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I have trouble understanding the answer to this question. I don't see how for normal subgroups $A,B$ of a group $G$ conditions $[A, A\cap B]=[B,A\cap B] = \{e\}$ and $A\cap B$ being abelian imply that $A\cap B \leq Z(G)$?

Edit: As David Hill pointed out, this statement is true for $G=AB$, which is acceptable assumption in the context of the liked post. So the question is, how $A\cap B \in Z(AB)$ implies that $[a,b]^k=[a,b^k]$ for $a\in A, b\in B$?

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Glinka
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  • The statement you write is false. Take $A=B=\langle r\rangle\leq D_8=\langle r,j\mid r^4=j^2=1,,jr=r^{-1}j\rangle$. I think you are misunderstanding what is being asserted in the linked post. – David Hill Nov 12 '15 at 19:26
  • @DavidHill, thank you! Though it seems like he meant exactly this... – Glinka Nov 12 '15 at 19:33
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    If one adds the assumption $AB=G$ this would be fine. In the context of the statement you are trying to prove, you may as well assume this as $G$ plays no role beyond being the ambient group $A$ and $B$ live in. – David Hill Nov 12 '15 at 19:48
  • @DavidHill, the goal there is to show that for $a \in A$, $b \in B$, $[a,b^k]=[a,b]^k$ for all $k$. As far as I understand, for this equality to stand we need to show that elements in $A$ commutes with elements with $B$. – Glinka Nov 12 '15 at 19:56
  • @DavidHill, sorry, not "elements in $A$ commutes with elements in $B$" but elements in $B$ commutes with each other. It is according to equalities I wrote in comments to the answer. – Glinka Nov 12 '15 at 20:05
  • In the post you link $b\in A\cap B$, which you have assumed is abelian. As $aba^{-1}\in A\cap B$ commutes with $b^{-1}\in A\cap B$ you have $[a,b]^k=((aba^{-1})b^{-1})^k=(aba^{-1})^kb^{-k}=ab^ka^{-1}b^{-k}=[a,b^k]$. – David Hill Nov 12 '15 at 20:08
  • @DavidHill, yes, thank you, this part is clear. I meant the same equality in the last indent. In there $a\in A, b\in B$ and we only have $[A, A\cap B] = [B, A\cap B] = {e}$ – Glinka Nov 12 '15 at 20:17
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    Why have posted this as a new question? I should have written $A \cap B \le Z(AB)$. – Derek Holt Nov 12 '15 at 20:35
  • @Derek Holt, to not have a discussion in a comments, but it happened anyway =) I still don't understand why the last equality holds. – Glinka Nov 12 '15 at 20:40
  • We proved $[A,A \cap B]=1$. That means every element of $A$ commutes with every element of $A \cap B$. Similarly, every element of $B$ commutes with every element of $A \cap B$. Hence every element of $AB$ commutes with every element of $A \cap B$, which is equivalent to $A \cap B \le Z(AB)$. – Derek Holt Nov 12 '15 at 20:44
  • Actually, you only need $[B,A \cap B]=1$ to deduce $[a,b^k]=[a,b]^k$, since $[a,b]$ is centralized by $b$. – Derek Holt Nov 12 '15 at 20:46

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Once you have gotten to the point that $A\cap B\leq Z(AB)$, you have $[A,B]\leq A\cap B\leq Z(AB)$. Now, the statement $[a,b^k]=[a,b]^k$ follows by induction: \begin{align*} [a,b]^k&=[a,b][a,b]^{k-1}\\ &=(aba^{-1}b^{-1})[a,b^{k-1}]\\ &=aba^{-1}[a,b^{k-1}]b^{-1}&[a,b^{k-1}]\in Z(AB)\\ &=(aba^{-1})(ab^{k-1}a^{-1}b^{1-k})b^{-1}\\ &=ab^ka^{-1}b^{-k}\\ &=[a,b^k] \end{align*}

David Hill
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