$G$ finite group, $H \trianglelefteq G$, $\vert H \vert = p$ prime, show $G = HC_G(a)$ $a \in H$ Let $G$ be a finite group. $H \trianglelefteq G$ with $\vert H \vert = p$ the smallest prime dividing $\vert G \vert$. Show $G = HC_G(a)$ with $e \neq a \in H$. $C_G(a)$ is the Centralizer of $a$ in $G$.
To start it off, I know $HC_G(a)\leq G$ by normality of $H$ and subgroup property of $C_G(a)$. So I made the observation that 
$\begin{align*} \vert HC_G(a) \vert &= \frac{\vert H \vert \vert C_G(a) \vert}{\vert H \cap C_G(a) \vert}\\&=\frac{\vert H \vert \vert C_G(a) \vert}{\vert C_H(a)\vert}\end{align*}$
But, from here on I never reach the result I'm looking for. Any help would be greatly appreciated!
Note: I posted a similar question earlier, except that one had the index of $H$ being prime, this has the order of $H$ being prime: Link

Since $N_G(H)/C_G(H)$ injects in $Aut(H)\cong C_{p-1}$ and $p$ is the smallest prime dividing $|G|$, it follows that $N_G(H)=C_G(H)$. But $H$ is normal, so $G=N_G(H)$ and we conclude that $H \subseteq Z(G)$. In particular $G=C_G(a)$ for every $a \in H$.
