Alternative proof: G group of order $p^2$, p prime $\Rightarrow$ $H$ is normal in $G$. Let $G$ be a group of order $p^2$ where $p$ is prime. If $H$ is a subgroup of order $p$, show that $H$ is normal in $G$.
I would like to prove this with the tools that the book has provided up to this point. Unfortunately that means proving the statement without:


*

*$|G|=p^2\Rightarrow G$ abelian.

*Group actions (or the class equation).
I know this is kind of repetitive but I haven't found the answer I was looking for. 
(The case where $G$ has an element of order $p^2$ is done: G is cyclic, therefore abelian, so any subgroup is normal.)
Thanks!
 A: First, as you have observed, either $G$ is cyclic (in which case the result is obvious), or every non-identity element has order $p$ (by Lagrange). We will assume that the latter holds. 
Now, prove that if $H$ and $K$ are any subgroups of $G$, then 
$$|HK|=\frac{|H||K|}{|H\cap K|}.$$
This is most easily proved using group actions, but can be done by adapting the proof of Lagrange's Theorem (count the number of cosets of the form $hK$).
Now, having proved this, suppose that $H$ is a subgroup of order $p$ which is not normal. Then, there exists $g\in G$ such that $gHg^{-1}=K$ for some subgroups $K\neq H$ of order $p$. Since $p$ is prime, we must have $|H\cap K|=1$ and therefore $HK=KH=G$. This means we can write $g=kh$ for some $k\in K$ and $h\in H$ and
$$gHg^{-1}=khHh^{-1}k^{-1}=kHk^{-1}=K.$$
But, the equality $kHk^{-1}=K$ implies $H=k^{-1}Kk=K$, a contradiction.
A: A basic fact about $p$-groups is that their center is nontrivial (guess we are allowed to use this, as you have not excluded it on your list). So if $|G| = p^2$ we either have $Z(G) = G$ (then it is abelian and every subgroup is normal) or $|Z(G)| = p$. Now either $H = Z(G)$, or $H \cap Z(G) = 1$ by order considerations for each subgroup $H$ of order $p$, in the second case we have $G = HZ(G)$ and $H$ must be normal, as every element $g \in G$ has the form $g = hz$ and hence $H^g = H^{hz} = H^z = H$.
