1.5 Automorphism Groups and the Semi-Direct Product

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Let G be a group, and define Aut(G) to be the group of automorphisms of G,

with function composition as the operation. Knowledge of the structure of

Aut(G) is frequently helpful, especially in the following situation. Suppose

that G is a group, and H /G. Then G acts on H by conjugation as a group

of automorphisms; thus there is a homomorphism G ! Aut(G). Note that

the kernel of this automorphism consists of all elements of G that centralize

every element of H. In particular, the homomorphism is trivial, i.e. G is

the kernel, precisely when G centralizes H.

In certain situations, it is useful to know the automorphism group of

a cyclic group Z = hxi, of order n. Clearly, any such automorphism is of

the form x 7! xa, where o(xa) = n. In turn, by Exercise 1 of Section 1.1,

o(xa) = n precisely when gcd(a, n) = 1. This implies the following.

Proposition 1.5.1 Let Zn = hxi be a cyclic group of order n. Then

Aut(Zn) _= U(Z/(n)), where U(Z/(n)) is the multiplicative group of residue

classes mod(n), relatively prime to n. The isomorphism is given by [a] 7!

(x 7! xa).

It is clear that if n = pe1

1 pe2

2 · · · per

r is the prime factorization of n, then

Aut(Zn) _= Aut(Zp1) × Aut(Zp2) × · · · × Aut(Zpr );

therefore to compute the structure of Aut(Zn), it suffices to determine the

automorphism groups of cyclic p-groups. For the answer, see Exercises 1

and 2, below.

1.5. AUTOMORPHISM GROUPS 17

Here’s a typical sort of example. Let G be a group of order 45 = 32 · 5.

Let P 2 Syl3(G), Q 2 Syl5(G); by Sylow’s theorem Q / G and so P acts

on Q, forcing P ! Aut(Q). Since |Aut(Q)| = 4 = _(5), it follows that

the kernel of the action is all of P. Thus P centralizes Q; consequently

G _= P ×Q. (See Exercise 6, below.) The reader is now encouraged to make

up further examples; see Exercises 13, 15, and 16.

Here’s another simple example. Let G be a group of order 15, and let

P,Q be 3 and 5-Sylow subgroups, respectively. It’s trivial to see that Q/G,

and so P acts on Q by conjugation. By Proposition 1.5.1, it follows that the

action is trivial so P,Q centralize each other. Therefore G _= P × Q; since

P,Q are both cyclic of relatively prime orders, it follows that P ×Q is itself

cyclic, i.e., G _= Z15. An obvious generalization is Exercise 13, below.

As another application of automorphism groups, we consider the semidirect

product construction as follows. First of all, assume that G is a group

and H,K are subgroups of G with H _ NG(K). Then an easy calculation

reveals that in fact, KH _ G (see Exercise 3 of Section 1.1. ). Now suppose

that in addition,

(i) G = KH, and

(ii) K \ H = {e}.

Then we call G the internal semi-direct product of K by H. Note that if G

is the internal semi-direct product of K by H, and if H _ CG(K), then G

is the (internal) direct product of K and H.

The above can be “externalized” as follows. Let H, K be groups and let

_ : H ! Aut(K) be a homomorphism. Construct the group K ×_ H, where

(i) K ×_ H = K × H (as a set).

(ii) (k1, h1) · (k2, h2) = (k1_(h1)(k2), h1h2).

It is routine to show that K ×_ H is a group, relative to the above binary

operation; we call K ×_ H the external semi-direct product of K by H.

Finally, we can see that G = K ×_ H is actually an internal semidirect

product. To this end, set K0 = {(k, e)| k 2 K}, H0 = {(e, h)| h 2 H}, and

observe that H0 and K0 are both subgroups of G. Furthermore,

(i) K0 _= K, H0 _= H,

(ii) K0 / G,

18 CHAPTER 1. GROUP THEORY

(iii) K0 \ H0 = {e},

(iv) G = K0H0 (so G is the internal semidirect product of K0 by H0),

(v) If k0 = (k, e) 2 K0, h0 = (e, h) 2 H0, then h0k0h0−1 = (_(h)(k), e) 2 K0.

(Therefore _ determines the conjugation action of H0 on K0.)

(vi) G = K ×_ H _= K × H if and only if H = ker _.

As an application, consider the following:

(1) Construct a group of order 56 with a non-normal 2-Sylow subgroup (so

the 7-Sylow subgroup is normal).

(2) Construct a group of order 56 with a non-normal 7-Sylow subgroup (so

the 2-Sylow subgroup is normal).

The constructions are straight-forward, but interesting. Watch this:

(1) Let P = hxi, a cyclic group of order 7. By Proposition 1.5.1 above,

Aut(P) _= Z6, a cyclic group of order 6. Let H 2 Syl2(Aut(P)), so H

is cyclic of order 2. Let Q = hyi be a cyclic group of order 8, and let

_ : Q ! H be the unique nontrivial homorphism. Form P ×_ Q.

(2) Let P = Z2 × Z2 × Z2; by Exercise 17, below, Aut(P) _= GL3(2). That

GL3(2) is a group of order 168 is a fairly routine exercise. Thus, let

Q 2 Syl7(Aut(P)), and let _ : Q ! Aut(P) be the inclusion map.

Construct P ×_ Q.

Let G be a group, and let g 2 G. Then the automorphism _g : G ! G

induced by conjugation by g (x 7! gxg−1) is called an inner automorphism

of G. We set Inn(G) = {_g| g 2 G} _ Aut(G). Clearly one has Inn(G) _=

G/Z(G). Next if _ 2 Aut(G), _g 2 Inn(G), then __g_−1 = __g. This

implies that Inn(G) / Aut(G); we set Out(G) = Aut(G)/Inn(G), the group

of outer automorphisms of G. (See Exercise 26, below.)

Exercises 1.5

1. Let p be an odd prime; show that Aut(Zpr ) _= Zpr−1(p−1), as follows.

First of all, the natural surjection Z/(pr) ! Z/(p) induces a surjection

U(Z/(pr)) ! U(Z/(p)). Since the latter is isomorphic with Zp−1,

1.5. AUTOMORPHISM GROUPS 19

conclude that Aut(Zpr ) contains an element of order p − 1. Next,

use the Binomial Theorem to prove that (1+p)pr−1

_ 1( mod pr) but

(1+p)pr−2

6_ 1( mod pr). Thus the residue class of 1+p has order pr−1

in U(Z/(pr)). Thus, U(Z/(pr)) has an element of order pr−1(p − 1) so

is cyclic.

2. Show that if r _ 3, then (1 + 22)2r−2

_ 1( mod 2r) but (1 + 22)2r−3

6_

1( mod 2r). Deduce from this that the class of 5 in U(Z/(2r)) has order

2r−2. Now set C = h[5]i and note that if [a] 2 C, then a _ 1( mod 4).

Therefore, [−1] 62 C, and so U(Z/(2r)) _= h[−1]i × C.

3. Compute Aut (Z), where Z is infinite cyclic.

4. If Z is infinite cyclic, compute the automorphism group of Z × Z.

5. Let G = KH be a semidirect product where K/G. If also H /G show

that G is the direct product of K and H.

6. Let G be a finite group of order paqb, where p, q are distinct primes.

Let P 2 Sylp(G), Q 2 Sylq(G), and assume that P, Q/G. Prove that

P and Q centralize each other. Conclude that G _= P × Q.

7. Let G be a finite group of order 2k, where k is odd. If G has more

than one involution, prove that Aut(G) is non-abelian.

8. Prove that the following are equivalent for the group G:

(a) G is dihedral;

(b) G factors as a semidirect product G = NH, where N / G, N is

cyclic and H is a cyclic subgroup of order 2 of G which acts on

N by inver ting the elements of N.

9. Let G be a finite dihedral group of order 2k. Prove that G is generated

by elements n, h 2 G such that nk = h2 = e, hnh = n−1.

10. Let N = hni be a cyclic group of order 2n, and let H = hhi be a

cyclic group of order 2. Define mappings _1, _2 : H ! Aut (N) by

_1(h)(n) = n−1+2n−1

, _2(h)(n) = n1+2n−1

. Define the groups G1 =

N ×_1 H, G2 = N ×_2 H. G1 is called a semidihedral group, and G2

is called a quasi-dihedral group . Thus, if G = G1 or G2, then G is a

2-group of order 2n+1 having a normal cyclic subgroup N of order 2n.

20 CHAPTER 1. GROUP THEORY

(a) What are the possible orders of elements in G1 − N?

(b) What are the possible orders of elements in G2 − N?

11. Let N = hni be a cyclic group of order 2n, and let H = hhi be a cyclic

group of order 4. Let H act on N by inverting the elements of N and

form the semidirect product G = NH (there’s no harm in writing this

as an internal semidirect product). Let Z = hn2n−1

h2i.

(a) Prove that Z is a normal cyclic subgroup of G or order 2;

(b) Prove that the group Q = Q2n+1 = G/Z is generated by elements

x, y 2 Q such that x2n = y4 = e, yxy−1 = x−1, x2n−1 = y2.

The group Q2n+1, constructed above, is called the generalized quaternion

group of order 2n+1. The group Q8 is usually just called the

quaternion group.

12. Let G be an abelian group and let N be a subgoup of G. If G/N is

an infinite cyclic group, prove that G _= N × (G/N).

13. Let G be a group of order pq, where p, q are primes with p < q. If

p/ (q − 1), prove that G is cyclic.

14. Assume that G is a group of order p2q, where p and q are odd primes

and where q > p. Prove that G has a normal q-Sylow subgroup. Give

a counter-example to this assertion if p = 2.

15. Let G be a group of order 231, and prove that the 11-Sylow subgroup

is in the center of G.

16. Let G be a group of order 385. Prove that its 11-Sylow is normal, and

that its 7-Sylow is in the center of G.

17. Let P = Zp ×Zp ×· · ·×Zp, (n factors) where p is a prime and Zp is a

cyclic group of order p. Prove that Aut(P) _= GLn(p), where GLn(p)

is the group of n × n invertible matrices with coefficients in the field

Z/(p).

18. Let H be a finite group, and let G = Aut(H). What can you say

about H if

(a) G acts transitively on H#?

(b) G acts 2-transitively on H#?

1.5. AUTOMORPHISM GROUPS 21

(c) G acts 3-transitively on H#?

19. Assume that G = NK, a semi-direct product with 1 6= N an abelian

minimal normal subgroup of G. Prove that K is a maximal proper

subgroup of G.

20. Assume that G is a group of order 60. Prove that G is either simple

or has a normal 5-Sylow subgroup.

21. Let G be a dihedral group of order 2p, where p is prime, and assume

that G acts faithfully on V = Z2 × Z2 × · · · × Z2 as a group of automorphisms.

If x 2 G has order p, and if CV (x) = {e}, show that for

any element _ 2 G of order 2, |CV (_ )|2 = |V |.

22. Assume that G = K1H1 = K2H2 where K1,K2 / G, K1 \ H1 =

K2 \H2 = {e}, and K1

_= K2. Show by giving a counter-example that

it need not happen that H1

_= H2.

23. Same hypotheses as in Exercise 22 above, except that G is a finite

group and that K1,K2 are p-Sylow subgroups of G for some prime p.

Show in this case that H1

_= H2.

24. Let G be a group. Show that Aut(G) permutes the conjugacy classes

of G.

25. Let G be a group and let H _ G. We say that H is characteristic in

G if for every _ 2 Aut(G), we have _ (H) = H. If this is the case, we

write H char G. Prove the following:

(a) If H char G, then H / G.

(b) If H char G then there is a homomorphism Aut(G) ! Aut(H).

26. Let G = S6 be the symmetric group on the set of letters X = {1, 2, 3, 4, 5, 6},

and let H be the stabilizer of the letter 1. Thus H _= S5. A simple

application of Sylow’s theorem shows that H acts transitively on the

set Y of 5-Sylow subgroups in G, and that there are six 5-Sylow subgroups

in G. If we fix a bijection _ : Y ! X, then _ induces an

automorphism of G via _ 7! _−1__. Show that this automorphism

of G must be outer.5 (Hint: this automorphism must carry H to the

normalizer of a 5-Sylow subgroup.)

5This is the only finite symmetric group for which there are outer automorphisms. See

D.S. Passman, Permutation Groups, W.A. Benjamin, Inc., 1968, pp. 29-35.

22 CHAPTER 1. GROUP THEORY

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