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I found the Proof Here Showing group with $p^2$ elements is Abelian , But here some proof are complicated , those who are beginner in algebra will difficult them to understand

Theorem : If ${\rm order}(G) = |G|= p^2 $, where $p$ is prime number, then $G$ is abelian.

My attempt : Possible of $|Z(G)|$ are $1 , p , p^2$

Step $1:$

We must know one theorem for this proof , that is if $order (G) =p^n $, where $p$ is prime number , then $Z(G) \neq (e) $or u can say that $ order(Z(G)) > 1$

Proof : We know that class equation of $G$ is

$|G| = \sum_{N(a) = |G|} \frac{|G|}{|N(a)|} + \sum_{N(a) \neq |G|} \frac{|G|}{|N(a)|}= |Z(G)| + \sum_{a \notin Z(G)} \frac{|G|}{|N(a)|} $

by Lagrange's Theorem ,$|N(a)|$ divide $|G|$ for each $a \notin Z$, Now take $|N(a) |=p^k$ where $0 < k <n$ , we will get $\frac{|G|}{|N(a)|}= p^{n-k}$

$p$ divide $\frac{|G|}{|N(a)|}$ whenever $a \notin Z(G)$ that implies $p$ divide $\sum_{a \notin Z(G)} \frac{|G|}{|N(a)|} .$

We also know that $|G|= p^n$ so $p$ will divide $|G| -\sum_{a \notin Z(G)} \frac{|G|}{|N(a)|} $---(from class equation $G$)

So $p$ will divide $|Z(G)|$

from the above modulo $p$, we have that $|Z(G)|\equiv 0\pmod{p}$, when $|Z(G)|>1$.

So Now theorem is proof, and $|Z(G)|= 1$ is rejected.

Step $ 2:$

Let $|Z(G)|= p^2$, then $|G|= |Z(G)| $ implies $G=Z(G)$ since $Z(G)$ is subgroup of $ G$.This show the behaviour of abelian Group because for any $a \in G $ implies $ a \in Z(G)$ that will hit $ax= xa$ for all $x, a \in G$ so ,$|G| = p^2$ is abelian.

So $|Z(G)|= p^2$ is accepted.

Step $3$:

Let $|Z(G)| = p $, we know that $Z(G)$ is a proper subgroup of $G$ so that there exist some $a \in G $ such that $a \notin Z(G)$.

We also know that $N(a) = \{ x \in G : xa = ax \}$ is a subgroup of $G$ . we also know that $Z(G) \subset N(a)$ because $Z(G) = \{x \in G : gx = xg \textrm{ } \forall g \in G \}$. Hence, if $x \in Z(G)$ then $xa = ax$ i.e. $x \in N(a)$.

Here we already said in first sentence that $a \notin Z(G)$ , so $a \in N(a)$.

By lagrange theorem , we have $|N(a)|$ divide $|G|$ .Its show that $|N(a)|$ must be greater than $p$ because $|G|= p^2$.

Consequently $|N(a)| = p^2$ implies $|N(a)|=|G|$ so $N(a)= G$ because $ N(a)$ is a subgroup of $G$. Now $x \in N(a)$ for all $x \in G$ its implies $ax = xa$ for all $ x\in G $.

Now from Step $:2$, it show that $a \in Z(G)$ which is contradiction because $a \notin Z(G) $ (read the first sentence again).

So $|Z(G)|= p$ is rejected.

Any suggestion for improving this proof is sufficient.

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jasmine
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    Looks fine but can be tidied up a bit. Step 2 feels out of order since it is your conclusion, so it’s better left to the end. In Step 3 you never explicitly state that $N(a) \ne Z(G)$, even though you justify this by pointing out $a$ belongs to one and not the other. This is at the heart of your argument so it deserves to be made clear. Minor problem in Step 1: the class equation sums over conjugacy classes but your summation index suggests summing over all elements outside of the center. – Erick Wong Oct 21 '19 at 15:23
  • @ErickWong thanks u for pointing out mistake – jasmine Oct 21 '19 at 15:27
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    In general, $N(S)$ is the normalizer of the subgroup generated by $S$. You should use $C(a)$, the centralizer of $a$, for the subgroup you are dealing with. – Arturo Magidin Oct 21 '19 at 16:41
  • In step 3, it is an easy exercise to show that if $G/Z(G)$ is cyclic, then $G$ is Abelian. This can shorten this step. Note that if $a$ is a generator for this factor group, then each $g \in G$ can be written as $$g=a^nh$$ for some $n$ and $h \in Z(G)$, and $g$'s of this type commute. – N. S. Oct 22 '19 at 20:18
  • @N.S. i thinks according to ur statement , then $|Z(G)|= p $ will also accepted and that will contradicts my step 3 which i have said that $ |Z(G)|=p$ is rejected – jasmine Oct 22 '19 at 20:21
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    @jasmine If $|Z(G)|=p$ then $|G/Z(G)|=p$ so this is cyclic. Then, by the above $G$ is abelian. But this means $G=Z(G)$ and hence $|Z(G)|=p^2$. In other words you prove that if $|Z(G)|=p$ then $|Z(G)|=p^2$, meaning it is not acceptable. – N. S. Oct 22 '19 at 21:26
  • @N.S. thanks u now got ur logics .. – jasmine Oct 22 '19 at 22:01

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