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I tried to prove this statement, and I would like to have some input if it's complete:

Consider:

$(aba^{-1})b^{-1} = b_1b^{-1} \in B$ $ $ (because $B\unlhd G$)

$a(ba^{-1}b^{-1}) = aa_1 \in A$ $ $ (because $A\unlhd G$)

Hence: $b_1b^{-1} = aa_1 = e$ $ $ (because $A\cap B = ${$e$})

This gives us: $aba^{-1}b^{-1} = e \rightarrow ab(ba)^{-1}=e \rightarrow ab = ba$

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    You should explain all of your statements. For example, since $B$ is normal in $G$, $aba^{-1}\in B$. Therefore, there exists $b_1\in B$ such that $aba^{-1}=b_1$. Moreover, you say $b_1b^{-1}=aa_1=e$. It would be better to say that since $b_1b^{-1}=aa_1$ is in $B$ and $A$, so $b_1b^{-1}\in A\cap B=\{e\}$, so $b_1b^{-1}=e$. – 2017-02-26
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    Thanks, added explanations on each line (You're right, I should've done this before) – 2017-02-26
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    Are you two doing the same homework, or passing the same exam? http://math.stackexchange.com/questions/2162605/proving-that-hk-kh – 2017-02-26
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    I don't know :) – 2017-02-27

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