TITLE: Normal subgroup of order $p$ is in the center
QUESTION [6 upvotes]: Prove that in a group of order $p^2$ ($p$ a prime), a normal subgroup of order $p$ lies in the center.   (Since this is Exercise 2.9.6 from Herstein's Topics in Algebra, there are some restrictions on what I can use.  This section is on Cayley's theorem.  I can show, using the class equation, that $Z(G)>1$ and $G$ is abelian, but I don't want to use the class equation yet.)  
If $H$ is a subgroup of order $p$, since $|G|$ does not divide $i(H)!$, $H$ contains a nontrivial proper subgroup $K$ of $G$. Since $p$ is a prime, $K$ must equal $H$.  Hence $H$ is normal.  Why is it in the center?

REPLY [2 votes]: If $G$ were cyclic it would be abelian, and we’d be done. Suppose it’s not. Then every element is of order $p$ by Lagrange’s theorem. And every subgroup of size $p$ is normal by Herstein’s lemma 2.9.1.
Subgroup $H$ must be cyclic having elements of order $p$. Then any two elements of $H$ commute.
Suppose $k\not\in H$. Then $k$ generates a group $K$ of size $p$, and this group is normal (using the first paragraph).
Now $H\cap K=\{e\}$ since it’s a subgroup of size $<p$. Then, for any $h\in H,$ $hkh^{-1}k^{-1} \in H\cap K$ using the normalcy of both subgroups (eg $hkh^{-1}\in K$ since $K$ is normal, and so on). But then
$$hkh^{-1}k^{-1} = e \Rightarrow hk = kh.$$