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. 
 A: First we need the following lemma:

If $G/Z(G)$ is cyclic then G is abelian.

You can check the prove in here: https://yutsumura.com/if-the-quotient-by-the-center-is-cyclic-then-the-group-is-abelian/.
Then you need this theorem:

If $p$ is a prime and $P$ is a group of prime power order $p^{\alpha}$ for some $\alpha \geq 1$, then $P$ has a nontrivial center: $Z(P) \neq 1$.

You can check the prove in the answer above: https://math.stackexchange.com/a/64374/435467.
Now is the prove of the question:

Since $Z(P) \neq 1$, and as the order of $P$ is $p^2$, there's no other choice that $|Z(P)| = p$. Then $|P/Z(P)| = \frac{p^2}{p} = p$, and as any group with prime order must be cyclic, $P/Z(P)$ is cyclic. By the above lemma, $G$ must be abelian.

A: Here is a way to show that the center of a group of order $p^2$ cannot be trivial without using the class equation. I think that the major drawback (and it is major) is that it is very specific for groups of order $p^2$. The class equation is much better because it is a much more general result
If $G$ has elements of order $p^2$, then the result follows because $G$ is cyclic. So suppose that all nontrivial elements of $G$ have order $p$.
Let $x\in G$ be of order $p$. 
Now, $\langle x\rangle$ is normal in $G$, since its index is the smallest prime that divides the order of $G$. Therefore, $yxy^{-1}\in\langle x\rangle$, so $yxy^{-1}=x^r$ for some $r$, $1\leq r \leq p-1$.
It is now easy to verify that $y^ixy^{-i} = x^{r^i}$. In particular, $y^{p-1}xy^{1-p} = x^{r^{p-1}}$. By Fermat's Little Theorem, $r^{p-1}\equiv 1 \pmod{p}$, so $y^{p-1}xy^{1-p} = x$. That is, $y^{p-1}$ centralizes $x$. But $y$ is of order $p$, so $y^{p-1}=y^{-1}$. Since $y^{-1}$ centralizes $x$, so does $y$. That is, $yx=xy$. Thus, the centralizer of $x$ contains at least $\langle x\rangle$ and $y$, hence is of order at least $p+1$. Since its order must divide $p^2$, the centralizer of $x$ is all of $G$, so $x$ is central.
Thus, $Z(G)$ is nontrivial.
A: So Deven's helpful comment shows why the center of any $p$-group is nontrivial, so look back at your first approach.
Hint: if $Z(G) = p$, consider an element $x \in G$ that is not in $Z(G)$.
What is the overlap between $Z(G)$ and the cyclic subgroup of $G$ generated by $p$? 
What must the centralizer of $x$ be? 
What does this imply about $Z(G)$?
Edit: This is an alternative to the path of showing that $G/Z(G)$ cyclic implies $G$ abelian; I consider this route to be also as illuminating and even elegant.
A: Your first approach is good; The center of a $p$-group is non-trivial: 
Proof: 
The center of any group is the union of the 1-element conjugacy classes in the group. For a $p-$group, the size of every conjugacy class is a power of p because the order of a conjugacy class must divide the order of the group. Then let $p^{n_i}$ be the order of the conjugacy classes, and the conjugacy class equation tells us that $|G| = p^n = |Z(G)| + \sum_i (p^{k_i})$, where $0 < k_i < n$ and thus $p$ must divide $|Z(G)|$ implying that the center is non-trivial. $\Box$ 
EDIT: Explaining class equation: 
The conjugacy classes partition the group, so we know that $|G| = \sum|cl(a)|$. But as I said in the proof above, the union of the singleton conjugacy classes is $Z(G)$, so we can rewrite this equality as 
$|G| = |Z(G)| + \sum|cl(a)|$ 
where we assume that each conjugacy class is represented only once (i.e we are not including the singletons in the second summand). However, since we know that the order of a conjugacy class divides the order of a group we can rewrite the second summand to be $\sum_i(p^{k_i})$ where $0 < k_i < n$ because clearly none of them can have the same order as $G$, and we finally get 
$|G| = |Z(G)| + \sum_i(p^{k_i})$ where $0 < k_i < n$
Now to see that $p$ must divide $|Z(G)|$ we see that we can move the second summand to left and replace $|G|$ with $p^n$ to get $p^n - \sum_i(p^{k_i}) = |Z(G)|$ and clearly we can factor out $p$. 
EDIT: Showing that the order of a conjugacy class must divide the order of a group: 
To prove this, we will show that the size of a conjugacy class of a, $cl(a)$ is the index of the centralizer of $a$, $C(a)$. 
Suppose $x$ and $y$ both make the same conjugate of $a$, or $xax^{-1} = yay^{-1}$. Then multiplying on the left by $y^{-1}$ and on the right by $x$ we can see that $y^{-1}xa = ay^{-1}x$ and hence, $y^{-1}x \in C(a)$. Thus we can also see that $x\in yC(a)$ and hence $xC(a) = yC(a)$. 
Similarly, suppose $xC(a) = yC(a)$ then it follows that $x \in yC(a)$ and hence $x = yz$ for some $z\in C(a)$. Thus $xax^{-1} = (yz)a(yz)^{-1} = yzaz^{-1}y^{-1}$. But we know that $z\in C(a)$ so we can rewrite this as $yazz^{-1}y^{-1}$, or $yay^{-1}$. Thus $x$ and $y$ make the same conjugate of $a$. 
Thus we have shown that the number of cosets of $C(a)$ equals the number of elements in $cl(a)$, or
$(G : C(a)) = |cl(a)|$ 
Since $C(a)$ is a subgroup of $G$, clearly $(G : C(a))$ divides $|G|$ and hence, $|cl(a)|$ divides $G$. $\Box$ 
A: Here is an alternative way to show this which does not show that the center of a finite $p$-group is non-trivial. The proof is far from elementary, though the results I will use are of more general use.
In fact, what I will show is that if $G$ is a finite $p$-group then $|G/G'|\equiv |G|\pmod {p^2}$ (so a finite group of order $p^n$ is solvable of derived length at most $\frac{n}{2}$ rounded up).
To do this we will be be using some results about complex characters of finite groups (so this answer will hopefully serve as a (slight) motivation to learn more about this topic). I will not go through the definitions of irreducible complex characters here, but if anyone wants to see them, they can start at http://en.wikipedia.org/wiki/Character_theory and work their way through the various parts involved.
The results we need for this are: 
If $Irr(G)$ denotes the set of irreducible complex characters of $G$ then $$|G| = \sum_{\chi\in Irr(G)}\chi(1)^2$$
If $\chi\in Irr(G)$ then $\chi(1)$ divides $|G|$.
And finally that $|G/G'| = |\{\chi\in Irr(G)\mid \chi(1) = 1\}|$.
Applying this to a finite $p$-group, we see that if $\chi\in Irr(G)$ with $\chi(1)\neq 1$ then $p$ divides $\chi(1)$, and we get $$|G| = \sum_{\chi\in Irr(G)}\chi(1)^2 = \sum_{\chi\in Irr(G),\, \chi(1) = 1}\chi(1)^2 + \sum_{\chi\in Irr(G),\, \chi(1)\neq 1}\chi(1)^2$$ $$= |\{\chi\in Irr(G)\mid \chi(1) = 1\}| + p^2m = |G/G'| + p^2m$$ for some natural number $m$, which proves the original claim.
A: I will present a different approach which only involves group actions.
If there is an element of order $p^2$ then the group is the cyclic group of order $p^2$ which is, of course, abelian.
Otherwise, suppose every element in $G$ has order less than $p^2$, it is equivalent to say that the order of every non-unit element in $G$ is $p$. Now we choose an arbitrary non-unit element $g\in G$, then we have $|\langle g \rangle|=p$. Suppose $\Omega=\{a_1\langle g \rangle,a_2\langle g \rangle,...,a_p\langle g \rangle\}$ is a left cosets partition of $G$. Let $\langle g \rangle$ acts on $\Omega$ by the natural left multiplication. Hence $$ |\text{Orbit}(a_i\langle g \rangle)|=\frac{|\langle g \rangle|}{|\text{Stab}(a_i\langle g \rangle)|}=\begin{cases}p,&\text{when } |\text{Stab}(a_i\langle g \rangle)|=1\\ 1,&\text{when }|\text{Stab}(a_i\langle g \rangle)|=p\end{cases} $$ But $|\text{Orbit}(\langle g \rangle)|=1$, so for every left coset $a_i\langle g \rangle$, $|\text{Orbit}(a_i\langle g \rangle)|=1$ which means for every $g^k,k\in\mathbb Z$ and $a_i$, $a_i^{-1}g^ka_i\in\langle g \rangle$. Thus for every $a_ig^m$, $(a_ig^m)^{-1}g^k(a_ig^m)\in\langle g \rangle$ which implies $\langle g \rangle$ is a normal subgroup of $G$. Since $g$ is an arbitrary non-unit element in $G$, it follows that every subgroup of order $p$ of $G$ is a normal subgroup.
Now take $g_1,g_2\ne e$ where $e$ is the unit and $\langle g_1 \rangle\cap\langle g_2 \rangle=\{e\}$. From what we have proved above, we know that $\langle g_1 \rangle$ and $\langle g_2 \rangle$ are normal subgroups. Hence
\begin{align}
g_1g_2g_1^{-1}&=g_2^{\ell_1}\tag 1\\
g_2g_1g_2^{-1}&=g_1^{\ell_2}\tag 2
\end{align}
for some $\ell_1,\ell_2\in\mathbb Z$. Note that $g_2g_1g_2^{-1}g_1^{-1}=g_1^{\ell_2-1}$ by $(2)$ but $g_2g_1g_2^{-1}g_1^{-1}=g_2^{1-\ell_1}$ by $(1)$. So $g_1^{\ell_2-1}=g_2^{1-\ell_1}\in\langle g_1 \rangle\cap\langle g_2 \rangle=\{e\}$. Therefore $g_1^{\ell_2}=g_1$ and $g_2^{\ell_1}=g_2$ and $g_1g_2g_1^{-1}=g_2$ for all $g_1,g_2\ne e$. Hence $G$ is abelian.
A: A slightly different way using actions and orbits.
Let $|G| = p^2$.  Show that $G$ is abelian.
If there exists an element $g \in G$ so that $|g| = p^2$, then $g$ generates $G$, 
making $G$ cyclic and thus abelian, and we are done.
So, assume there is no element of $p^2$ power.  Since the order of each element 
must divide $|G|$, all elements (except the identity) must be of $p$ power.
Let $a \in G$ be arbitrary.  We want to show that $a$ commutes with every other 
element of $G$.
If $a$ is the identity, we are done, so assume otherwise.  Then $|a| = p$.
Define $H = \langle a \rangle$, so $|H|=p$.  Since $|G:H|=p$, and $G$ is a $p$-group, $H$ is normal in $G$.  So, let $G$ act on $H$ by conjugation, 
and let $H_G$ denote the union of the single-element orbits of this action.
Since orbits partition, we now have $$ |H| = |H_G| + \sum_{|\mathcal{O}| > 1} |\mathcal{O}|, $$ where the sum on the right counts all elements of $H$ in non-singleton orbits (similar to the class equation, but we are only acting on the subgroup $H$).
Since the size of orbits must divide the size of the group doing the acting (in 
this case $G$), we can infer that every non-singleton orbit has size $p$ (since $p^2$ would be too big).
Looking at the union of singleton orbits, we see $H_G = \left\lbrace h \in H : \forall g \in G : g^{-1}hg = h\right\rbrace$.  In other words, $H_G$ contains all elements $h$ that commute with every element of $G$.
Since the identity is in $H$, and the identity commutes with everything, 
$|H_G| \geq 1$, but since the smallest a "multi-element" orbit can be
is $p$, there are no elements in multi-element orbits, and we have that
$|H_G| = |H| = p$, implying that every element of $H$ (in particular, its generator $a$) commutes with every element
of $G$.
Since $a$ was arbitrary, every element of $G$ must commute with every other element of $G$, and we are done.
A: 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.
