TITLE: Group of order $p^2$ is abelian.
QUESTION [7 upvotes]: Yes, I know that there are tons of solutions of this up here, but I, essentially, wanted to try it a different way and ah, well.
Let $|G| = p^2$ for some prime $p$.
Consider $x \in G$.
So, $|x| = p, $ or $ p^2$.
If it is the latter, then, we are done.
So, if $|x|= p$, then, let, $H = [h|  h=x^i, i \in Z] \Rightarrow |H|= p$.
As $p$ is the smallest prime dividing $|G|$, $H$ is a normal subgroup.
Now, consider some $y \in G$, but not in $H \Rightarrow y^p= e$.
Then, its easy to show that $G = \langle x,y \rangle$.
Now, consider a homomorphism $f:G \rightarrow \operatorname{Aut}(H), g \to c_g $, wherein $c_g$ represents conjugacy by $g$. So, for example, $c_g(x) = gx g^{-1}$.
Then, it is clear that, $H \subset \operatorname{Ker}(f) $, as it is cyclic and therefore abelian.
So, we need only look at $G/H$.
Basically, I am asking for a hint as to why $c_y(x) = yxy^{-1} = x = c_e(x)$, where $x$ and $y$ are the generators of $H$ and $G/H$, respectively?
Because, if I establish it for the generators, then it follows that it applies for for all elements, i.e, that $\operatorname{Im}(f)$ is trivial and therefore that $G$ is abelian.
Thank You!

REPLY [3 votes]: Hint: The coset of $y$ is of order $p$ in $G/H$. This implies that $c_y$ has order that is a factor of $p$ in $Aut(H)$. But $Aut(C_p)$ has order $p-1$.

REPLY [2 votes]: $H$ is cyclic of order $p$, and its automorphism group is cyclic of order $p-1$, so contains no elements of order $p$.
$y$ has order $p$, so the action of $y$ on $H$ has period which is a factor of $p$. So we can have $1$ or $p$. $p$ is not available, so the action is trivial.