Finite groups and Normal subgroups - $G$ is a finite group such that $|G|=n$. Let $p$ be the smallest prime such that $p|n$. $G$ is a finite group such that $|G|=n$. Let $p$ be the smallest prime such that $p|n$. If $H\triangleleft G$ such that $|H|=p$, then prove that $H\subset Z(G)$.
My Work:
Since $|H|$ is a prime $H\cong \mathbb{Z}/(p)$. Hence $H$ is cyclic and abelian. I was stuck afterwards. Can anyone please give me a hint?
 A: Let $N$ be the normalizer of $H$ in $G$ and let $C$ be the centralizer. We are given that $N=G$, and we want to show that $C=G$ as well.
Note that $N/C$ acts by conjugation on $H$, so $N/C$ is isomorphic to a subgroup of $Aut(H)$. $N=G$, so in fact, $G/C$ is isomorphic to a subgroup of $Aut(C_p)$, which has order $p-1$. The order of $G/C$ divides both $|G|$ and $p-1$, but since $p$ is the smallest prime dividing $|G|$, this order must in fact be $1$, so $C=G$, as desired.
A: I write down a proof using the action of $G$ on $H$. I'm new to group theory so that I'm not quite sure if there is any gap in my proof. Any critisism and correction is welcome. Please do not merely down vote.
Proof:
Let $G$ act on $H$ by conjugation. By the orbit-stabilizer theorem, any orbit $Gh$ must have length dividing $|G|$. Since also $|Gh| \leq |H| =p$ where $p$ is the smallest prime divisor of $|G|$, there cannot be any prime divisor of $|G|$ larger than $p$ who also divides $|Gh|$. We conclude that $|Gh| = 1$ or $p$.
If $|Gh| = 1$ for all $h \in H$, then we are done, because that means $g^{-1}hg = h$ for every $h \in H$ and $g \in G$. Suppose that $|Gh| = p$ for some $h \in H$. In that case, the whole $|H|$ forms a single orbit. Therefor the action is transitive. In particular, there exists some $g \in G$ such that $g^{-1}hg = 1$ for some $h \in A \neq 1$, which is absurd, since the only conjugate of $1$ is itself.
