I try to prove the following:
Let $G$ be a group and $N$ be a normal subgroup. Suppose $N$ is abelian and $G/N$ is abelian. Then $G = N \cdot (G/N)$ is abelian.
Proof:
For the first assertion that $G = N \cdot (G/N)$ let $g \in G$, then $g = g n n^{-1} \in gN \cdot N$ and then $gN \in G/N$ and $n^{-1} \in N$.
For the second assertion, let $g_1,g_2 \in G$. Then
$$ g_1 g_2 = g_1 n_1^{-1} n_1 g_2 n_2^{-1} n_2 = g'_1 n_1 g'_2 n_2 $$
where $g'_i = g_i N$. Then $$g'_1 n_1 g'_2 n_2 = g'_1 g'_2 n'_2 n_2$$ since $N$ is normal, and
$$ g'_1 g'_2 n'_2 n_2 = g'_2 g'_1 n'_2 n_2$$
since $G/N$ is abelian, and
$$ g'_2 g'_1 n'_2 n_2 = g_2 n_2^{-1} g_1 n^{-1}_1 n'_2 n_2$$
and here I am stuck.
Best Answer
The claim is false, $C_3=A_3$ is normal in $S_3$, since $|S_3:C_3|=2$, and $S_3/C_3\simeq C_2$ is also abelian, but $S_3$ is not. Note that $C_2\rtimes C_3\simeq S_3$. In general if $|C_q|=q\mid p-1=\varphi(p)=|{\rm Aut}\, C_p|$ the semidirect product $C_p\rtimes C_q$ is a counterexample.