The following is a direct proof that any extension of $G$ by $V$ splits. It is taken from [a joint paper of mine with Tim Dokchitser][1], where the proof starts in the last paragraph of page 12.

First, note that for any $n\in\mathbb{N}$, $H^n(H,V)=0$, since it is killed by $|H|$ and $|V|$, which are coprime. This implies that the inflation map
$$
H^2(G/H,V^H)\longrightarrow H^2(G,V)
$$
is an isomorphism. But also, either $V$ is trivial, in which case what we want to prove is obvious, or $V^H=0$. Indeed, since $H$ is normal, $V^H$ is a subrepresentation. But $C$ is a $p$-group, so it has a fixed vector in $V^H$, which is then fixed under all of $G$. This contradicts irreducibility of $V$, unless $V$ is the trivial representation.
So $H^2(G,V)=0$, as required.


  [1]: http://arxiv.org/abs/1103.2047