1.7 The Commutator Subgroup and Iterated Constructions

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For any group G there is the so-called commutator subgroup , G0 (or sometimes

denoted [G,G]) which is defined by setting

G0 = hxyx−1y−1| x, y 2 Gi.

Note that G is a normal subgroup of G, since the conjugate of any commutator

[x, y] := xyx−1y−1 is again a commutator. If you think about the

following long enough, it becomes very obvious.

Proposition 1.7.1 Let G be a group, with commutator subgroup G0.

(a) G/G0 is an abelian group.

(b) If _ : G ! A is a homomorphism into the abelian group A, then there

is a unique factorization of _, according to the commutativity of the

diagram below:

G

_ - A

@

@

@

@

@

_

R 􀀀

􀀀

􀀀

􀀀

􀀀

¯_

_

G/G0

The following concept is quite useful, especially in the present context.

Let G be a group, and let H _ G. H is called a characteristic subgroup of

G (and written H char G) if for any automorphism _ : G ! G, _(H) = H.

Note that since conjugation by an element g 2 G is an automorphism of G,

it follows that any characteristic subgroup of G is normal. The following

property is clear, but useful:

H char N char G =) H char G.

In particular, since the commutator subgroup G0 of G is easily seen to be

a characteristic subgroup of G, it follows that the iterated commutators

1.7. THE COMMUTATOR SUBGROUP 29

G(1) = G0,G(2) = (G(1))0, . . . are all characteristic, hence normal, subgroups

of G.

By definition, a group G is solvable if for some k, G(k) = {e}. The historical

importance of solvable groups will be seen later on, in the discussions

of Galois Theory in Chapter 2.

The following is fundamental, and reveals the inductive nature of solvability:

Theorem 1.7.2 If N / G, then G is solvable if and only if both G/N and

N are solvable.

There is an alternative, and frequently more useful way of defining solvability.

First, a normal series in G is a sequence

G = G0 _ G1 _ · · · ,

with each Gi normal in G. Thus, the commutator series

G = G(0) _ G(1) _ G(2) · · ·

is a normal series. Note that G acts on each quotient Gi/Gi+1 in a normal

series by conjugation (how is this?). A subnormal series is just like a normal

series, except that one requires only that each Gi be normal in Gi−1 (and

not necessarily normal in G). The following is often a useful characterization

of solvability.

Theorem 1.7.3 A group G is solvable if and only if it has a subnormal

series of the form

G = G0 _ G1 _ · · · _ Gm = {e}

with each Gi/Gi+1 abelian.

A subnormal series G = G0 _ G1 _ · · · _ Gm = {e} is called a composition

series if each quotient Gi/Gi+1 is a non-trivial simple group. Obviously,

any finite group has a composition series. As a simple example, if n _ 5,

then Sn _ An _ {e} is a composition series. For n = 4 one has a composition

series for S4 of the form S4 _ A4 _ K _ Z _ {e}, where K _= Z2 × Z2

and Z _= Z2.

While it seems possible for a group to be resolvable into a composition

series in many different ways, the situation is not too bad for finite groups.

30 CHAPTER 1. GROUP THEORY

Theorem 1.7.4 (Jordan-H¨older Theorem.) Let G be a finite group,

and let

G = G0 _ G1 _ · · · _ Gh = {e},

G = H0 _ H1 _ · · · _ Hk = {e}

be composition series for G. Then h = k and there is a bijective correspondence

between the sets of composition quotients so that these corresponding

quotients are isomorphic.

Let G be a group, and let H,K _ G with K/G. We set [H,K] =h[h, k]| h 2

H, k 2 Ki, the commutator of H and K. Note that [H,K] _ K. In particular,

set L0(G) = G,L1(G) = [G,L0(G)], . . .Li(G) = [G,Li−1(G)], . . .. Now

consider the series

L0(G) _ L1(G) _ L2(G) · · · .

Note that this series is actually a normal series. This series is called the

lower central series for G. If Li(G) = {e} for some i, call G nilpotent .

Note that G acts trivially by conjugation on each factor in the lower central

series. In fact,

Theorem 1.7.5 The group G is nilpotent if and only if it has a finite normal

series, with each quotient acted on trivially by G.

The descending central series is computed from “top to bottom” in a

group G. There is an analogous series, constructed from “bottom to top:”

Z0(G) = {e} _ Z1(G) = Z(G) _ Z2(G) _ Z3(G) _ · · · ,

where Zi+1 = Z(G/Zi(G)) for each i. Again this is a normal series, and it

is clear that G acts trivially on each Zi(G)/Zi+1(G). Thus, the following is

immediate:

Theorem 1.7.6 G is nilpotent if and only if Zm(G) = G for some m _ 0.

The following should be absolutely clear.

Theorem 1.7.7 Abelian =) Nilpotent =) Solvable

The reader should be quickly convinced that the above implications cannot

be reversed.

We conclude this section with a characterization of finite nilpotent groups;

see Theorem 1.7.10, below.

1.7. THE COMMUTATOR SUBGROUP 31

Proposition 1.7.8 If P is a finite p-group, then P is nilpotent.

Lemma 1.7.9 If G is nilpotent, and if H is a proper subgroup of G, then

H 6= NG(H). Thus, normalizers “grow” in nilpotent groups.

The above ahows that the Sylow subgroups in a nilpotent group are all

normal, In fact,

Theorem 1.7.10 Let G be a finite group. Then G is nilpotent if and only

if G is the direct product of its Sylow subgroups.

Exercises 1.7

1. Show that H char N / G ) H / G.

2. Let H be a subgroup of the group G with G0 _ H. Prove that H /G.

3. Let G be a finite group and let P be a 2-Sylow subgroup of G. If

M _ P is a subgroup of index 2 in P and if _ 2 G is an involution not

conjugate to any element of M, conclude that _ 62 G0 (commutator

subgroup). [Hint: Look at the action of _ on the set of left cosets of

M in G. If _ an even permutation or an odd permutation?]

4. Show that any subgroup of a cyclic group is characteristic.

5. Give an example of a group G and a normal subgroup K such that K

isn’t characteristic in G.

6. Let G be a group. Prove that G0 is the intersection of all normal

subgroups N / G, such that G/N is abelian.

7. Give an example of a subnormal series in A4 that isn’t a normal series.

8. Let G be a group and let x, y 2 G. If [x, y] commutes with x and y,

prove that for all positive integers k, (xy)k = xkyk[y, x](k

2).

9. A sequence of homomorphisms K _! G

_!

H is called exact (at G)

if im _ = ker _ . Prove the following mild generalization of Theorem

1.7.2. If K _! G

_!

H is an exact sequence with K,H both

solvable groups, so is G.

32 CHAPTER 1. GROUP THEORY

10. Let K ! G1 ! G2 ! H be an exact sequence of homomorphisms of

groups (meaning exactness at both G1 and G2.) If K and H are both

solvable, must G1 and/or G2 be solvable? Prove, or give a counterexample.

11. Let F be a field and let

G =

__

_ _

0

_

| _, _,  2 F, _ 6= 0

_

.

Prove that G is a solvable group.

12. Let G be a finite solvable group, and let K / G be a minimal normal

subgroup. Prove that K is an elementary abelian p-group for some

prime p (i.e., K _= Zp × Zp × · · · × Zp).

13. Show that the finite group G is solvable if and only if it has a subnormal

series

G = G0 _ G1 _ · · · _ Gm = {e},

with each Gi/Gi+1 a group of prime order.

14. By now you may have realized that if G is a finite nonabelian simple

group of order less than or equal to 200, then |G| = 60 or 168. Using

this, if G is a nonsolvable group of order less than or equal to 200,

what are the possible group orders?

15. Let q be a prime power; we shall investigate the special linear group

G = SL2(q).

(a) Show that G is generated by its elements of order p, where q = pa.

This is perhaps best done by investigating the equations

_

_ _

 _

_

=

_

1 (_ − 1)−1

0 1

_ _

1 0

 1

_ _

1 (_ − 1)−1

0 1

_

, if  6= 0,

_

_ _

 _

_

=

_

1 0

(_ − 1)_−1 1

_ _

1 _

0 1

_ _

1 0

(_ − 1)_−1 1

_

, if _ 6= 0,

_

_ 0

0 _−1

_

=

_

1 0

_−1 − 1 1

_ _

1 1

0 1

_ _

1 0

_ − 1 1

_ _

1 −_−1

0 1

_

.

1.7. THE COMMUTATOR SUBGROUP 33

(b) Show that if q _ 4, then G0 = G. Indeed, look at

_

1 _

0 1

_ _

μ 0

0 μ−1

_ _

1 −_

0 1

_ _

μ−1 0

0 μ

_

=

_

1 _(1 − μ2)

0 1

_

.

(c) Conclude that if q _ 4, then the groups GL2(q), SL2(q), PSL2(q)

are all nonsolvable groups.

16. If p and q are primes, show that any group of order p2q is solvable.

17. More generally, let G be a group of order pnq, where p and q are

primes. Show that G is solvable. (Hint: Let P1 and P2 be distinct

p-Sylow subgroups such that H := P1 \ P2 is maximal among all

such pairs of intersections. Look at NG(H) and note that if Q is a

q-Sylow subgroup of G, then Q _ NG(H). Now write G = Q · P1 and

conclude that the group H_ = hgHg−1| g 2 Gi is, in fact, a normal

subgroup of G and is contained in P1. Now use induction together

with Theorem 1.7.2.)

18. Let P be a p-group of order pn. Prove that for all k = 1, 2, . . . n, P

has a normal subgroup of order pk.

19. Let G be a finite group and let N1,N2 be normal nilpotent subgroups.

Prove that N1N2 is again a normal nilpotent subgroup of G. (Hint:

use Theorem 1.7.10.)

20. (The Heisenberg Group.) Let V be an m-dimensional vector space over

the field F, and assume that F either has characteristic 0 or has odd

characteristic. Assume that h , i : V × V ! F is a non-degenerate,

alternating bilinear form. This means that

(i) hv,wi = −hw, vi for all v,w 2 V , and

(ii) hv,wi = 0 for all w 2 V implies that v = 0.

Now define a group, H(V ), the Heisenberg group, as follows. We set

H(V ) = V × F, and define multiplication by setting

(v1, _1) · (v2, _2) = (v1 + v2, _1 + _2 +

1

2hv1, v2i),

where v1, v2 2 V , and _1, _2 2 F. Show that

(i) H(V ), with the above operation, is a group.

34 CHAPTER 1. GROUP THEORY

(ii) If H = H(V ), then Z(H) = {(0, _)| _ 2 F}.

(iii) H0 = Z(H), and so H is a nilpotent group.

21. The Frattini subgroup of a finite p-group. Let P be a finite p-group,

and let _(P) be the intersection of all maximal subgroups of P. Prove

that

(i) P/_(P) is elementary abelian.

(ii) _(P) is trivial if and only if P is elementary abelian.

(iii) If P = ha, _(P)i, then P = hai.

(Hint: for (i) prove first that if M _ P is a maximal subgroup of

P, then [M : P] = p and so M / P. This shows that if x 2 P, the

xp 2 M. By the same token, as P/M is abelian, P0 _ M. Since M

was an arbitrary maximal subgroup this gives (i). Part (ii) should be

routine.)

22. Let P be a finite p-group, and assume that |P| = pk. Prove that the

number of maximal subgroups (i.e., subgroups of index p) is less than

or equal to (pk−1)/(p−1) with equality if and only if P is elementary

abelian. (Hint: If P is elementary abelian, then we regard P as a

vector space over the field Z/(p). Thus subgroups of index p become

vector subspaces of dimension k − 1; and easy count shows that there

are (pk − 1)/(p − 1) such. If P is not elementary abelian, then _(P)

is not trivial, and every maximal subgroup of P contains _(P). Thus

the subgroups of P of index P correspond bijectively with maximal

subgroups of P/_(P); apply the above remark.)

23. Let P be a group and let p be a prime. Say that P is a Cp-group if

whenever x, y 2 P satisfy xp = yp, then xy = yx. (Note that dihedral

and generalized quaternion groups of order at least 8 are definitely

not C2-groups.) Prove that if P is a p-group where p is an odd prime,

and if every element of order p is in Z(P), then P is a Cp-group, by

proving the following:

(a) Let P be a minimal counterexample to the assertion, and let

x, y 2 P with xp = yp. Argue that P = hx, yi.

(b) Show that (yxy−1)p = xp.

(c) Show that yxy−1 2 hx, _(P)i; conclude that hx, yxy−1i is a proper

subgroup of P. Thus, by (b), x and yxy−1 commute.

1.7. THE COMMUTATOR SUBGROUP 35

(d) Conclude from (c) that if z = [x, y] = xyx−1y−1, then zp =

[xp, y] = [yp, y] = e. Thus, by hypothesis, z commutes with x

and y.

(e) Show that [y−1, x] = [x, y]. (Conjugate z by y−1.)

(f) Show that (xy−1)p = e. (Use Exercise 8.)

(g) Conclude that x and y commute, a contradition.7

24. Here’s an interesting simplicity criterion. Let G be a group acting

primitively on the set X, and let H be the stabilizer of some element

of X. Assume

(i) G = G0,

(i) H contains a normal solvable subgroup A such that G is generated

by the conjugates of A.

Prove that G is simple.

25. Using the above exercise, prove that the groups PSL2(q) q _ 4 are all

simple groups.

26. Let G be a group and let Z = Z(G). Prove that if G/Z is nilpotent,

so is G.

27. Let G be a finite group such that for any subgroup H of G we have

[G : NG(H)] _ 2. Prove that G is nilpotent.

7The above have been extracted from Bianchi, Gillio Berta Mauri and Verardi,

Groups in which elements with the same p-power permute, LE MATHEMATICHE, Vol.

LI (1996) - Supplemento, pp. 53-61. The authors actually show that a necessary and

sufficient condition for a p-group to be a Cp-group is that all elements of order p are

central. In the general case when G is not a p-group, then the authors show that G is a

p-group if and only if G has a normal p-Sylow sugroup which is also a p-group.

36 CHAPTER 1. GROUP THEORY

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