Pretty straightforward:
If $H$ is a normal subgroup of $G$ and if both $H$ and $G/H$ are abelian, is $G$ abelian?
Pretty straightforward:
If $H$ is a normal subgroup of $G$ and if both $H$ and $G/H$ are abelian, is $G$ abelian?
No, consider $G=S_3$ and $H=A_3=\{e,(123),(132)\}$. Then $H\triangleleft G$, and $H\cong \mathbb{Z}/3\mathbb{Z}$ and $G/H\cong\mathbb{Z}/2\mathbb{Z}$ are both abelian, but $G$ is not abelian.
As a more general class of examples, consider the group $D_n$ of symmetries on a regular $n$-gon for $n\geq 3$ (of course $D_3 \cong S_3$, so we're really only interested in n>3). Let $R$ denote the element of $D_n$ corresponding to a counter-clockwise rotation of $2\pi/n$, and let $H$ be the subgroup of $D_n$ generated by $R$. Then $H \triangleleft D_n$ (since $\vert D_n\,:\,H\vert=2$), and clearly both $H$ and $D_n/H\cong \mathbb{Z}_2$ are abelian. However, $D_n$ is not abelian.