# $p$ is prime and $p^2\large\mid\normalsize|G|$

Hints needed:

Let $p$ be a prime and $G$ a finite group such that $p^2\large\mid\normalsize|G|$ then $p\large\mid\normalsize|\text{Aut}(G)|$.

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So we can certainly start by noticing that if $G$ contains a non-central element of order $p$, then we are done, as conjugation by such an element is an automorphism of order $p$. So maybe one should assume all elements of order $p$ are central and then maybe look at the quotient by the subgroup generated by these. – Tobias Kildetoft Sep 3 '12 at 8:52
Possible duplicate: math.stackexchange.com/questions/3849/… – user641 Sep 3 '12 at 19:47

We know there is a sylow $p$-group with order $p^{n}$, $n\ge 2$. Name this group to be $H$. Consider the inner automorphism of $G$: $$f: H\rightarrow Aut(G): h\rightarrow hgh^{-1}$$

The rest details are "easy" to fill out(see the comments or deleted parts for help). Note that $n\ge 2$ is necessarily since otherwise $\mathbb{Z}_{p}$ has an automorphism group of order $p-1$, so has no automorphisms of order $p$.

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What kind of hint is that? – Gigili Sep 3 '12 at 9:19
I don't get what you wrote. – Bombyx mori Sep 3 '12 at 9:23
The OP asked for hints and you gave them the complete solution. – Gigili Sep 3 '12 at 9:24
Given that the Sylow $p$-subgroup is in the centre, you need the Schur-Zassenhaus Theorem to show that it has a complement and is therefore a direct factor. – Derek Holt Sep 3 '12 at 11:33
To a great extent I agree with Marc. The deleted parts give more hints. But in the end one needs to deal with the cases $C_{p^2}$ and $C_p\times C_p$ "by hand". In addition to the problem of splitting a $p$-Sylow as a direct factor. – Jyrki Lahtonen Sep 3 '12 at 11:33

I don't think you can prove this without distinguishing some cases.

First suppose $G$ is Abelian; then by the structure theorem it contains either a cyclic factor of order $p^k$ with $k\geq2$ or a factor $(\mathbf Z/p\mathbf Z)^2$ (this is where $p^2\mid |G|$ is used), and you can compute the order of their automorphism groups, which inject into $\mathop{\mathrm{Aut}} G$.

If $G$ is not Abelian, and has a non-central element of order a power of $p$, then conjugation is your friend.

You can show that every element of $G$ is a product (within the cyclic group it generates) of an element of order a power of $p$ and an element of order indivisible by $p$. In the case that remains, you may (I think) use this to prove that you've got a direct product decomposition of $G$, and apply the first case.

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This is a cleaner proof, though it boils down to the same thing it is much easier to understand. – Bombyx mori Sep 3 '12 at 11:39
Thanks for your attepmt. Thanks. I am reading yours and above answer right now. :) – Babak S. Sep 3 '12 at 15:18