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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 $|G'|<\infty$?

Glinka
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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.

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Derek Holt
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  • Could you explain why we have $[a,b^k]=[a,b]^k$ equality? – Glinka Nov 12 '15 at 17:08
  • Got it. By induction. It's true for $k=1$. Let $[a,b^{k-1}]=[a,b]^{k-1}$, then $[a,b^k]=ab^ka^{-1}b^{-k}=ab^{k-1}b\underbrace{a^{-1}b^{1-k}a}{\in B}a^{-1}b^{-1}=ab^{k-1}\underbrace{a^{-1}b^{1-k}a}{\in B}ba^{-1}b^{-1}=[a,b^{k-1}][a,b]=[a,b]^k$ – Glinka Nov 12 '15 at 18:25
  • Could you explain why if $[A, A\cap B]=[B,A\cap B] = {e}$ and $A\cap B$ is abelian, then $A\cap B \leq Z(G)$? Posted this question here – Glinka Nov 12 '15 at 19:16
  • Sorry, I should have written $A \cap B \le Z(AB)$. In fact we can assume that $G=AB$ and I had done that without mentioning it! – Derek Holt Nov 12 '15 at 20:34
  • It is not straightforward btw that if $[A,B]$ is generated by finitely many elements of finite order then it's finite. Though it is true for commutators, it's false for arbitrary group $G$. – Glinka Nov 13 '15 at 06:42
  • It's true for abelian groups (and more generally for nilpotent groups), and $[A,B] \le A \cap B$, which we are assuming is abelian. – Derek Holt Nov 13 '15 at 08:58
  • oh, right. Of cause) This fact is also true for non-abelian commutators $[A,B]$, since $[a,b][x,y] = [x^{[a,b]},y^{[a,b]}][a,b]$, so we can accumulate powers of the same commutator with the length of the product intact. So if we have $n=o([a,b])$ items of $[a,b]$ in the product, we can accumulate it in $[a,b]^n = e$ and so decrease the length of the product. So the length cannot be more then $o([a_1,b_1]) \cdot\ldots\cdot o([a_k, b_k])$ and thus the finiteness. – Glinka Nov 13 '15 at 09:09