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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!

Cortizol
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AlpArslan
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  • Showing $f$ has trivial image means that $G$ centralizes $H$, in other words $H \subseteq Z(G)$. From here, there is another (nontrivial IMO) step to deduce $G$ itself must be abelian. – Cihan Jun 29 '13 at 09:43

3 Answers3

3

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 $p-1$.

Jyrki Lahtonen
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  • Thanks, but, well, how exactly is $|Aut(H)|= p-1$? – AlpArslan Jun 29 '13 at 09:53
  • @Mwarsi: As $H$ is cyclic, generated by say $g$, an automorphism $f$ will be completely determined after you know $f(g)$. There are $p-1$ choices for that image, as the neutral element is already taken. Therefore there are at most $p-1$ automorphisms, which is already good enough for your purposes. It is fairly easy to see that we do get exactly $p-1$ automorphisms, because we can use any non-identity elements of $H$ as $f(g)$. – Jyrki Lahtonen Jun 29 '13 at 09:57
2

$H$ 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.

Mark Bennet
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  • Forgive me for sounding obtuse, but how can I show that the automorphism group is of order p-1? – AlpArslan Jun 29 '13 at 09:53
  • @MWarsi Well the available images for $x$ are $x, x^2, x^3 \dots x^{p-1}$ and each defines an automorphism (check). $x^p=1$ does not give an automorphism. – Mark Bennet Jun 29 '13 at 09:55
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More simple: $\langle y \rangle$, similarly to $\langle x \rangle$, is normal in $G$. Since $\langle x \rangle\cap\langle y \rangle =1$, $x$ and $y$ commute.

Boris Novikov
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