Show that If $|G|=p^2$ and $H\leq G$ with $|H|=p$, for $p$ any prime, then $H$ is normal in $G$ If $|G|=p^2$ and $H\leq G$ with $|H|=p$, for $p$ any prime, then $H$ is normal in $G$.
I am sort of stuck with this proof and I would appreciate a hint (not a full solution, please!). Preferably, please don't make use of the general case for the smallest prime dividing $|G|$.
Thanks!
 A: You know what $G$ is, either $\mathbb{Z}_{p^2}$ or $\mathbb{Z}_p \times \mathbb{Z}_p$ (go ahead and prove this). Both are abelian. Every subgroup of an abelian group is normal (since everything commutes). 
A: Let $g\in G\setminus H$. Then $g$ has order either $p$ or $p^2$. For the first case we have $\langle g\rangle\cap H=1$. In the latter case $G$ is cyclic. 
A: Let $G/H=\{x_0H,...,x_{p-1}H \}$ (with $x_0=1$) and let $G$ act on $G/H$ by left-multiplication, that is $\phi : \left\{ \begin{array}{ccc} G & \to & \text{Bij}(G/H) \\ g & \mapsto & \varphi_g \end{array} \right.$ where $\varphi_g : x_iH \mapsto gx_iH$. Show that $\text{ker}(\phi)=H$.
A: You can easily prove that $G$ is abelian.  Hint :Use the class equation to deduce that the center of $G$ is nontrivial.  Then suppose $a \notin Z(G)$. Derive a contradiction by finding suitable bound for $|C(a)|$, where $C(a)$ is the centralizer of $a$.  The full solution is given below.

 Nontrival center: The class equation gives $|G| = |Z(G)| + \sum_{i=1}^n [G : C(a_i)]$, where $a_i \notin Z(G)$.  Then, $[G : C(a_i)] \ne 1$, so $p$ divides each index.  Since $p$ divides $G$, it follows that $p$ divides $Z(G).$



 $G$ is abelian: Suppose $a \in G$, and $a \notin Z(G)$.  Since the centralizer of $a$, $C(a)$ contains both $a$ and $Z(G)$ it is bigger than $Z(G)$.  However, $|C(a)|$ must be smaller than $|G| = p^2$ since $a$ does not commute with every element.  Hence, $|Z(G)| < |C(a)| < p^2$, a contradiction since $|Z(G)|$ is either $p$ or $p^2$.

A: $\operatorname{core}_G(H)$ must be trivial if it is proper in $H$.  Thus if $G$ is nonabelian, then $G$ (a group of order $p^2$) injects into $\operatorname{Sym}(G\backslash H)$ (a group of order $p!$), which is impossible.
