Showing group with $p^2$ elements is Abelian I have a group $G$ with $p^2$ elements, where $p$ is a prime number. Some (potentially) useful preliminary information I have is that there are exactly $p+1$ subgroups with $p$ elements, and with that I was able to show $G$ has a normal subgroup $N$ with $p$ elements. 
My problem is showing that $G$ is abelian, and I would be glad if someone could show me how. 
I had two potential approaches in mind and I would prefer if one of these were used (especially the second one). 
First: The center $Z(G)$ is a normal subgroup of $G$ so by Langrange's theorem, if $Z(G)$ has anything other than the identity, it's size is either $p$ or $p^2$. If $p^2$ then $Z(G)=G$ and we are done. If $Z(G)=p$ then the quotient group of $G$ factored out by $Z(G)$ has $p$ elements, so it is cylic and I can prove from there that this implies $G$ is abelian. So can we show theres something other than the identity in the center of $G$?
Second: I list out the elements of some other subgroup $H$ with $p$ elements such that the intersection of $H$ and $N$ is only the identity (if any more, due to prime order the intersected elements would generate the entire subgroups). Let $N$ be generated by $a$ and $H$ be generated by $b$. We can show $NK= G$, i.e every element in G can be wrriten like $a^k b^l $. So for this method, we just need to show $ab=ba$ (remember, these are not general elements in the set, but the generators of $N$ and $H$). 
Do any of these methods seem viable? I understand one can give very strong theorems using Sylow theorems and related facts, but I am looking for an elementary solution (no Sylow theorems, facts about p-groups, centrailzers) but definitions of centres and normalizers is fine. 

Here is a proof of $|Z(G)|\not=p$ which does not invoke the  proposition
"if $G/Z(G)$ is cyclic then $G$ is abelian".
Suppose $|Z(G)|=p$. Let $x\in G\setminus Z(G)$. Lagrange's theorem and $p$ being prime dictate that the centralizer $Z(x)$ of $x$ has order $1,p$,or $p^2$. Note that $\{x\}\uplus Z(G)\subset Z(x)$, $Z(x)$ has at least $p+1$ elements, so $|Z(x)|=p^2$, but this implies $x\in Z(G)$, a contradiction.
