TITLE: $G$ group, $H \trianglelefteq G$, $\vert H \vert$ prime, then $H \leq Z(G)$
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: We use the following fact.
If $G$ is group and $H \le G$ , $N_G(H)/ C_G(H) $ is isomorphic to a subgroup of $\text{Aut}(H)$. ($N_G(H)$ is the normalizer of $H$ and $C_G(H)$ is the centralizer of $H$ in $G$. )
Since $p$ is a prime, $H$ is cyclic. So $|\text{Aut}(H)| = \varphi(p) = p-1 $. So by Lagrange's theorem $|N_G(H)|/ |C_G(H)|$ divides $p-1 $. Since $N_G(H) = G$, $\frac{|G|}{C_G(H)}$ divides $p-1$. Since $p$ is the smallest prime dividing $G$, we must have $|C_G(H)| = |G|$.So $G = C_G(H)$. This is same as saying $H \le Z(G)$.