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5.4. Group Actions and Group Structure

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5. ACTIONS OF GROUPS

(b)

Another way is to recall that Z2 Z2 can be described as a group with four elements e; a; b; c, with each nonidentity element of order 2 and the product of any two nonidentity elements equal to the third. Show that any permutation of fa; b; cg determines an automorphism and, conversely, any automorphism is given by a permutation of fa; b; cg.

5.3.8. Describe the automorphism group of Zn Zn. (The description need not be quite as explicit as that of Aut.Z2 Z2/.) Can you describe the automorphism group of .Zn/k ?

5.4. Group Actions and Group Structure

In this section, we consider some applications of the idea of group actions to the study of the structure of groups.

Consider the action of a group G on itself by conjugation. Recall

that the stabilizer of an element is called its centralizer and the orbit of an element is called its conjugacy class. The set of elements z whose conjugacy class consists of z alone is precisely the center of the group. If G is finite, the decomposition of G into disjoint conjugacy classes gives the equation

X

jGj

jGj D jZ.G/j C

;

jCent.g/j

g

where Z.G/ denotes the center of G, Cent.g/ the centralizer of g, and the sum is over representatives of distinct conjugacy classes in G nZ.G/. This is called the class equation.

Example 5.4.1. Let’s compute the right side of the class equation for the group S4. We saw in Example 5.1.17 that S4 has only one element in its center, namely, the identity. Its nonsingleton conjugacy classes are of sizes 6, 3, 8, and 6. This gives 24 D 1 C 6 C 3 C 8 C 6.

Consider a group of order pn, where p is a prime number and n a

positive integer. Every subgroup has order a power of p by Lagrange’s theorem, so for g 2 G nZ.G/, the size of the conjugacy class of g, namely,

jGj

;

jCent.g/j

is a positive power of p. Since p divides jGj and jZ.G/j 1, it follows that p divides jZ.G/j. We have proved the following:

Proposition 5.4.2. If jGj is a power of a prime number, then the center of G contains nonidentity elements.

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We discovered quite early that any group of order 4 is either cyclic or

isomorphic to Z2 Z2. We can now generalize this result to groups of order p2 for any prime p.

Corollary 5.4.3. Any group of order p2, where p is a prime, is either cyclic or isomorphic to Zp Zp.

Proof. Suppose G, of order p2, is not cyclic. Then any nonidentity element must have order p. Using the proposition, choose a nonidentity element g 2 Z.G/. Since o.g/ D p, it is possible to choose h 2 G n hgi. Then g and h are both of order p, and they commute.

I claim that hgi \ hhi D feg. In fact, hgi \ hhi is a subgroup of hgi,

so if it is not equal to feg, then it has cardinality p; but then it is equal to hgi and to hhi. In particular, h 2 hgi, a contradiction.

It follows from this that hgihhi contains p2 distinct elements of G,

hence G D hgihhi. Therefore, G is abelian.

Now hgi and hhi are two normal subgroups with hgi \ hhi D feg and

hgihhi D G. Hence G Š hgi hhi Š Zp Zp.

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Look now at Exercise 5.4.1, in which you are asked to show that a

group of order p3 (p a prime) is either abelian or has center of size p.

Corollary 5.4.4. Let G be a group of order pn, n > 1. Then G has a normal subgroup feg N G. Furthermore, N can be chosen so that

¤

every subgroup of N is normal in G.

Proof. If G is nonabelian, then by the proposition, Z.G/ has the desired properties. If G is abelian, every subgroup is normal. If g is a nonidentity element, then g has order ps for some s 1. If s < n, then hgi is a proper subgroup. If s D n, then gp is an element of order pn 1, so hgpi is a proper subgroup.

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Corollary 5.4.5. Suppose jGj D pn is a power of a prime number. Then G has a sequence of subgroups

feg D G0 G1 G2 Gn D G

such that the order of Gk is pk, and Gk is normal in G for all k.

Proof. We prove this by induction on n. If n D 1, there is nothing to do. So suppose the result holds for all groups of order pn0 , where n0 < n.

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Let N be a proper normal subgroup of G, with the property that every subgroup of N is normal in G. The order of N is ps for some s, 1 s < n. Apply the induction hypothesis to N to obtain a sequence

feg D G0 G1 G2 Gs D N

with jGkj D pk. Apply the induction hypothesis again to N

G D G=N to

obtain a sequence of subgroups

feg D N

G0 N

G1 N

G2 N

Gn s D N

with j N

Gkj D pk and N

Gk normal in N

G. Then put GsCk D 1 N

Gk, for

1 k n

s, where W G ! G=N is the quotient map. Then the

sequence .Gk/0kn has the desired properties.

n

We now use similar techniques to investigate the existence of sub

groups of order a power of a prime. The first result in this direction is Cauchy’s theorem:

Theorem 5.4.6. (Cauchy’s theorem). Suppose the prime p divides the order of a group G. Then G has an element of order p.

The proof given here, due to McKay,1 is simpler and shorter than other

known proofs.

Proof. Let X be the set consisting of sequences .a1; a2; : : : ; ap/ of elements of G such that a1a2 : : : ap D e. Note that a1 through ap 1 can be chosen arbitrarily, and ap D .a1a2 : : : ap 1/ 1. Thus the cardinality of X is jGjp 1. Recall that if a; b 2 G and ab D e, then also ba D e. Hence if .a1; a2; : : : ; ap/ 2 X, then .ap; a1; a2; : : : ; ap 1/ 2 X as well. Hence, the cyclic group of order p acts on X by cyclic permutations of the sequences.

Each element of X is either fixed under the action of Zp, or it belongs

to an orbit of size p. Thus jXj D n C kp, where n is the number of fixed points and k is the number of orbits of size p. Note that n 1, since .e; e; : : : ; e/ is a fixed point of X . But p divides jXj

kp D n, so X has

a fixed point .a; a; : : : ; a/ with a ¤ e. But then a has order p.

n

Theorem 5.4.7. (First Sylow theorem). Suppose p is a prime, and pn divides the order of a group G. Then G has a subgroup of order pn.

1J. H. McKay, Amer. Math. Monthly 66 (1959), p. 119.

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Proof. We prove this statement by induction on n, the case n D 1 being Cauchy’s theorem. We assume inductively that G has a subgroup H of order pn 1. Then ŒG W H is divisible by p.

Let H act on G=H by left multiplication. We know that ŒG W H

is equal to the number of fixed points plus the sum of the cardinalities of nonsingleton orbits. The size of every nonsingleton orbit divides the cardinality of H , so is a power of p. Since p divides ŒG W H , and p divides the size of each nonsingleton orbit, it follows that p also divides the number of fixed points. The number of fixed points is nonzero, since H itself is fixed.

Let’s look at the condition for a coset xH to be fixed under left mul

tiplication by H . This is so if, and only if, for each h 2 H , hxH D xH . That is, for each h 2 H , x 1hx 2 H . Thus x is in the normalizer of H in G (i.e., the set of g 2 G such that gHg 1 D H ).

We conclude that the normalizer NG.H / H . More precisely, the

¤

number of fixed points for the action of H on G=H is the index ŒNG.H / W H , which is thus divisible by p. Of course, H is normal in NG.H /, so we can consider NG.H /=H , which has size divisible by p. By Cauchy’s theorem, NG.H /=H has a subgroup of order p. The inverse image of this subgroup in NG.H / is a subgroup H1 of cardinality pn.

n

Definition 5.4.8. If pn is the largest power of the prime p dividing the order of G, a subgroup of order pn is called a p-Sylow subgroup.

The first Sylow theorem asserts, in particular, the existence of a p–

Sylow subgroup for each prime p.

Theorem 5.4.9. Let G be a finite group, p a prime number, H a subgroup of G of order ps, and P a p–Sylow subgroup of G. Then there is a a 2 G such that aHa 1 P .

Proof. Let X be the family of conjugates of P in G. According to Exercise 5.4.2, the cardinality of X is not divisible by p. Now let H act on X by conjugation. Any nonsingleton orbit must have cardinality a power of p. Since jXj is not divisible by p, it follows that X has a fixed point under the action of H . That is, for some g 2 G, conjugation by elements of H fixes gP g 1. Equivalently, H NG.gP g 1/ D gNG.P /g 1, or g 1Hg NG.P /. Since jg 1Hgj D jH j D ps, it follows from Exercise 5.4.3 that g 1Hg P .

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5. ACTIONS OF GROUPS

Corollary 5.4.10. (Second Sylow theorem). Let P and Q be two p–Sylow subgroups of a finite group G. Then P and Q are conjugate subgroups.

Proof. According to the theorem, there is an a 2 G such that aQa 1 P . Since the two groups have the same size, it follows that aQa 1 D P

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Theorem 5.4.11. (Third Sylow theorem). Let G be a finite group and let p be a prime number. The number of p–Sylow subgroups of G divides jGj and is congruent to 1 .mod p/. In other words, the number of p–Sylow subgroups can be written in the form mp C 1.

Proof. Let P be a p–Sylow subgroup. The family X of p–Sylow subgroups is the set of conjugates of P , according to the second Sylow theorem. Let P act on X by conjugation. If Q is a p–Sylow subgroup distinct from P , then Q is not fixed under the action of P ; for if Q were fixed, then P NG.Q/, and by Exercise 5.4.3, P Q. Therefore, there is exactly one fixed point for the action of P on X , namely, P . All the nonsingleton orbits for the action of P on X have size a power of p, so jXj D mp C 1.

On the other hand, G acts transitively on X by conjugation, so

jGj

jXj D jNG.P/j

divides the order of G.

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We can summarize the three theorems of Sylow as follows: If pn is

the largest power of a prime p dividing the order of a finite group G, then G has a subgroup of order pn. Any two such subgroups are conjugate in G and the number of such subgroups divides jGj and is conjugate to 1 .mod p/.

Example 5.4.12. Let p and q be primes with p > q. If q does not divide p

1, then any group of order pq is cyclic. If q divides p

1, then any

group of order pq is either cyclic or a semidirect product Zp Ì Zq. In this

˛

case, up to isomorphism, there is exactly one nonabelian group of order pq.

Proof. Let G be a group of order pq. Then G has a p–Sylow subgroup P of order p and a q–Sylow subgroup of order q; P and Q are cyclic, and since the orders of P and Q are relatively prime, P \ Q D feg. It follows from the third Sylow theorem that P is normal in G, since 1 is the only

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5.4. GROUP ACTIONS AND GROUP STRUCTURE

255

natural number that divides pq and is congruent to 1 .mod p/. Therefore, PQ D QP is a subgroup of G of order pq, so PQ D G.

According to Corollary 3.2.5, there is a homomorphism ˛ W Zq !

Aut.Zp/ such that G Š Zp Ì Zq. Since Zq is simple, ˛ is either trivial

˛

or injective; in the latter case, ˛.Zq/ is a cyclic subgroup of Aut.Zp/ of order q. But, by Corollary 5.3.4, Aut.Zp/ Š Zp 1. Therefore, if q does not divide p

1, then ˛ must be trivial, so G Š Zp Zq Š Zpq.

On the other hand, if q divides p

1, then Aut.Zp/ Š Zp 1 has a

unique subgroup of order q, and there exists an injective homomorphism ˛ of Zq into Aut.Zp/. Thus there exists a nonabelian semidirect product Zp Ì Zq.

˛

It remains to show that if ˛ and ˇ are non-trivial homomorphisms of

Zq into Aut.Zp/, then Zp Ì Zq Š Zp Ì Zq. Since ˛ and ˇ are injective,

˛

ˇ

˛.Zq/ and ˇ.Zq/ are both equal to the unique cyclic subgroup of order q in Aut.Zp/ Š Zp 1. Write ˛1 D ˛.Œ1q/ and ˇ1 D ˇ.Œ1q/. Then ˛1 and ˇ1 are two generators of the same cyclic group of order q, so there exist integers r and s such that ˛1 D ˇr1, ˇ1 D ˛s1, and rs 1 .mod q/. Then for all t , ˛.Œt q/ D ˛t1 D ˇrt

1

D ˇ.Œrtq/. Likewise, ˇ.Œtq/ D ˛.Œstq/.

Now we can define an isomorphism from Zp Ì Zq to Zp Ì Zq, by

˛

ˇ

.Œap; Œtq/˛ 7! .Œap; Œrtq/ˇ :

Here we have decorated a pair .Œap; Œt q/ with an ˛ if it represents an element of Zp Ì Zq, and similarly for ˇ.

˛

I leave as an exercise for the reader to check that this formula does

give an isomorphism from Zp Ì Zq to Zp Ì Zq; see Exercise 5.4.4.

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˛

ˇ

Example 5.4.13. Since 3 does not divide .5

1/, the only group of order

15 is cyclic. Since 3 divides .7

1/, there is a unique nonabelian group of

order 21, as well as a unique (cyclic) abelian group of order 21.

Example 5.4.14. We know several groups of order 30, namely, Z30, D15, Z3 D5, and Z5 D3. We can show that these groups are mutually nonisomorphic; see Exercise 5.4.5.

Are these the only possible groups of order 30? Let G be a group of

order 30. Then G has (cyclic) Sylow subgroups P , Q, and R of orders 2, 3, and 5.

By the third Sylow theorem, the number n5 of conjugates of R is con

gruent to 1 (mod 5), and divides 30. Hence n5 2 f1; 6g. Likewise the number n3 of conjugates of Q is congruent to 1 (mod 3) and divides 30. Hence nr 2 f1; 10g. I claim that at least one of Q and R must be normal.

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5. ACTIONS OF GROUPS

If R is not normal, then R has 6 conjugates. The intersection of any two distinct conjugates is trivial (as the size must be a divisor of the prime 5). Therefore, the union of conjugates of R contains 6 4 D 24 elements of order 5. Likewise, if Q is not normal, then the union of its 10 conjugates contains 20 elements of order 3. Since G has only 30 elements, it is not possible for both R and Q to be non-normal.

Since at least one of R and Q is normal, N D RQ is a subgroup of G

of order 15. Now N is normal in G, since it has index 2, and cyclic, since any group of order 15 is cyclic.

We have G D NP and N \ P D feg, so according to Corollary 3.2.5,

there is a homomorphism ˛ W Z2 ! Aut.Z15/, such that G Š Z15 Ì Z2.

˛

To complete the classification of groups of order 30, we have to classify such homomorphisms; the nontrivial homomorphisms are determined by order 2 elements of Aut.Z15/.

We have Aut.Z15/ Š Aut.Z5/Aut.Z3/ Š ˚.5/˚.3/ Š Z4 Z2.

In particular, if is an automorphism of Z5 Z3, then there exist unique automorphisms 0 of Z5 and 00 of Z3 such that for all .Œa; Œb/ 2 Z5Z3, ..Œa; Œb// D .0.Œa/; 00.Œb//. The reader is asked to check the details of these assertions in Exercise 5.4.7.

It is easy to locate one order 2 automorphism of Zn for any n, namely,

the automorphism given by Œk 7! Œ k. Since Aut.Z5/ and Aut.Z3/ are cyclic, each of these groups has exactly one element of order 2. Then Aut.Z5/ Aut.Z3/ has exactly three elements of order 2, namely,

'1..Œa; Œb// D .Œ a; Œb/; '2..Œa; Œb// D .Œa; Œ b/; and '3..Œa; Œb// D .Œ a; Œb/:

We will also write 'i for the homomorophism from Z2 to Aut.Z5/

Aut.Z3/ whose value at Œ12 is 'i . It is straightforward to check that

(a)

.Z5 Z3/ Ì Z2 Š D5 Z3,

'1

(b)

.Z5 Z3/ Ì Z2 Š Z5 D3, and

'2

(c)

.Z5 Z3/ Ì Z2 Š D15;

see Exercise 5.4.6.

Thus these three groups are the only nonabelian groups of order 30

(and Z30 is the only abelian group of order 30).

Example 5.4.15. Let us determine all groups of order 28, up to isomorphism. Let G be such a group. The number n7 of 7–Sylow subgroups of G is congruent to 1 (mod 7) and divides 28, so n7 D 1. Therefore G has a unique 7–Sylow subgroup N , which is cyclic of order 7 and normal in G. Let A denote a 2–Sylow subgroup, of order 4. Then N \ A D feg

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jN jjAj

and NA D G, because jNAj D

D 28. Thus G is the semidirect

jN \ Aj

product of N and A.

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

classify the non-abelian groups of order 28, we have to classify the nontrivial homomorphisms from groups of order 4 into Aut.Z7/ Š Z6.

Aut.Z7/ has a unique subgroup of order 2, generated by the automor

phism j W Œx7 7! Œ x7. Any non-trivial homomorphism from a group of order 4 into Aut.Z7/ must have image hj i, since the size of the image is a common divisor of 4 and 6. So we are looking for homomorphisms from a group of order 4 onto hj i Š Z2.

Z4 has a unique homomorphism ˛ onto hj i determined by ˛ W Œ14 7!

j . Therefore, up to isomorphism, Z7 Ì Z4 is the unique non-abelian group

˛

of order 28 with 2–Sylow subgroup isomorphic to Z4. This group is generated by elements a and b satisfying a7 D b4 D 1 and bab 1 D a 1. See Exercise 5.4.10.

I claim that there is also, up to isomorphism, a unique non-abelian

group of order 28 with 2-Sylow subgroup isomorphic to Z2 Z2. Equivalently, there exist homomorphisms from Z2 Z2 onto hj i Š Z2, and if ˇ and are two such homomorphisms, then Z7 Ì .Z2 Z2/ Š

ˇ

Z7 Ì .Z2 Z2/: One homomorphism ˇ from Z2 Z2 onto hj i is de

termined by ˇ..Œx; Œy// D j x. One can show that if is another such homomorphism, then there is an automorphism ' of Z2 Z2 such that D ˇ ı '. It follows from this that Z7 Ì .Z2 Z2/ Š Z7 Ì .Z2 Z2/:

ˇ

See Exercises 5.4.13 and 5.4.14.

Note that D14 and D7 Z2 are models for the non-abelian group of

order 28 with 2–Sylow subgroup isomorphic to Z2 Z2. In particular, these two groups are isomorphic. See Exercise 5.4.11.

Exercises 5.4

5.4.1. Suppose jGj D p3, where p is a prime. Show that either jZ.G/j D p or G is abelian.

5.4.2. Let P be a p–Sylow subgroup of a finite group G. Consider the set of conjugate subgroups gP g 1 with g 2 G. According to Corollary 5.1.14, the number of such conjugates is the index of the normalizer of P in G, ŒG W NG.P /. Show that the number of conjugates is not divisible by p.

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5. ACTIONS OF GROUPS

5.4.3. Let P be a p–Sylow subgroup of a finite group G. Let H be a subgroup of NG.P / such that jH j D ps. Show that H P . Hint: Refer to Exercise 5.1.10, where it is shown that HP is a subgroup of NG.P / with

jP j jH j

jHP j D

:

jH \ P j

5.4.4. Let p > q be prime numbers such that q divides p

1. Let ˛ and ˇ

be two injective homomorphisms of Zq into Aut.Zp/ Š Zp 1. Complete the proof in Example 5.4.12 that Zp Ì Zq Š Zp Ì Zq.

˛

ˇ

5.4.5. Show that the groups Z30, D15, Z3 D5, and Z5 D3 are mutually nonisomorphic.

5.4.6. Verify the following isomorphisms:

(a)

.Z5 Z3/ Ì Z2 Š D5 Z3

'1

(b)

.Z5 Z3/ Ì Z2 Š Z5 D3

'2

(c)

.Z5 Z3/ Ì Z2 Š D15

5.4.7. Verify the assertion made about Aut.Z15/ in Example 5.4.14.

5.4.8. Show that an abelian group is the direct product of its p–Sylow subgroups for primes p dividing jGj.

5.4.9. We have classified all groups of orders p, p2, and pq completely (p and q primes). Which numbers less than 30 have prime decompositions of the form p, p2, or pq? For which n of the form pq does there exist a non-abelian group of order n?

5.4.10. Let ˛ be the unique non-trivial homomorphism from Z4 onto hj i Aut.Z7/. Show that Z7 Ì Z4 is generated by elements a and b satisfying

˛

a7 D b4 D 1 and bab 1 D a 1, and conversely, a group generated by elements a and b satisfying these relations is isomorphic to Z7 Ì Z4.

˛

5.4.11. Show that D14 and D7 Z2 are both groups of order 28 with 2– Sylow subgroups isomorphic to Z2 Z2. Give an explicit isomorphism D14 Š D7 Z2.

5.4.12. Is D2n isomorphic to Dn Z2 for all n? For all odd n?

5.4.13. Let N and A be groups, ˇ W A ! Aut.N / a homomorphism and ' 2 Aut.A/. Show that N Ì A Š N Ì A.

ˇ

ˇ ı'

5.4.14. Let W Z2 Z2 ! Z2 be a surjective group homomorphism.

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5. ACTIONS OF GROUPS

(b)

5.3.8. Describe the automorphism group of Zn Zn. (The description need not be quite as explicit as that of Aut.Z2 Z2/.) Can you describe the automorphism group of .Zn/k ?

5.4. Group Actions and Group Structure

In this section, we consider some applications of the idea of group actions to the study of the structure of groups.

Consider the action of a group G on itself by conjugation. Recall

X

jGj

jGj D jZ.G/j C

;

jCent.g/j

g

where Z.G/ denotes the center of G, Cent.g/ the centralizer of g, and the sum is over representatives of distinct conjugacy classes in G nZ.G/. This is called the class equation.

Consider a group of order pn, where p is a prime number and n a

positive integer. Every subgroup has order a power of p by Lagrange’s theorem, so for g 2 G nZ.G/, the size of the conjugacy class of g, namely,

jGj

;

jCent.g/j

is a positive power of p. Since p divides jGj and jZ.G/j 1, it follows that p divides jZ.G/j. We have proved the following:

Proposition 5.4.2. If jGj is a power of a prime number, then the center of G contains nonidentity elements.

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We discovered quite early that any group of order 4 is either cyclic or

isomorphic to Z2 Z2. We can now generalize this result to groups of order p2 for any prime p.

Corollary 5.4.3. Any group of order p2, where p is a prime, is either cyclic or isomorphic to Zp Zp.

I claim that hgi \ hhi D feg. In fact, hgi \ hhi is a subgroup of hgi,

so if it is not equal to feg, then it has cardinality p; but then it is equal to hgi and to hhi. In particular, h 2 hgi, a contradiction.

It follows from this that hgihhi contains p2 distinct elements of G,

hence G D hgihhi. Therefore, G is abelian.

Now hgi and hhi are two normal subgroups with hgi \ hhi D feg and

hgihhi D G. Hence G Š hgi hhi Š Zp Zp.

n

Look now at Exercise 5.4.1, in which you are asked to show that a

group of order p3 (p a prime) is either abelian or has center of size p.

Corollary 5.4.4. Let G be a group of order pn, n > 1. Then G has a normal subgroup feg N G. Furthermore, N can be chosen so that

¤

every subgroup of N is normal in G.

n

Corollary 5.4.5. Suppose jGj D pn is a power of a prime number. Then G has a sequence of subgroups

feg D G0 G1 G2 Gn D G

such that the order of Gk is pk, and Gk is normal in G for all k.

Proof. We prove this by induction on n. If n D 1, there is nothing to do. So suppose the result holds for all groups of order pn0 , where n0 < n.

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5. ACTIONS OF GROUPS

feg D G0 G1 G2 Gs D N

with jGkj D pk. Apply the induction hypothesis again to N

G D G=N to

obtain a sequence of subgroups

feg D N

G0 N

G1 N

G2 N

Gn s D N

with j N

Gkj D pk and N

Gk normal in N

G. Then put GsCk D 1 N

Gk, for

1 k n

s, where W G ! G=N is the quotient map. Then the

sequence .Gk/0kn has the desired properties.

n

We now use similar techniques to investigate the existence of sub

groups of order a power of a prime. The first result in this direction is Cauchy’s theorem:

Theorem 5.4.6. (Cauchy’s theorem). Suppose the prime p divides the order of a group G. Then G has an element of order p.

The proof given here, due to McKay,1 is simpler and shorter than other

known proofs.

Each element of X is either fixed under the action of Zp, or it belongs

kp D n, so X has

a fixed point .a; a; : : : ; a/ with a ¤ e. But then a has order p.

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Theorem 5.4.7. (First Sylow theorem). Suppose p is a prime, and pn divides the order of a group G. Then G has a subgroup of order pn.

1J. H. McKay, Amer. Math. Monthly 66 (1959), p. 119.

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Proof. We prove this statement by induction on n, the case n D 1 being Cauchy’s theorem. We assume inductively that G has a subgroup H of order pn 1. Then ŒG W H is divisible by p.

Let H act on G=H by left multiplication. We know that ŒG W H

Let’s look at the condition for a coset xH to be fixed under left mul

We conclude that the normalizer NG.H / H . More precisely, the

¤

n

Definition 5.4.8. If pn is the largest power of the prime p dividing the order of G, a subgroup of order pn is called a p-Sylow subgroup.

The first Sylow theorem asserts, in particular, the existence of a p–

Sylow subgroup for each prime p.

Theorem 5.4.9. Let G be a finite group, p a prime number, H a subgroup of G of order ps, and P a p–Sylow subgroup of G. Then there is a a 2 G such that aHa 1 P .

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Corollary 5.4.10. (Second Sylow theorem). Let P and Q be two p–Sylow subgroups of a finite group G. Then P and Q are conjugate subgroups.

Proof. According to the theorem, there is an a 2 G such that aQa 1 P . Since the two groups have the same size, it follows that aQa 1 D P

n

On the other hand, G acts transitively on X by conjugation, so

jGj

jXj D jNG.P/j

divides the order of G.

n

We can summarize the three theorems of Sylow as follows: If pn is

Example 5.4.12. Let p and q be primes with p > q. If q does not divide p

1, then any group of order pq is cyclic. If q divides p

1, then any

group of order pq is either cyclic or a semidirect product Zp Ì Zq. In this

˛

case, up to isomorphism, there is exactly one nonabelian group of order pq.

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natural number that divides pq and is congruent to 1 .mod p/. Therefore, PQ D QP is a subgroup of G of order pq, so PQ D G.

According to Corollary 3.2.5, there is a homomorphism ˛ W Zq !

Aut.Zp/ such that G Š Zp Ì Zq. Since Zq is simple, ˛ is either trivial

˛

or injective; in the latter case, ˛.Zq/ is a cyclic subgroup of Aut.Zp/ of order q. But, by Corollary 5.3.4, Aut.Zp/ Š Zp 1. Therefore, if q does not divide p

1, then ˛ must be trivial, so G Š Zp Zq Š Zpq.

On the other hand, if q divides p

1, then Aut.Zp/ Š Zp 1 has a

unique subgroup of order q, and there exists an injective homomorphism ˛ of Zq into Aut.Zp/. Thus there exists a nonabelian semidirect product Zp Ì Zq.

˛

It remains to show that if ˛ and ˇ are non-trivial homomorphisms of

Zq into Aut.Zp/, then Zp Ì Zq Š Zp Ì Zq. Since ˛ and ˇ are injective,

˛

ˇ

1

D ˇ.Œrtq/. Likewise, ˇ.Œtq/ D ˛.Œstq/.

Now we can define an isomorphism from Zp Ì Zq to Zp Ì Zq, by

˛

ˇ

.Œap; Œtq/˛ 7! .Œap; Œrtq/ˇ :

Here we have decorated a pair .Œap; Œt q/ with an ˛ if it represents an element of Zp Ì Zq, and similarly for ˇ.

˛

I leave as an exercise for the reader to check that this formula does

give an isomorphism from Zp Ì Zq to Zp Ì Zq; see Exercise 5.4.4.

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˛

ˇ

Example 5.4.13. Since 3 does not divide .5

1/, the only group of order

15 is cyclic. Since 3 divides .7

1/, there is a unique nonabelian group of

order 21, as well as a unique (cyclic) abelian group of order 21.

Example 5.4.14. We know several groups of order 30, namely, Z30, D15, Z3 D5, and Z5 D3. We can show that these groups are mutually nonisomorphic; see Exercise 5.4.5.

Are these the only possible groups of order 30? Let G be a group of

order 30. Then G has (cyclic) Sylow subgroups P , Q, and R of orders 2, 3, and 5.

By the third Sylow theorem, the number n5 of conjugates of R is con

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Since at least one of R and Q is normal, N D RQ is a subgroup of G

of order 15. Now N is normal in G, since it has index 2, and cyclic, since any group of order 15 is cyclic.

We have G D NP and N \ P D feg, so according to Corollary 3.2.5,

there is a homomorphism ˛ W Z2 ! Aut.Z15/, such that G Š Z15 Ì Z2.

˛

To complete the classification of groups of order 30, we have to classify such homomorphisms; the nontrivial homomorphisms are determined by order 2 elements of Aut.Z15/.

We have Aut.Z15/ Š Aut.Z5/Aut.Z3/ Š ˚.5/˚.3/ Š Z4 Z2.

It is easy to locate one order 2 automorphism of Zn for any n, namely,

'1..Œa; Œb// D .Œ a; Œb/; '2..Œa; Œb// D .Œa; Œ b/; and '3..Œa; Œb// D .Œ a; Œb/:

We will also write 'i for the homomorophism from Z2 to Aut.Z5/

Aut.Z3/ whose value at Œ12 is 'i . It is straightforward to check that

(a)

.Z5 Z3/ Ì Z2 Š D5 Z3,

'1

(b)

.Z5 Z3/ Ì Z2 Š Z5 D3, and

'2

(c)

.Z5 Z3/ Ì Z2 Š D15;

see Exercise 5.4.6.

Thus these three groups are the only nonabelian groups of order 30

(and Z30 is the only abelian group of order 30).

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jN jjAj

and NA D G, because jNAj D

D 28. Thus G is the semidirect

jN \ Aj

product of N and A.

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

classify the non-abelian groups of order 28, we have to classify the nontrivial homomorphisms from groups of order 4 into Aut.Z7/ Š Z6.

Aut.Z7/ has a unique subgroup of order 2, generated by the automor

Z4 has a unique homomorphism ˛ onto hj i determined by ˛ W Œ14 7!

j . Therefore, up to isomorphism, Z7 Ì Z4 is the unique non-abelian group

˛

of order 28 with 2–Sylow subgroup isomorphic to Z4. This group is generated by elements a and b satisfying a7 D b4 D 1 and bab 1 D a 1. See Exercise 5.4.10.

I claim that there is also, up to isomorphism, a unique non-abelian

ˇ

Z7 Ì .Z2 Z2/: One homomorphism ˇ from Z2 Z2 onto hj i is de

ˇ

See Exercises 5.4.13 and 5.4.14.

Note that D14 and D7 Z2 are models for the non-abelian group of

order 28 with 2–Sylow subgroup isomorphic to Z2 Z2. In particular, these two groups are isomorphic. See Exercise 5.4.11.

Exercises 5.4

5.4.1. Suppose jGj D p3, where p is a prime. Show that either jZ.G/j D p or G is abelian.

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jP j jH j

jHP j D

:

jH \ P j

5.4.4. Let p > q be prime numbers such that q divides p

1. Let ˛ and ˇ

be two injective homomorphisms of Zq into Aut.Zp/ Š Zp 1. Complete the proof in Example 5.4.12 that Zp Ì Zq Š Zp Ì Zq.

˛

ˇ

5.4.5. Show that the groups Z30, D15, Z3 D5, and Z5 D3 are mutually nonisomorphic.

5.4.6. Verify the following isomorphisms:

(a)

.Z5 Z3/ Ì Z2 Š D5 Z3

'1

(b)

.Z5 Z3/ Ì Z2 Š Z5 D3

'2

(c)

.Z5 Z3/ Ì Z2 Š D15

5.4.7. Verify the assertion made about Aut.Z15/ in Example 5.4.14.

5.4.8. Show that an abelian group is the direct product of its p–Sylow subgroups for primes p dividing jGj.

5.4.10. Let ˛ be the unique non-trivial homomorphism from Z4 onto hj i Aut.Z7/. Show that Z7 Ì Z4 is generated by elements a and b satisfying

˛

a7 D b4 D 1 and bab 1 D a 1, and conversely, a group generated by elements a and b satisfying these relations is isomorphic to Z7 Ì Z4.

˛

5.4.11. Show that D14 and D7 Z2 are both groups of order 28 with 2– Sylow subgroups isomorphic to Z2 Z2. Give an explicit isomorphism D14 Š D7 Z2.

5.4.12. Is D2n isomorphic to Dn Z2 for all n? For all odd n?

5.4.13. Let N and A be groups, ˇ W A ! Aut.N / a homomorphism and ' 2 Aut.A/. Show that N Ì A Š N Ì A.

ˇ

ˇ ı'

5.4.14. Let W Z2 Z2 ! Z2 be a surjective group homomorphism.

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