TITLE: Prove H is in the center Z(G)
QUESTION [7 upvotes]: Exercise from Artin's 2nd edition of Algebra.
Let $H$ be a normal subroup of prime order $p$ in a finite group $G$. Suppose that $p$ is the smallest prime that divides the order of $G$. Prove that $H$ is in the center $Z(G)$.
My attempt:
 Choose any $e\neq h \in H$, since $h$ is normal, the conjugacy class including $h$ lies in $H$, so this class has size less than $p$. The only positive integer less than $p$ which divides $|G|$ is one. So the class containing $h$ has size one, and $h \in Z(G)$.
Correct?

REPLY [4 votes]: Correct. Alternatively stated: $H$ is normal so it is a union of conjugacy classes; all those ccls have order dividing $G$, and hence have size either $1$ or bigger than $p$ (by definition of $p$ as the smallest prime dividing $|G|$). Therefore it is a union of size-1 ccls.