I faced the following problem:
Suppose $G$ is a group, $A,B$ are normal subgroups of $G$ and set $\{[a,b], a\in A\lhd G, b\in B\lhd G\}$ is finite. Then the order of subgroup $[A,B]$ is also finite.
We know that since $A, B$ are normal, $[A,B] \leq A\cap B \lhd G$. Also $[A,B] \lhd A$ and $[A,B] \lhd B$ and so $[A,B] \lhd A\cap B$. It is as well clear that $(A\cap B)^{(1)} \leq [A,B]$ and since $(A\cap B)^{(1)} \lhd A\cap B$ we have $$ (A\cap B)^{(1)} \lhd [A,B] \lhd A\cap B \lhd G $$ and $(A\cap B)^{(1)}$ and $[A,B]$ are finitely generated. Apart from that I have nothing. How do I approach this problem?
Maybe there's a way to use here the fact (non-trival, but proven) that if $[G:Z(G)]<\infty$ then $|G'|<\infty$?
It is proved here that if a group $G$ has only finitely many commutators, then $[G,G]$ is finite.
This certainly applies to $A \cap B$ so that has finite derived group, which we can factor out and hence assume that $A \cap B$ is abelian.
Now, for $a \in A$ and $b \in A \cap B$, $[a,b^k]=[a,b]^k$ for all $k$ and, since there are only finitely many such elements, $[a,b]$ has finite order. Hence $[A,A \cap B]$ is generated by finitely many elements of finite order, so $[A,A \cap B]$ is finite and we can factor it out and assume it trivial. Similarly for $[B,A \cap B]$.
So now we have $A \cap B \le Z(AB)$. So, for $a \in A$, $b \in B$, $[a,b^k]=[a,b]^k$ for all $k$. Hence all elements of $[a,b]$ have finite order, so $[A,B]$ is finite.