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 $a\in A, b\in B$?
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*}