Prove that if $A,B\unlhd G, A\cap B = ${$e$}$\rightarrow \forall a\in A $ and $\forall b\in B $ ab=ba

Question

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