Let $G$ be a finite group. Let $H \trianglelefteq G$, with $\vert H \vert = p$, a prime, where $p$ is the smallest prime dividing $\vert G \vert$. Prove that $H \leq Z(G)$. (Hint: If $a \in H$, by normality, its conjugacy class lies inside $H$.)
My approach so far:
Let $a \in H$. Then $C(a) = \{gag^{-1} : g \in G\} \subseteq H$. Now, since $\vert H \vert$ is prime, it follows that $H$ is cyclic and moreover $H$ is abelian, hence $C_H(a) = H$ for all $a \in H$ and equivalently $[H: C_H(a)] = 1$ for all $a \in H$.
Try to show $\vert C(a) \vert = [G:C_G(a)] = [H:C_H(a)] = 1$.
So, since $\vert C(a) \vert = 1$ for all $a \in H$, we have $C(a) = \{a\}$ for all $a \in H$ and equivalently $a \in Z(G)$ for all $a \in H$, so we can say $H \subseteq Z(G)$ and moreover $H \leq Z(G)$.
My main question is, how to show $[G:C_G(a)] = [H:C_H(a)]$. But also, I haven't used the fact that $p$ is the smallest prime dividing $\vert G \vert$, so is my reasoning wrong anywhere?