I am working on a problem on finite group theory, and would like asking a question on the correct operation of normal subgroup.
Suppose that $H$ is normal subgroup of $G$ and the factor group $G/H$ is abelian: $\forall a,b \in G$ and $aH, bH \in G/H$:
$$\begin{align} aHbH &= bHaH \qquad \qquad &&(1) \\ abH &= baH &&(2) \\ ab(ba)^{−1} &\in H &&(3a) \\ ab(ba)^{−1} &=H &&(3b) \\ \langle ab(ba)^{−1} \rangle &=H &&(3c) \\ aba^{−1}b^{−1} &= H &&(4)\\ [a, b] &= H &&(5)\\ G' &= H &&(6) \\ \end{align} $$
My question is: Which one is correct, the (3a), (3b) or (3c)? Regardless of which one, I would like you showing me how to get to my final goal, which is (6) $G' = H$. Please advise and thank you for your time.
part $3c$ is not true since that would imply that there is only one $H$ so that $G/H$ is abelian, and this is not true since in particular in an abelian group $G$, $H$ can be any of the subgroups of $G$.
part $3b$ is also not true since the left is an element and the right is a set (good job)
part $3a$ is true.We need $H$ to contain all of the commutators, to see this write $abH=baH\implies abH(baH)^{-1}=H\implies ab(ba)^{-1}H=H\implies ab(ba)^{-1}\in H$.
Part $5$ is not true since the thing in the left is an element. And finally part $6$ is not true since we can use the same counterexample as for part $3c$.
What is true and you should prove is that if $\frac{G}{H}$ is abelian then $H<G'$.(This should be simple since by $3a$ $H$ contains all the commutators and $G'$ is the subgroup generated by the commutators.