If a normal subgroup and its factor group are both abelian, then what can be said about the group?

Question

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?

Answer 1

To answer the question of your title, groups with this property are called metabelian. Some metabelian groups are indeed abelian, but there are plenty of examples found in that link which are not abelian.

Answer 2

Hint. The symmetric group $S_3$ has a nontrivial normal subgroup.

Rethink your "proof".