If a normal subgroup and its factor group are both abelian, then what can be said about the group? The textbook I am using to self-study Abstract Algebra has the following problem.  

Prove or disprove: If $H$ is a normal subgroup of $G$ such that $H$ and $G/H$ are abelian, then $G$ is abelian.  

My attempt:
Since $H\triangleleft G$ and $H,G/H$ are abelian, then $g_1g_2H=g_2g_1H\text{ ,  }\forall g_1,g_2\in G$
$\implies g_1g_2h_1=g_2g_1h_2 \text{ for some }h_1,h_2\in H$
$\implies g_1g_2h_1h_2^{-1}=g_2g_1$
$\implies g_1^{-1}g_2^{-1}g_1g_2h_1h_2^{-1}=e$
How do I conclude if $G$ is abelian or not? 

Hint. The symmetric group $S_3$ has a nontrivial normal subgroup.
Rethink your "proof".
