Question

(*) Let G be a group. Let G, G denote the smallest subgroup of G containing S = {xyr-ly-1: 2, YEG}. (The subgroup (G,G] is called the commutator subgroup of G.) (a) Show that u-zyr-?-?u= (u-cu)(u-yu)(u--xu)-1(u-yu)-1 for all 2, 4, U E G. Deduce that (G,G| 4G. (b) Show that the quotient group G/[G,G] is abelian. (c) If N 4G and G/N is abelian, show that (G,G] C N. (In other words, G/(G,G) is the largest abelian quotient of G.) (d) Show that if K <G and (G,G] CK then K 1G.

Answer 1

(*) Solution:

(a)
Suppose that u is in G and consider an element in [G,G] of the form xyx^-1y^-1 .
Then, u^-1xyx^-1y^-1u = (u^-1xu)(u^-1yu)(u^-1x^-1u)(u^-1y^-1u) = (u^-1xu)(u^-1yu)(u^-1xu)^-1(u^-1yu)^-1 which is in S $\subset$ [G,G], by definition of S.

Hence, for every element s in S(s is of the form xyx^-1y^-1) and for every u in G, u^-1su is in [G,G].
Since [G,G] is the smallest subgroup of G containing S or is equivalently, the subgroup of G generated by S, hence every element in [G,G] is a finite product of elements of S (note that S is closed under the operation of taking inverses).

Thus, if s₁s₂...s_n is an arbitrary element in [G,G] where s_i is in S for all i between 1 and n, and if u is in G, then
u^-1s₁s₂...s_nu = (u^-1s₁u)(u^-1s₂u)...(u^-1s_nu) which is in [G,G] since each u^-1s_iu is in [G,G] by the previous paragraph and the fact that [G,G] is a subgroup.

Consequently, [G,G] is normal in G.


(b)
The quotient group G/[G,G] makes sense since [G,G] is a normal subgroup of G.
Let [G,G]x and [G,G]y be two arbitrary elements in the quotient where x,y are in G.
Observe that xyx^-1y^-1 is in S $\subset$ [G,G]. Thus, the coset [G,G]xyx^-1y^-1 contains the identity of G.
Equivalently, [G,G]xyx^-1y^-1 = [G,G] and thus, [G,G]xy . [G,G]x^-1y^-1 = [G,G] . Thus, [G,G]xy . [G,G](yx)^-1 = [G,G] and hence,
[G,G]xy = [G,G]yx.
Therefore, [G,G]x . [G,G]y = [G,G]y . [G,G]x for all x,y in G.
Consequently, G/[G,G] is abelian.


(c)
Suppose that N is a normal subgroup of G such that G/N is abelian.
Suppose that s is in S. By definition, s is of the form xyx^-1y^-1 for some x,y in G.
Observe that Nxyx^-1y^-1 = Nx . Ny . Nx^-1 . Ny^-1 = Nx . Nx^-1 . Ny . Ny^-1 (since G/N is abelian) and thus, Nxyx^-1y^-1 = N.
Thus, s = xyx^-1y^-1 is in N. Consequently, S $\subset$ N. Since [G,G] is the smallest subgroup of G containing S and N is a subgroup of G that contains S, hence [G,G] $\subset$ N.


(d)
Suppose that K is a subgroup of G and [G,G] is contained in K.
Suppose that k is in K and u is in G.
Then, uku^-1k^-1 is in S $\subset$ [G,G] $\subset$ K. Also, k is in K. Thus, uku^-1 = uku^-1k^-1 k is in K as K is a subgroup.
Since k in K and u in G are arbitrary, hence, K is a normal subgroup of G.


NOTE THAT THE QUOTIENT GROUP G/[G,G] IS CALLED THE ABELIANIZATION OF G, FOR PARTS (b) and (c).