In the following problem, the only parts I didn't understand were c) and e). The remaining I did. Please, help me!
Let $A,B$ be subgroups of $G$ that normalize each other. Assume that the set $X=\{[a,b]:a\in A,b\in B\}$ is finite. We will show that $[A,B]$ is finite. Note first that, without loss of generality, we may assume that $G=AB$. With this assumption $A$ and $B$ are normal subgroups of $G$. Let $U=[A,B]\subset $ $A\cap B$. Clearly $U\vartriangleleft $ $G$.
a) Show that $C_{G}(X)$ is a normal subgroup of finite index in G. Show that $C_{G}(X)$ centralizes $U$.
b) Deduce from part (a) that $C_{G}(X)\cap U$ is a central subgroup of $U$ and has finite index in $U$. Schur theorem implies that $U'$ is finite.
c) Show that, without loss of generality, we may assume that $U^{\prime }=1$.
d) Clearly the subset $\{[a,u]:a\in A,u\in U\}$ of $X$ is finite and these elements commute with each other. Show that $[a,u]^{2}=[a,u^{2}]$. Conclude that $[A,U]$ is finite.
e) Show that, without loss of generality, we may assume that $[A,U]=1$. Conclude that, without loss of generality $U$ is central in $G$.
f) Show that $X$ is closed under the squaring map $x\rightarrow $ $x^{2}$. Conclude that $[A,B]$ is finite.
proof of a)
$A,B\vartriangleleft G\Rightarrow X^{g}=X,\forall g\in G\Rightarrow N_{G}\left( X\right) =G$ e $C_{G}\left( X\right) \vartriangleleft G$ $ \Rightarrow G/C_{G}\left( X\right) \simeq Aut\left( X\right) $ is finite, because $X$ is.
As $U=\left[ A,B\right] =\left\langle X\right\rangle $ then $C_{G}\left( X\right) =C_{G}\left( U\right) .$ Therefore, $C_{G}\left( X\right) $ centralizes $U,$that is, $\left[ C_{G}\left( X\right) ,U\right] =1.$
\bigskip Proof of b)
As $C_{G}\left( X\right) \cap U=C_{U}\left( X\right) $, then $\left[ C_{U}\left( X\right) ,U\right] \leq \left[ C_{G}\left( X\right) ,U\right] =1. $ Therefore, $C_{U}\left( X\right) \leq Z\left( U\right) \leq U.$
As $\left( G:C_{G}\left( X\right) \right) $ is finite and $UC_{G}\left( X\right) /C_{G}\left( X\right) \simeq U/C_{U}\left( X\right) ,$ then $% U/C_{U}\left( X\right) $ is finite. Thereofore, by Schur Theorem $U^{\prime } $ is finite, as required.
I do not know how to prove that $U/U^{\prime }$ is finite$.$ I know that $ U/U^{\prime }$ is abelian and finitely generated because $X$ is finite (by hypothesis).
Excuse me, my English is not very good.