f Ovi g form a right-handed orthonormal basis with the help of two further unit vectors.

R should be a polyhedron.

The dodecahedron has six 5-fold axes through the centroids of the oppo site faces.

The dodecahedron has 15 2-fold axis through the center of the oppo site edges.

The dodecahedron has 60 tries.

Consider any edge of the dodecahedron.

One of our families has six edges.

There is a cube in the dodecahedron.

By explicit computation, we could do this.

The rotation group has 60 elements and must be A5.

The rotation groups of the dodecahedron and icosahedron are similar to the group of even permutations A5.

The data of the previous exercise and the method of Exercises 4.1.2 and 4.1.3 can be used to calculate the matrices for the icosahedron rotation.

To be a unit, we need to be a unit with a P.

I mean the inner product in R3.

Symmetry of the figure is a type of reflection that sends a geometric figure onto itself.

It is easy to check the multiplication table under multiplication and inverse.

We have 14 matrices that allow us to compute a lot of products.

These are new matrices.

There are sixteen symmetries of the square tile, eight rotation, five reflections, and two reflection rotation.

A group is the set of sixteen symmetries.

The formula for the reflection J needs to be verified.

J is linear.

The standard basis of R3 is found with its matrix.

There is a plane that doesn't pass through the origin.

There are eight non rotation matrices in the set of eight rotation matrices.

The sixteen symmetries form a group.

It preserves distance, d.a/, b/, for all.

Let's use an isometry of Rn such as.0/ D 0.

Another characterization of matrices is given by exercise 4.4.3

F D ffi g is an orthonormal basis.

There is a linear map on Rn.

There is a matrix with respect to some orthonormal basis.

Every orthonormal basis has a matrix.

U is an isometry.

The determinant is.1.

Let A be a matrix.

A change of basis matrix is called V.

A linear isometry's determinant is.1.

The SO.n; R/ is a set of n-by-n matrices with a determinant equal to 1.

A matrix is a reflection matrix.

The role of reflections and rotation in SO.2; R/ should be sorted out next.

Consider the situation in three dimensions.

C1 is an eigenvalue for any element of SO.3.

jP must be a reflection.

A 2 SO.3 would be R/.

A has an eigenvector v with eigenvalue 1.

jP is a rotation of P.

Suppose A 2 O.3; R/ n SO.3; R/.

So is a reflection.

It is a reflection.

Show that j is not a straight line.

Show that j, 1 D j., if it's a linear isometry.

If A is a matrix with a single digit number, show it.

The columns of A are normal.

An orthonormal basis is the rows of A.

There are three reflection matrices that make up a matrix of rotation-reflection.

Every symmetry is implemented by a linear isometry of R3 for "reasonable" figures.

I don't distinguish between linear isome tries and their standard matrices.

R is an index 2 subgroup of G and G D.

The full symmetry group of the cube has 48 elements.

The figure must come in two versions that are mirror images of each other.

A band with symmetry.

Observe that i2 D 1 and that for any rotation.

Chapter 5.

Let's do an action of G on X.

x0 is sent to x00, then x1 is sent to x00.

O.x/ is the orbit of x.

Group G acts on its own by left multiplication.

Let G be any group and H any subgroup.

The action is not straight forward.

If they have the same cycle structure, they are conjugate.

The set of subgroup of G should be any group and X.

G acts on X by conjugate, cg.H / D gHg 1.

Let G do something.

Stab.x/ D fg 2 G W gx D xg.

Stab.x/ is a subgroup of G.

Let G act on X and x 2 X.

This calculation shows that is defined.

G is finite.

Consider the actions of a group G.

The partition of n determines the parameters of cycle structures.

The identity is the only element of cycle structure 14.

There are three elements.

There are three-cycles.

There are four-cycles.

Let X be the family of k element subsets.

The proof is the same as before.

Let D r1 C r2 C rk.

Group G can act on a set.

Gx D fgx W g 2 Gg.

On the set f1; 2; : : : ;ng, the symmetric group Sn acts naturally.

H D fe; 2/g S3.

H is normal.

There are two blue and two white beads.

There are four arrangements and two other arrangements.

Consider the set F D f.g; x/ 2 G X W gx D xg.

Every arrangement is fixed.

The rotation of the nonagon is 2.

3 and hrki have three different paths.

A white bead is required for any fixed arrangement.

There are 12 fixed points in X.

There are many ways to color the faces of a dodecahedron.

A mathematical object is a set.

Aut.G/ is a subgroup of Int.G/.

W. R. Scott wrote Group Theory.

Z, and let r D.

The previous result and corollary 3.6.26 have been followed.

Several abelian groups are deter mined in the exercises.

Aut.Z2 Z2/ S S3 can be shown in two different ways.

Consider the actions of a group G.

24 D 1 C 6 C 3 C 8 C 6 is given.

There are nonidentity elements in G.

G, of order p2, is not a constant.

It is possible to choose h 2 G n hgi since o.g/ D p.

They commute when g and h are both of order p.

I say that hgi hhi D feg.

G is abelian.

Let G be a group of order.

Every subgroup is normal if G is abelian.

The power of a prime number is called jGj D pn.

Suppose the result holds for all groups of order pn0, where n0 is.

There is a quotient map.

The existence of sub groups of order a power of prime is being investigated.

If b 2 G and b D e, then also ba D e.

If ap/ 2 X, then ap 1/ 2 X as well.

The fixed point of X is e.

OEG W H is divided by p.

H should act by left multiplication.

H is fixed.

Let's look at the condition for a coset xH to be fixed.

This is possible for each h 2 H, hxH D xH.

That is the number of h 2 H x 1hx 2 H.

The set of g 2 G is called gHg 1 D H.

The normalizer is NG.H / H.

We can considerNG.H, which has a size that is divisible by p.

P is a p-Sylow subgroup.

The set of conjugates of P is found in the family X of p-Sylow subgroups.
P should act on X.

There is a subgroup of order.

Let p and q be primes.

Any group of order pq is cyclic.

Group of order pq.

The subgroup of G of order pq is called PQ D QP.

G S Zp Zq S Zpq must be trivial.

The number n5 of conjugates of R is con gruent to 1 and divides 30. n5 2 f1; 6g.

R has 6 conjugates if it is not normal.

We have Aut.Z15/S Aut.Z5/Aut.Z3/S S Z4 Z2.

The value of Aut.Z3/ is 'i'.

We can determine all groups of order 28, up to isomorphism.

Let G be a group.

G is congruent to 1 and divides 28, so n7 D 1.

A subgroup of order 4 is called a 2-Sylow subgroup.

The abelian groups of order 28 are Z7 Z4 and Z2 Z2.

A7 D b4 D 1 and bab 1 D a 1 are elements that make up this group.

There are two exercises called 5.4.13 and 5.4.14.

The subgroup is similar to Z2 Z2.

Z2 is a surjective group.

The show is determined by the kernel.

The non-abelian group of order 20 should be categorized.

The non-abelian group of order 18 should be categorized.

The non-abelian group of order 12 should be categorized.

Without loss of continuity, this section can be omitted.

The techniques of the previous Sec tions will be used to classify the transitive subgroup of S5.

A5 is the only normal subgroup of S5 other than S5 and feg.

The subgroup of S5 that is not equal to S5 or A5 is called G.

Then OES5 W G 5; that is, jGj 24.

I claim that this action is faithful because of the morphism of S5 into Sym.X and S Sd.

The kernels is a subgroup of S5.

2 G is an element of order 5.

A 5-cycle is the only possible structure.

D.1 2 3 4 5/.

That is normal in G.

We know that for any 2 S5.

The group's cardinality is no more than 20.

We have G h; I.

Let H be a subgroup and let G be a finite group.

Group G acts on sets X and Y.

On a set X, let G act transitively.

The subgroup generated by.1234/ and.14/.23/ is D4 S4.

There is a set of conjugates of D4 in S4.

The resulting homomorphism of S4 to S3 should be explicitly computed.

P should be a prime.

The power of p is equal to G.

N is a normal subgroup of a group.

Let G be a finite group with a prime and a subgroup.

P should be a 7-Sylow subgroup.

Chapter 6.

The definitions of rings and fields were encountered by us.

R is an abelian group.

A multiplication is associative.

That is, Ra.nb/ D nLa.b.

The number systems Z, Q, R, and C are rings.

In Section 1.8, we have talked about polynomials in one variable over a field.

QOEx is an element of 7xyz C 3x2yz2 C 2yz3.

The set ROEx1; is a commutative ring with multiplicative identity.

Let R be a field and let X be any set.

The element of S is the opposite of x.

If no proper subring of R contains S, R is generated as a ring.

Let S be a subset of EndK.V.

The following construction is inspired by the previous example.

Let G be a finite group.

You are asked to confirm that ZG is in the exercises.

Rn is the direct sum of R1 and R2.

There is a field.

The group Z2 is written as fe; g, where 2 D e.

It is not automatic.

R can be any ring with multiplicative identity.

First, for any 2 R and 2 Z,.n 1/a D and a.

It is not always injective.

Consider the ring C.R/ of continuous real-valued functions on R.

There are more examples of ring homomorphisms in the exercises.

We define 'a' by this formula.

R0 is a ring.

xn to R0 extending'and sending each xj to aj is what we would like.

If the elements aj are mutually commuted, it makes sense.

R0 is a ring.

The proof is the same as the one variable substitution principle.

We usually use p.a/.

Set a D x.

Let T 2 EndK.V be a vector space over K.

What does this mean, and how does it follow from the proposition EndK.V?

Moreover, 'T.Pi i xi / D Pi.i I D Pi i T i.

We usually write 'T.p/'.

R is a ring with a multiplicative identity element.

The map is in the other direction.

S is a ring ker.

Since it is a subgroup, it is a Homomorphism of abelian groups.

If R and x are the same.

R x 2 ker.

Similarly, xr 2 ker.

nZ is the name of the person.

R can be any ring with a multiplicative identity element.

The latter does not.

Consider the situation of Corollary.

K can be a field.

If K is finite, it is never injected.

Any field is a simple ring.

Part (a) is an exercise.

There are 2 N such as x 2 Ik and y 2 I.

x C y 2 in I

If x 2 I and r 2 R are combined, then there is 2 N such as x 2 In.

I'm an ideal.

R is a left ideal of RS.

R has an identity element.

It's easy to check thatRS is a left ideal.

hSi CRS is a left ideal.

J is a left ideal.

The ideal is two-sided.

If R has an identity element, then hSi C RSR D RSR.

Every ideal is a principal.

The ideal in the field is the principal.

S is a principal ideal of Z.

The proof of (c) is similar to the proof of 2.2.21

g D qf 2 koExf.

Consider a direct sum of rings.

Ri is an ideal person.

The map is a ring isomorphism.

Show that the ring is a unital one.

S should be a part of a set X.

Since y is a matrix, it is possible to write it down.

A nonzero homomorphism of a field is injective.

M is an ideal in a ring R with identity and a 2 R n M.

R should be a ring without identity.

The multiplication in R is well defined.

It is easy to check the ring axioms once we have checked that the multiplication is defined.

If 1 is the multiplicative identity in R, then 1 C I is the multiplicative identity in R.

The ring Z is the quotient of the ring Zn.

G.x/ C.f / D r.x/ C.f.

The example is K D R and f.x/ D x2 C 1.

The following is the basic homomorphism theorem.

It also respects multiplication.

All of the polynomials are in the kernel.

x2 C 1 is a multiple of g.

Since C is a field, ROEx is a field.

The rule of multiplication in C is the same as the rule of ROEx.

Let B D i Bi go.

There is a problem with showing that is surjective.

The subgroup of R and the subgroup of R containing J are both represented by 1.B/.

Each of the next three results is an analogue for the rings of a homomor phism theorem.

Let me be an ideal of R and D.

R is a ring isomorphism onto R.

'.x/ C I.

The quotient map is marked by R and J.

Is it R or J?

The group homo morphism from R to R is defined by '.x/.

R be a surjective homomorphism of rings.

I D fa C r W a 2 A and I g.

It's simple.

If x 2 R is a nonzero element, then R is simple.

Since R is simple, R D Rx is nonzero.

This follows from the previous ones.

Rather than appealing to the result of the proposition.

If.n is a prime, show that nZ is maximal ideal in Z.

The Chinese remainder theorem is given a version by this exercise.

State and prove a generalization of the result for the ring of polynomials.

The ring's nonzero elements can be zero.

You are asked to verify the examples.

An integral domain is the ring of Z.

Any field is an inseparable part of it.

If R is an integral domain, then ROEx is an integral domain.

There are two common constructions of fields.

One construction is treated in two places.

The field of fractions1 is a construction.

R should be an integral domain.

The quotient of S should be indicated by the relation between Q.R/ and S.

The field of fractions and quotient ring should not be confused with the quotient field.

R can be seen as a subring of Q.R/.

Q.R/ is a field.

R is injected into Q.R/.

There are two exercises 6.4.11 and 6.4.12.

Q.R/ is a field containing R as a subring if R is an integral domain.

The field of rational functions is called Q.KOEx.

The field is called K.x/.

This field is called K.x1.

It is said that R has a characteristic p.

J in a ring R has no nontrivial zero divi sors that can be easily characterized.

Every maximal ideal is prime in a commutative ring.

If M is a maximal ideal, then R is an integral domain.

M is a prime ideal.

Every ideal in Z has a form called d Z.

R should be a commutative ring.

The set N of nilpotent elements of R is ideal.

Show that there are no nilpotent elements.

S is a synonym for N ker.

An integral domain is the ROEx with coefficients in R.

The results of the previous problem should be generalized.

Q.R/ is a field.

Q.R/ is a field containing R.

The function for the numbers is d.n.

The function is d.f for the poly nomials.

If it admits a Euclidean function, call it an integral domain R.

Take d.z/ D jzj2 for the "degree" map.

We have d.w/.

Let the nonzero elements be in ZOEi.

A nonzero, nonunit element is in an integral domain.

Proper factors of a are said to be elements b and c.

The irreducible elements are not always prime.

Two elements are said to be associates if they divide the other.

If d divides each ai, then d divides c.

There is a gcd of fai g.

R should be a Euclidean domain.

The ideal in R is the main one.

Every Euclidean domain is a prin cipal ideal domain and a unique factorization domain.

It's natural to ask if the implications can be reversed.

ZOEx is a unique factorization domain.

This means that ZOEx is a UFD.

There are only a few possibilities.

5 W x 2 Zg.

These are two different factorizations.

5 is not a unique factorization domain because some elements are irreducible.

A proper factorization is 1 x/.

x does not have factorization by irreducibles.

A principal ideal domain is a unique factorization domain.

We show that an element can't have two different factorizations.

Suppose that R is an ideal domain.

Let me D. Since R is a PID, there exists an element b 2 I such that I D bR.

Let R be the ideal domain.

Let R be the ideal domain.

A proper factorization of a D bc is admitted.

There is a sequence that is contrary to Lemma 6.5.15.

R should be an integral domain.

R is an integral domain and p 2 R is a nonzero nonunit element.

The "pR prime" is tautological.

Sup pose p is prime and p has a factorization.

P divides a D pr.

b is a unit.

Suppose that R is an ideal domain.

If r is a unit, then J D aR D pR.

There is at least one factorization by irreducibles.

If r D 0 reads 1 D q1 qs.

It's not invertible since irreducibles are not.

p1 is prime according to Lemma.

The ideal in R is the main one.

Try N.z/ D jzj2 for the function.

You can try N.z/ D jzj2 for the function.

There is a unit called a D ub.

A principal ideal domainR is irreducible.

Exercise 6.3.11 is a generalization of the Chinese remainder theorem.

The elements x and y are in good shape.

Show that R is not a ideal domain.

R is a field if it is irreducible.

Show that KOEOEx is an ideal domain.

Divisors in a unique factor ization domain are discussed in the first part of this section.

Write a letter B.

If c is not a unit, consider irreducible factorizations of b and c.

Let a1; : : ; as be nonzero elements in a unique factorization domain.

Let m.j / D mini fNJ be used for each j.

D is a common divisor of fa.

For each j, k.j, and m.j. E divides d.

If b is a unit, then p divides it.

P divides a or b.

R is a factorization domain.

If p is irreducible, there is no need for maximal pR.

ZOEx is a UFD.

If the coefficients are relatively prime, call an element of ROEx primitive.

This discussion can be extended to elements of F OEx.

Such as d1b2 D ud2b1.

Take R D Z.

The product of two primitive elements is primitive.

F.x/ is primitive.

The existence of irreducible factorizations is implied by one property.

The number of irreducible factors appearing in any irreducible factorization of a.

If b is a proper factor, then m.b/ m.a/.

The sequence is finite if it follows.

This is what is shown in the proof.

This is what was shown in the proof.

The ascending chain condition for principal ideals is satisfied by R.

Every irreducible in R is prime.

Lemma 6.6.4 and Lemmas 6.6.13 are the previous ones.

There are elements in 5 that are not prime.

In R, irreducibles are prime.

R is a unique factorization domain.

Where f1 is primitive, let f.x/ 2 ROEx.

The previous exercise should be generalized over a unique factorization domain.

Let R be a UFD.

Show that.R is anintegrative domain for all irreducible p.

The ascending chain condition requires 5 to have elements that are not prime.

Without loss of continuity, this section can be skipped.

The rings ZOEx and KOEx are not ideal.

The ascend ing chain condition for ideals is an equivalent condition to the finite generation property.

The ideal is generated by a subset of a ring R.

Every ideal of R is finite.

The ascending chain condition is satisfied by R.

Every ideal of R is finitely generated.

This shows that R is good for ideals.

R has an ideal that is not finitely generated.

R doesn't meet the ideals of the ACC.

Noetherian rings are named after someone.

The minimum degrees of nonzero elements of J are given.

Noetherian is the name of ROEx.

J D J0 is what we claim.

F 2 J has a degree and leading term.

The Hilbert basis theorem is followed by this.

R is a commutative ring with an identity element.

5 is not a UFD.

In this section, we will look at some elementary techniques for determining whether a polynomial is irreducible.

The problem of determining whether a polynomial is irreducible is limited.

Reducing the polyno Mial modulo a prime is a basic technique for testing for irreducibility.

F is irreducible over Q.

F is irreducible over Q.

Over Q, f.x/ is irreducible.

The last equation shows that,0 is a contradiction.

The Eisenstein criterion says that x3 C 14x C 7 is irreducible.

This is irreducible by Eisenstein's criterion.

First, then f is irreducible.

The details are provided for example 6.8.6.

Chapter 7.

The solution of polynomial equations is the most traditional concern of algebra.

During the Middle Ages, Arabic scholars preserved and augmented this knowledge.

For another 250 years, the matter stood.

Neither of them had time to celebrate his success.

The structure of the field and the symmetry of the roots are what we will be concerned with.

The element i2 is called a discriminant.

The method described in this section can be used to find the roots of f.

The fields in this section are general.

The exercises ask you to check this.

L has something over K.

The basis of M over K is fi j g.

L is related to K.

The set of numbers is a countable union of sets.

On dimK.L/, the second statement is proved.

There is nothing left to prove.

There is an element a1 2 L n K.

We have I, D f.x/KOEx, where f.x/ is an element of minimum degree in I.

This is part a.

This shows both parts.

To find the in verse of an element.

It follows that.

The desired inverse is D 1, so r.

There is a converse to the proposition.

L is a finite extension of K.

We have KiC1 D Ki.aiC1, where aiC1 is a combination of Ki and K.

Ki D KOEa1; : : ; ai for all I and, in par ticular, L DKOEa1; : : ; an.

L is over K.

We have the following propo sition.

K L should be a field extension.

It suffices that K.a; b/ A.

The set of numbers is count able, so it is a countable field.

Show that the finiteness of dimK.L/ is not true.

Let D e2i be the number.

The dimensions are 6 over Q.

Refer to Exercise 7.2.8.

Over Q, 2x C 2 is irreducible.

There are elements of the form a0 C a1 C a22 C a33 C a44.

K L should be a field extension.

I have 62 K and dimK.E/ D 6.

If I have 2 K, then dimK.E/ D 3.

Otherwise, dimK.E/ D 6.

K.,1, and E have dimensions over K.

So dimK.E/ D 3 would be 2 K.

Mapwise and pointwise, 1 to 2.

Aut.L/ is a group of automorphisms.

Aut.L/ is a subgroup of F automorphisms of L.

An automorphism that fixes,j and interchanges the other two roots can be found in AutK.E/.

Fix.H / D fa 2 E W.a/ D for all 2 H g.

K Fix.H is a field for each subgroup.

Part a is an exercise.

Let K D fix.AutK.E.

There are no proper subgroup of AutK.E/ S Z3 because of multiplicativity of dimensions.

The case dimK.E/ D 6 needs to be considered.

K M E is an intermediate field.

H is one of the groups just listed.

h.x/ splits into linear factors in EOEx.

The splitting field for h.x/ is E.

There is an automorphism of E that takes a to b.

K M E is an intermediate field.

Let M D Fix.H.

If H is one of the Hi, then M D M D K.

K.,i

Let K be a subfield of C, let f.x/ 2KOEx be an irreducible cubic polynomial, and let E be the splitting field of f.x/ in C.

The result is remarkable.

There is an infinite set with the split ting field E.

It is irreducible over Q.

K C should be a field.

If f.x/ 2KOEx is an irreducible polynomial, what would it be?

Suppose f.x/ 2KOEx is a irreducible square root.

Why is the denominator not zero?

K.,i

Show that if M is any of the fields.

If M is K.i, then AutM.E/ D A3.

K.0/ that fixes K pointwise.

There is a priori that K Fix.AutK.E.

Since dimK.E/ is finite, E is related to K.

Let V 2 E and p.x/ show the minimal polynomial of V over K.

Let V D V1 be the distinct elements of f.V/ W 2 AutK.E/g.

It follows that p.x/ divides g.x/ since V is a root of g.x/.

The extension is called Galois over K.

Section will give the details of the proof of the equality.

The range of AutM.E/ is all of the subgroups of AutK.E/.

Their roots are 3/.

The fixed field of V is Q.

This result is also proved in a general setting.

The roots are in C.

Degree 4 is over Q and E D Q.

The automorphism is restricted to E.

It is of order 4 and 2.

D4 has 10 groups.

There are ten intermediate fields, including the ground field and the splitting field, in the Galois correspondence.

It's possible to determine each field explicitly.

The field extensions of Q are called Galois.

Chapter 8.

X was fined for g 2 G and x 2 X.

From R to End.M is a ring Homomorophism.

R has a multiplicative identity and M is unital.

V is made into a unital koEx-module.

The corresponding homomorphism is End.V.

There is a structure arising from the linear map T.

An R-module is also known as a left R-module.

A right R-module can be defined similarly.

A ring R is a right R module.

R is the ring of n-by-n matrices.

M should be an R-module and R should be a ring.

By left multiplication, let R act on itself.

M is an R-module.

Any collection of submodules of M should be called fM.

Let M be an increasing sequence of submodules.

Two submodules of M should be A and B.

M and S M are modules.

M is a submodule of RS.

M is finite.

R-module isomorphism is a form of R-module homomorphism.

An R-module homomorphism from M to M is called an R-module endomorphism.

M stands for EndR.M.

R is a commutative ring.

Rn is the set of n-by-1 matrices over R.

R is non-commutative.

R is the ring of n-by-n matrices.

In the next section, we will discuss module homomorphisms.

The direct sum of M1 and M2 is called M1 and M2 respectively.

There is an equivalency between (a), (c), and (d).

The basis for M is a linearly independent set.

The field K is free as a K-module.

Modules over other rings are not free.

xng is a basis of M.

The map in (b) is an R-module homomorphism.

If the map is surjective, B will generate M. This shows that there is no difference between (a) and (b).

A minimal generatiing set is any basis of M.

The elements of B0 are R- linear.

B0 does not generate M.

R can be any ring with multiplicative identity.

R always means a ring and M is an R-module.

The submodule of M is calledIM.

M is a submodule of N.

V is an n-dimensional space over a field K.

V n is a free module over EndK.V.

Let V be a finitedimensional space over a field.

M is a submodule of N.

HomR.M; N is an abelian group.

EndR.M is a ring with addition and multiplication.

M is an R-module and N is an R-submodule.

The quotient map is an R-module homomorphism.

M is an R-module and N is an R-submodule.

R-modules is a homomorphism.

There are ana logues for modules in all of the homomorphism theorems.

The analogous theorem for abelian groups is used to prove each of the theorems.

There is a surjective homomorphism of R-modules.

M is the quotient homomorphism.

Also respects the R actions.

Let R be any ring, M any R-module, and x 2 R.

Ann.x/ is a submodule of R and is a left ideal.

R-module homomorphism of M onto M is allowed.

The quotient map is marked by M.

K is M and N is N.

M is an R-module.

The Factorization Theorem can be proved.

The Diamond Isomorphism Theorem needs to be proved.

R should be a ring with an identity element.

R is a ring with a multiplicative identity element.

There is a special case that all the Mi are equal.

We have to show that to make sure.

Then also y.j and D x.j.

R-bi linear and alternating:.x; x/ D 0 for all x 2 M.

A permutation can be written as a product of transposi tions.

A multi linear function is a multi linear function.

S.'/ D P2S is a multi linear functional.

It is an alternating multi linear functional.

A is shown in a similar argument.

Is alternating.

Let x1, x2, and xn 2 M be used.

I claim that the sum is zero.

This shows that A. is correct.

A sequence of elements of Rn.

Lemma 8.3.5 states that it is an alternating multi linear function.

: ; an/ D

: ; an/ D.a1;

The properties are given in the proposition.

An alternating map.

Let A and B be matrices over R.

Adding a multiple of one column will get B from A.

The formula used to compute the determinants is very inefficient.

When R is a field, the previous proposition provides an efficient method of computing determinants.

The sign of the determinant can be changed by operations of the first and second types.

The number of row interchanges performed in the reduction is 1/k det.B.

The proof of the other is the same as the proof of the other.

The determinant is called the.i; j; minor of A.

The cofactor matrix of A is called 1/iCj det.Ai;j.

The expansion of the determinant is called the cofactor expansion.

Moreover, D ak;j.

1/iCj det.

The cofactor matrix of At is C t.

If the determinant is a unit in R, an element of Matn.R/ is non-invertible.

Det.A/ is nonzero and R is an integral domain.

The following trick can eliminate the additional hypotheses.

Consider the matrix X D.xi;j /1i;j n in Matn.R0/.

It's easy to check that.

Look at that '.

Mi;j / D '.M /i;j.
It follows that '.,.M // D,' using these two observations.

The trick is worth remembering.

If k > n, show that.Rn/k has no nonzero alternating multi linear functions with values in R.

Replacing the j -th column of A with a matrix called Aj is how it is obtained.

We will determine the struc ture of finitely generated modules over a principal ideal domain in the following section.

Represent elements of the R-module M n by 1-by-n matrices.

If fv1 is linearly independent over R and OEv1, then C is the zero matrix.

R is a commutative ring with an identity element.

We will show that.

A D.ai;j is the n-by-m matrix.

The matrix B D.bi;j is a m-by-n matrix.

The vj's linear independence makes it necessary for us to haveAB D En.

We have A0B0 D.

Any two basis have the same qualities.

R is a commutative ring with an identity element.

The zero module is free of rank zero.

This is more than just a convention, it follows from the definitions.

A principal ideal domain is what R means for the rest of this section.

F 0 D span.ff1; : : ; fn 1g/

Let N 0 D N F 0 be a submodule of F.

Every element has a unique expansion.

D y C rh 2 N 0 C Rh.

N D Rh C N 0.

Suppose that M has a finite set.

N is free.

If ai;j D 0 is diagonal, then D j is.

There is a Smith normal form on my page.

A version of Gauss ian elimination is used to Diagonalize the matrix A.

We will only be interested in the case of R in applications.

Two rows are interchanged in the third type of elementary row operation.

The inverse of Pi;j is Matm.R/.

I j is inverse U.

I; j.

Left multiplication by U.

Elementary row oper ations are similar to elementary column operations.

We need another measure of size if R is not Euclidean.

Each nonzero element can be factored as a D up1p2 p since R is a unique factorization domain.

The number of irreducibles appearing in a factorization is unique.

The length of a to be is defined.

A has a nonzero entry in the.1; 1/ position.

Proceed as before if, divides V.

We are done.

Lemma 3.5.10 was replaced by 8.4.8.

It's a field.

A means of computing the greatest common divisor d.x/ is provided by the diagonalization procedure.

The row matrix A D.a1.x/ should be marked with the letters A.

The first column of Q should be marked with.t1.x/.

The matrix is C D.ci;j.

The matrix D.di;j should be marked with D.

We conclude that CD D En is the n-by-n identity matrix.

R and C are in Matn.R/.

In Matn.R/, C is inverse C 1.

N is a free module.

All the dj are nonzero.

R should be a PID.

R is a principal ideal domain in the remaining exercises.

The proof of Theorem 8.4.12 shows that N is free.

A module is a module.

The previous exercise should be retained.

dswsg is a basis of range.

The kernels can be found by finding the one in Zn.

To find a description of ker.A/, use the diagonalization A0 D PAQ.

Let x1 be a set of generators of minimal cardinality.

Dsvg is a basis of N and di divides dj for i j.

The be R-modules and Bi Ai submodules are included.

M is a finitely generated module.

For all I, then s D t, D k and.ai.

The idea of torsion is introduced before addressing the unique statement in the theorem.

0 and r1r2.sx Cty/ D 0.

M is a module if M D Mtor is correct.

There is a chance that M is free of torsion.

G is a finite abelian group.

G is a finitely generated module.

Since it is finite, G is finitely generated.

V is a finitedimensional space over a field.

Let T 2 end.

0 and.Pi,i xi /v D Pi,i T i.v/ D 0.

If a 2 R is a period of 2 M, then Rx S R is a period of 2 M.

Any submodule of M is finitely generated.

If M is free, it is a free module.

Part a is an exercise.

A is a free submodule and B is a free submodule.

The rank of B is determined.

This proves something.

Any module that is free is free.

Let x 2 M, let ann.x/ D.a/, and let p 2 R be irreducible.

If p divides a, then Rx is S R.

If p does not divide, then p Rx.

.p/ is the prescription drug name.

It's that Rx D.

Suppose p 2 R and pM D are irreducible.

R is the quotient map.

A surjective R-module homomorphism is called M.

We are ready for the proof of being unique.

The unique rank is Lemma rank.A0/ D rank.B0/.

Periods of M are as and bt.

We can assume it's D bt D m.

Fix a problem.

We have.ai / D.bi / D.p/.

The invariant factors of M are called the Structure Theorem.

Up to multiplication by units is how they are determined.

There is a unique sequence of exponents.

It is easy to check that M is a submodule.

The period m of M is divided by pr.

If p does not divide, M OEp D f0g.

There is an irreducible factorization of a period of M.

The most common divisor is rsg.

Suppose that xj 2 M OEpj for 1 j s and Pj xj D 0.

It follows that xi D 0.

Let R be the ideal domain.

The period of OEri is pmi.

R should be a principal idea domain.

Set,1.x/ D x2 C 1,,2.x/ D.x 1/, and,3.x/ D.x 3/3.

We can get a procedure for resolving any number of simultaneous congruences using the same considerations.

Take ri D a=,i as a first approximation.

Let a.x/ D,1.x/,2.x/,3.x/.

We have to find ui.x/ such that ui.x/ri.x/ 1mod for each.

We can take ui.x/ to be ri.x/.

There is a list of irreducibles appearing in an irreducible factorization of a period m of M.

AOEpj D M OEpj is the internal direct product of the submodules.

M is an R-module and S is a subset of R.

R is an integral domain with a non-principal ideal J.

Let R be the ideal domain.

P is an irreducible of R.

V is finitely generated as a K-module.

T v 2 V1 for all.

The invariant factors of T are called the polynomials ai.x/.

The companion matrix is by Ca.

The matrix of T is the companion matrix of a.x/.

Both conditions are equivalent.

J / D xd C J D.a0 C ad 1xd 1/ C J.

The generator v0 and a.x/ 2 ann.v0/ are both cyclic with V.

This shows that condition is implied.

The invariant factors of T are V.

The rational canonical form is an invariant for similarity of linear transformations.

T2 D U T1U 1 follows.

The rational canonical form of A is a unique matrix that is similar to a matrix A.

The invariant factors of A are what they are called.

The rational canonical form in Matn.F is the same as the one in Matn.K.

If A and B are the same in Matn.K/, they are the same in Matn.F/.

If they are similar in Matn.F /, then they have the same form in Matn.F.

K F should be fields.

K F should be fields.

Let A 2 Matn.K/ Matn.F.

We need to find the source of the code.

The matrix A can be used to "lift" the transformation T.

D Pi ai;j fi.

We have all f 2 F.

Finally, if Pi, i 2 ker.

Pi, I fi / D Pi, Iei.

D 0 is for all.

T is a part of the module.

As shown in Equation (8.6.1).

V1 has period aj.x/.

Let's say the degree is aj.x/.

We are going to look at how to find a basis for the transformation T determined by multiplication by A is.

The matrix has columns in v1, v2, v3 and v4.

Let A 2 Matn.K.

Let T 2 EndK.V /.

The characteristic polynomial of T is called T.x/.

Let A be the matrix of T with some basis of V.

The product of A is a 2 Matn.K.

The result is related to the proposition.

T has a degree at most dim.V.

There is a relation between the mini mal polynomial and the characteristic polynomial.

The roots of T 2 EndK.V have an important characterization.

Let T 2 end.

If, and only if, T has an eigenvector in V with eigenvalue.

The computation of the rational canonical form of T was made in the text.

All of the coefficients are related to A.x/.

T 2 EndK.V is where V is n-dimensional.

Let A 2 Matn.K be any field.

It is followed by A.A/ D 0.

The analysis continues from the previous section.

There are m blocks and d basis elements in the ordered basis B.

Write p.x/ D xd C ad 1xd 1 C a0

Cp is the companion matrix of p.x/.

Consider the special case that p.x/ is linear.

Up to permutation of the Jordan blocks, the matrix of T in Jordan is unique.

The elementary divisors of T are V.

A 2 Matn.K/ Matn.F is allowed.

Let A 2 Matn.K.

We can give a typical application of these ideas to matrix theory.

The matrix in Matn.K/ is similar to the transpose.

It is similar to Jm.0/.

It's similar to its transpose.

If Ai is similar to Bi, then A is similar to B.

The Jordan matrix is similar to the transpose.

The proof of the proposition is let a 2 Matn.K/.

It is possible to show that A and At are similar.

Similarity in Matn.F is defined in the following way.

J is similar to Jt.

The V is a module with a period power of.x.

x 1 D x 0, x 2 D x 1, etc.

There are no solutions for /x D x r.

The number of Jordan blocks is the number of independent eigenvectors of A.

1 doesn't have a solution.

There are two possible Jordan forms.

Consider one of the submodules.

Call the generator w0, the sub module W, and the period a.x/.

A has a Jordan form in Mat5OEQ.

We can see that the Jordan form of A is J2.1/.

The second method from the text is used to repeat the previous exercise.

There is a basis consisting of eigenvectors.

The Jordan form of A is diagonal.

The divisors of A are linear.

A0 C N, where A0 is diagonalizable, N is nilpotent, and A0N D NA0

N 2 Matn.K should be a nilpotent matrix.

There is a characteristic N.x/ D xn.

The trace of N is zero.

S 1AS is a form of Jordan.

Chapter 9.

The fields in this chapter are not subfields of the complex numbers.

It is a finite field extension.

In particular, b is algebraic over K.

Part (b) is related to part (a).

The smallest subfield of L containing E and F is E F.

If E is over K and F, then E is over F.

Let a 2 E F.

The dimensions of Q.

We have a different point of view regarding the extensions.

K pointwise is fixed by isomorphisms.

There is a technical variation of part (c) of the proposition.

The splitting field is unique to isomor phism.

The idea is to use both of them.

The required isomorphism is L.

L are the roots of p1.x/.

Q, and extending.

Over Q, 3 is irreducible.

A finite-dimensional extension of a finite field is a finite field.

In this section, we look at multiple roots of polynomials and their relation to the formal derivative.

A multiple root is a root of more than one.

The rule from calcu lus is that we define D.xn/ D nxn 1 and extend linearly.

If, and only if, f.x/ is a constant polynomial, Df.x/ D 0 will be.

If K is a field, then f.x/ 2KOEx is irreducible.

If the characteristic of a field K is zero, any irreducible polynomial in KOEx has only simple roots in any field extension.

The map is a combination of two things.

K is a finite field.

Any irreducible polynomial has simple roots in any field extension.

AutK.L/ is a subgroup.

It fixes K pointwise and sends it to V.

L and K.V/ L give the result.

The left coset of AutM.L/ has 1 and 2 in it.

If 2 AutK.L/, then permutes the roots of f.x/.

.,/ is a root of f.x/.

There are only two possibilities for the Galois group.

We want to get similar results in general.

H is a subgroup of Aut.L/.

H is Fix.H and D is D for all 2 H g.

Fix.H is a subfield of L.

If K1 K2 L are fields, then K01.

An element in a field extension of K is said to be separable over K if it has a minimal polynomial.

There is a claim that a 2 M0 D K.

There is nothing to show if Mj 1 D Mj is the case.

L is over K because dimK.L/ is finite.

Both are monic.

The extension is called Galois.

The extension is separable and can be divided into linear factors over L.

M L is an extension of K M L.

L is finite by the hypothesis applied to K.,/, jAutK.,/,L/j D dimK.

We have jAutK.M, jIsoK.M, and L/j D dimK M.

There is an element 2 L that is called L D K.

If the char acteristic is zero, then K is infinite.

V/ are different.

For the moment, put D k, C V.

D jIsoK.K./; E/j n D dimK L.

If the characteristic is zero, Separability is automatic.

Put F D Fix.H.

L is Galois over F.

First note that jH j jAutK.L/j dimK.L/ is finite.

Then F D Fix.AutF.L is by a proposition so L is Galois over F.

There is a V such that L D F.V/.

K L should be a Galois field extension.

AutK.L/, H D AutFix.H/.L/.

M is an intermediate field.

If AutM.L/ is normal, M is invariant.

This is the final part of the proof.

We need the following variant of the proposition.

L is a Galois field extension.

H D AutF.L/ and dimF.L/ D jH j.

The minimal polynomial of V over F is marked by p.x/.

Both are monic.

D deg.p/ jH j is dimF.F.V.

With K D F, we can reach the con clusions.

L was generated by A.

An important technical result is used in the sequel.

Without loss of continuity, the remainder of this section can be omitted.

L should be a field.

Any collection of L automorphisms is linearly independent.

There is a collection of distinct automorphisms of L.

We have n D 0.

Let K L be a field extension.

nC1g is a subset of AutK.L/.

The collection of j is linearly dependent.

The j can't be all distinct by the previous proposition.

L and M are finite-dimensionalg.

The proof of the proposition is given by this exercise.

Show that A is Galois over A.

Let xn be variables, and let K be any field.

Kd OEx1; xn KS OEx1; xn for d 0.

xn/ is a subfield of K.x1.

The Galois group AutKS.x1; is called xn/.K.x1.

By Exercise 9.6.1, xn is the fixed field.

The extension is called Galois and the group is called the Galois group.

If j > n, we put j.x1; : : xn/ D 0

The i is the same as the monic monomials.

A partition is a finite decreasing sequence of nonnegative numbers.

The partition's size is jj D P i.

We need to produce a more obvious basis in order to do this.

The person is positive.

For any permutation, note that.

This proves the parts.

Take n D 4 and D 3.

For any ring A, the ring of symmetric polynomials in AOEx1, xn equals AOE1, and so on.

Go ahead and do this procedure.

P holds.

The least partition is in P.d.

This shows that P holds.

An example is used to illustrate the algorithm.

Consider the numbers in three variables.
Take p D x3 C y3 C.

It is possible for such computations to be automated.

For each d, Kd OEx1; : : : ; xn is invariant.

Let's use variables.

Un denote the roots of this polynomial in a splitting field E.

The Galois group AutK.t1; is a symmetric group.

Fix K and take it to fi.

We know that I differentiate odd and even permutations.

Even if, and only if,.i/ D i, every permutation is satisfying.

Suppose that f is irreducible and separable.

The irreducible separable polynomial is f 2KOEx.

The discriminant has a square root in K.

We don't need to know the roots of f to compute the dis criminant.

F.x/ is irreducible over Q.

The matrix has ai and n columns.

If the matrix R.f has a determinant zero, f and g have a nonconstant common divisor.

We want to show that.

We need to change our point of view at this point.

So has a summand.

1/n.n 1//2 R.f; f 0/.

Check to see if x4 C x2 C x C 1 is irreducible over Z3.

One such is that f.x/.x/ D g.x/'.x/.

The Galois groups of quartic polynomials are determined in this section.

The Galois group of g is over K.

We distinguish cases based on whether the resolvent is irreducible.

It is not possible to distinguish between these cases.

If G is D4 or Z4 then 2 is either 4 or 2.

If H.x/ splits over K.i/, the Galois group of g is Z4.

Let L D K.i/ D K.1; 1 and 3.

The preceding poly nomial is H.x/.

Suppose that H.x splits over L.

OEE W L D 2 and G D Z4 were discussed before the lemma.

The Galois group G is Z4.

We have.1 D.

Soon, some examples of the use of this lemma will be given.

There are two square roots of 3.

It is possible to choose the square roots so that their product is q.

It's not a square in Q.

The group is called Galois.

Take K D Q and f.x/ D 21 C 12 x C 6 x2 C 4 x3 C x4.

If g is not irreducible, it must factor into two irreducible quadratics.

G is either Z4 or D4.

The nonzero roots are 8.

The Galois group is D4.

The coefficients of f are invariant because of the automorphism of C.

i2 is assumed to be negative.

The Galois group G of f is in order 4.

The extension of K is not fixed pointwise.

This is not the same as the other one.

The ground field should be Q and f.x/ D x4 C 5x C 5.

The Eisenstein criterion shows the irreducibility of f.

The alternating group contains G of f.

The resolving field is divided into Q.i/ D Q.

The root of the resolvent is 5.

H.x splits over Q.

The Galois group is Z4.

x4 C px C p is irreducible with the Galois group S4.

The Galois group should be used for the following polynomials.

The computation of Galois groups of polynomials in QOEx of degree 5 or more will be discussed in this section.

The degree partition is m1, m2 and m3.

F is an irreducible polynomial with coefficients in Z.

We won't prove Theorem 9.9.1 here.

The alternating group A5 contains the Galois group G of f.

3 is irreducible and the reduction modulo 5 has a degree partition.

We have to show that some cycle types don't show up.

The group is called D5 rather than A5.

The situation is interesting.

It works and is based on mathematics that you know.

To find the probable Galois groups for each of the quintic polynomials, look at the reductions of the modulo primes.

The article was written by Soicher and McKay.

Chapter 10.

The section deals with a decomposition of a finite group into simple pieces.

Simple is what N is called.

A composition series is a finite group.

A composition series and a group of order 1 are both simple.

A finite group can be solved if it has a composition series in which the successive quotients are in order.

The section develops by means of exercises a description of solvability of groups.

The commutator subgroup OEG; G is normal.

There is a solution to G.

The trivial subgroup feg is a natural number.

G is easy to solve.

We show that G.k/ Hk for all k, in particular, G.r/ D feg.

Exercise 10.2.4 that G.1/ H1 follows since it is abelian.

This is the end of the argument.

You can use whatever criterion you want in the following exercises.

The alternating groups An for n 5 are not solvable in this section.

An is a simple group for n 5.

Conjugacy classes in the symmetric group are mined by cycle structure.

This fact is frequently used.

Two-cycles can be written as a product of one or two 3-cycles.

The elements are a result of two disjoint 2-cycles.

There is a 3-cycle.

All 3-cycles are conjugate in An.

Any 3-cycle is conjugate in An to.123/.

0 D.45/ is even, and 0.123/0 1 D is not.

There is a 3-cycle that does not commute.

N has a 3-cycle in the last two cases.

Since n 5, An has an element.a2a4a5.

N has a 3-cycle.

These computations may look disorganized.

It is necessary to prove that A5 and A6 are easy.

If the centralizer of N is An, then X has 2 elements.

N is the other conjugate of N.

N M D feg is assumed to be minimality.

M N is a subgroup of An.

M 1N 1 D NM D is normal.

A4 is normal.

In G, Z.A/ of A is normal.

The center of An is trivial for all 4 people.

The center of A4 is feg.

An is not abelian.

A normal subgroup of An is 10.3.4.

If you observe that a simple nonabelian group is not solvable, then you can conclude that An and Sn are not.

By inspection, the assertion is valid.

Q is a field extension of degree '.n/ over Q.

n.x/ is irreducible over Q.

Suppose p is not a root of f.

The injective group homomorphism from AutK.u/.E/ into the cyclic group was created by U.

AutK.u/.E/ is a group with an order that divides.

The length of the tower is called r.

L is over K.

The statement is proved by the act of inducting.

The root is f.x/.

There is a splitting field of f.x/ over F.

L is a radical extension of F.

The length of the radical tower is used to prove this.

The Galois group is feg if r D 0 is true.

AutK.K1 is a Galois over K group.

L is Galois over K./ and fur thermore, AutK./.L., S AutK./L.L/.

S AutK./L.L/ is solvable.

L is abelian, so solvable.

Both N and G are solvable.

Without loss of continuity, this section can be omitted.

There is a converse to the proposition.

There is a subtle proof of this.

K E should be a Galois extension.

A multiplicative 1-cocycle is called E.

This nonzero element is called a 1.

D f 1f

This gives me something.

K E should be a Galois extension.

The Galois extension is called E.

We can be a generator of AutK.E.

The order of the group is AutK.E/.

K E is a Galois field extension.

AutK./.E.

will be solved.

E./ is an extension of K./.

Chapter 11.

The norms are preserved.

Affine hyperplanes are not necessarily passing through the origin.

P is always of the form 2 Rn W hx.

There are two distinct points in Rn.

This is not a positive of the proposition.

Every extension in part a is linear if 0 2 R and.0/ D 0 are included.

Any isometry of Rn is affine.

The special case that R contains 0 and.0/ D 0 should be considered first.

Let V D span.R and let fa; akg be a basis of V contained in R.

I claim that T1 is not a straight line.

Rn is a measurement.

This is the final proof of part a.

Rn that ends 0.

There is only one linear extension of 0.

This proves it.

Suppose that 0 2 R and.0/ D 0.

L.x/ C b0 is an affine extension.

Part (d) follows by taking R D Rn.

Any element of O.n; R/ is a reflection matrix.

p D has the desired properties if the orthogonal reflection is in this hyperplane.

The structure of the isometries of Rn is being worked on.

Define the translation for b 2 Rn.

Let be an isometry of Rn and D.

A translation and a linear isometry are what D b is.

A linear isometry is called B.

The isometry is a translation.

It's affine reflection.

Otherwise h j, D tV.s,j is a glide-reflection.

A, B, C, and D are noncoplanar points.

The identity map is called D.

Show that for any two-cycles, b/, T, and b/ is the reflection in the hyperplane.

If, and only if, det.T is even if.

Show that any element of Sn is a product of 2-cycles.

That 2p C 2q > qp follows.

There are five regular polyhedra.

The figure shows a graph.

A tree is a graph that admits no cycles.

The next page shows a tree.

The number of connected components of S n G is G.

G 0 is still connected.

Hence,.G / D.G 0/ D 2.

Hence,.G / D.G 0/ D 2.

The proof has been completed by this.

The sphere has a dodecahedron projected on it.

On the facing page, Figure 11.2.3 shows the projection of a dodecahedron onto the sphere.

For two dimensions, it's easy to get the corresponding result: a fi nite subgroup of SO.

The rotation group SO.3 has a finite subgroup called G.

The number of G's acting on P should be marked by M.

Let xM be a representative of the M.

The exercises give a guide.

The analysis of the last exercise should be extended to the case n D 2.

The choice is a v 2 O nfu.

Consider the size of the O P.

O must contain with any of its elements.

Let the sun set at 2 o'clock.

The choice is a v 2 O nfu.

Consider the size of the O P.

Let v 2 O satisfy all y 2 Onfug.

The subgroup of O.3 is called iH S H Z2.

Define G D H.

The groups are classified by their mode of action, which is more than a classification up to group isomorphism.

V D R3 is a set of linear combinations.

Physical crystals are what we have defined as a crystal in three dimensions.

Solid table salt has a huge array of sodium and chlorine atoms in a ratio of one to one.

The unit cube is a fundamental domain of this lattice.

The translations of the crystal are called L.

The point group is defined by this example.

There is a point group in a crystal.

Take x 2 L and A 2 G0 into account.

Under G0, L is invariant.

The rotation order in G0 is 2, 3, 4, or 6.

This is followed by Exercise 11.3.

L is a two-dimensional lattice.

The basis for L is the linear span of fa.

For x 2 L1, there is a @ 2 GL.R2/ such that '.x/ D@.x/ for x 2 L1.

This proves something.

This proves something.

The statements will be verified in the exercises.

The entire symmetry groups are found in the lattice L.

A line through the origin.

L B is where.v/ is located.

L0 is the lattice generated by L .A.

The semidirect product of L and D1 is 2 G. This isomor phism class is called D1m.

Then G is generated by L.

Put v D.u/.

G is a product of L and D1.

The New York Times has a type D1g crystal.

There is no restriction on the lattice of the group G.

The point group D1 has to consider two cases.

L0 is the lattice generated by L .A.

D2mg is a isomorphism class.

D2c is a isomorphism class.

The other two are not.

The hexagonal lattice is generated by two equal lengths at an angle of 3.

G0 contains reflections in two lines.

There are reflections in the lines spanned by and b.

The case is called D4m.

The case is called D4g.

L and Z6 are semidirect products of G.

The group G has a rotation of order 6 and the lattice is hexagonal.

This is the final proof of the Theorem.

It is possible to classify three-dimensional crystals.

L is a two-dimensional lattice.

By using jja.

L0 is the linear span of fa.

Show that this means v D 0.

Let G be a subgroup.

This is a definite, bi linear function.

If g 2 G, then hhgxjgyii D.

T is an element of GL.n; R/.

T g D '.g/T for g 2 G1 should be verified.

G1 and G2 are conjugate in O.n.

These are considered in turn.

I'm satisfied with your explanation.

I'm not happy with your explanation.

The decision was supported by all of the committee members.

Some committee members did not agree with the decision.

One of the committee members did not support the deci sion.

The decision was opposed by one of the committee members.

The following is not a correct expression.

The decision was not supported by all of the committee members.

We might try to combine the negation with the con junction or the disjunction.

"not(not(A))" is equivalent to A.

Write out the truth table for A and B.

To verify this, write out the truth tables.

"if" is used in the sentences.

The picnic will be ruined if it rains tomorrow.

We notice something amiss when we form the negation.

Tomorrow's rain will not ruin our picnic.

Uncertainty is mitigated by the negation.

It's sometimes used instead of for emphasis.

We are going to discuss quantifiers in a more systematic way.

The quantity x2 is positive for every nonzero x.

If f is differentiable, then it is continu ous.

A positive square is a real number.

Here are a few examples.

How can we form the negation of sentences with tifiers?

Statements and deductions are not the only concerns of logic.

Every subgroup of an abelian group is normal.

3Z is a subgroup of Z.

Every car is going to end up in a pile of rust.

Miata is a vehicle.

Most statements require proof.

We have to assume A and prove B under this assumption.

The language of sets is the subject of this essay.

A set is a collection of objects.

If each element of A is also an element of B, we write A B.

Two sets are the same if they have the same elements.

W x 2 A and x 2 Bg.

There is a relation between intersection and logical conjunction.

Q.x/g, then B D 2 C W P.x/ or Q.x/g.

Every set has a subset of the empty set.

The properties of set operations are discussed here.

A B D B A.

It's just a matter of checking definitions.

Show that A n B D A C.B/.

The Cartesian product is an important construction.

The set of such pairs is called A B.

The surjective function is shown in Figure B.1.

If it is both injective and surjective, f is said to be a synonym.

The image of A under f is referred to as f.A/.

If B is a subset of Y, we write f 1.B/ for 2 X W and 2 Bg.

It is obvious to you that Tn2N Xn D X1 is the set of all nonneg ative real numbers.

A set is not finite.

It is clear that every set is finite or infinite, but it is not possible to determine which is which.

A subset of a set is finite.

A union of finitely many sets is not.

A countable set is 0g.

You need to climb a ladder.

There is a unique composition for all n and for all permutations.

You can get from one rung to the next.

P.k/ means P.k C 1/ for all k 2 N.

This is the final proof of the identity.

I prefer to leave out the proof of the equivalence because it is more abstract than the actual subject matter.

The properties just listed are satisfied by an defined procedure.

A group, a ring, and a field are elements of an associative multiplication.

1 gets us onto the grid.

As intuitive as this picture may be, it seems to require a new principle of induction on two variables.

The following situation is useful to consider.

This suffices to show that for all 2 N and 2 Tn.

Since each t 2 T belongs to some Tn, P.t holds for all t 2 T.

P.1; n/ holds for all 2 N; that is, P.m; n/ holds for all 2 T1.

P.m; n/ holds for all.m; n/ in Tk [ f.k C 1; n/ W n 2 Ng D TkC1.

R is used to solve the equation.

To show that C is a field, you have to check that nonzero Ele ments have multiplicative inverses.

In the form reit, where r > 0 and t 2 R.

The polar form is the form for a complex number.

It's easy to multiplication two numbers written in polar form: r1eit r2eis D r1r2ei.sCt/.

For a complex number, we have Zn D rneint.

A quick review of linear algebra is provided in this appendix.

S is dependent on other things.

A linear independent set can't contain the zero.

V is its own span.

F0g and Kn are subspaces.

V is a subspace of Kn.

R3 is the basis of 5g.

The equation is equivalent to 1 D 0, 2 D 0, and 3 D 0.

M x is a combination of the columns.

Matm;n.K/ and HomK.Kn; Km/ are related to the same thing.

The reader is left to check it.

The identity transformation on Kn is based on En.

M and M 0 are two matrices and M x D M 0x is for all x 2 Kn.

M j D M Oej D M 0 Oej D.M 0/j for all j, so M D M 0.

The person is injective.

The person is surjective.

This proves it.

We need to compute T to do that.

The kernel of T is 2 V W T.x/ D 0g.

Km is a subspace of Kn.

There is a nonzero vectors in the kernels.

S is dependent on something.

Aa D 0 doesn't have zero solutions.

Let fv1 be a set of s vectors.

The span should be all of Kn.

The identity matrix has columns.

Every basis of Kn has elements.

V satisfy.PsiD1,ivi/ C Vv D 0.

Then vj D Pi.

If the span is equal to V, call a set of vectors.

Every set contains a finite set.

Let V be a subspace of Kn.

Any basis of V has no more than n elements.

The basis of V is the linearly independent set.

The basis of V is a subset of any spanning set.

T 1 is the inverse map from V to Ks.

The dimensions of V are called the cardinality of a basis.

The empty set is the basis of V.

The dimensions of f0g are 0.

If V is not the zero subspace, then any basis of V has car dinality greater than 0.

Such that I R D idKm.

The identity matrix is marked by E.

The columns of M are independent.

The M columns have the same span as Kn.

T is a linear isomorphism.

There is a left inverse.

T has an inverse.

M is not spherical.

There is a left inverse.

M has an inverse.

M t is not spherical.

The rows of M are independent.

The rows of M have the same span.

The equivalency of (a) and (c) is given.

T is injective if it has a left inverse.

We have (f) H, (c) H, (e) H.

W be a linear map.

Let S be a set of numbers.

It is also linear.

T.S is independent.

If T.S is a basis of W, then show that S is a basis of V.

The basis of f0g is the empty set.

The map should be a linear one.

Show that if T is injective.

Show the dimensions of the range of T.

If T is surjective, show that.

The standard inner product on Rn is hx.

The norm of x 2 Rn is defined by jjxjj D hx.

If their inner product is zero, the two vectors are orthogonal.

Since fvi, vng is a basis of Rn.

There are many normal bases of Rn.

The Gram-Schmidt procedure goes as follows.

We can assume that the ai are not zero.

Put B1 D fa1

Then wj is related to the other.

If that's the case, put BjC1 D Bj.

A person named Bj C1 is orthonormal.

There are a few suggestions.

One of the most useful topics in algebra is linear and multi linear.

Both group theory and field theory are related to num ber theory.

The books are challenging, but they are accessible with a knowledge of this course.

Document Outline

Preface

Chapter 1. Algebraic Themes 1.1. What Is Symmetry? 1.2. Symmetries of the Rectangle and the Square 1.3. Multiplication Tables 1.4. Symmetries and Matrices 1.5. Permutations 1.6. Divisibility in the Integers 1.7. Modular Arithmetic 1.8. Polynomials 1.9. Counting 1.10. Groups 1.11. Rings and Fields 1.12. An Application to Cryptography

Chapter 3. Products of Groups 3.1. Direct Products 3.2. Semidirect Products 3.3. Vector Spaces 3.4. The dual of a vector space and matrices 3.5. Linear algebra over Z 3.6. Finitely generated abelian groups

Chapter 5. Actions of Groups 5.1. Group Actions on Sets 5.2. Group Actions---Counting Orbits 5.3. Symmetries of Groups 5.4. Group Actions and Group Structure 5.5. Application: Transitive Subgroups of S_5 5.6. Additional Exercises for Chapter 5

Chapter 7. Field Extensions -- First Look 7.1. A Brief History 7.2. Solving the Cubic Equation 7.3. Adjoining Algebraic Elements to a Field 7.4. Splitting Field of a Cubic Polynomial 7.5. Splitting Fields of Polynomials in C[x]

Chapter 9. Field Extensions -- Second Look 9.1. Finite and Algebraic Extensions 9.2. Splitting Fields 9.3. The Derivative and Multiple Roots 9.4. Splitting Fields and Automorphisms 9.5. The Galois Correspondence 9.6. Symmetric Functions 9.7. The General Equation of Degree n 9.8. Quartic Polynomials 9.9. Galois Groups of Higher Degree Polynomials

Chapter 11. Isometry Groups 11.1. More on Isometries of Euclidean Space 11.2. Euler's Theorem 11.3. Finite Rotation Groups 11.4. Crystals