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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.2012-09-03
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    Possible duplicate: http://math.stackexchange.com/questions/3849/let-p-be-a-prime-then-does-p-alpha-mid-g-longrightarrow-p-mid-autg2012-09-03

2 Answers 2

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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?2012-09-03
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    I don't get what you wrote.2012-09-03
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    The OP asked for hints and you gave them the complete solution.2012-09-03
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    Fine. Let me modify it.2012-09-03
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    Good, thanks and +1.2012-09-03
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    It doesn't work quite this way. For example, if $G$ is abelian, then the image of $f$ consists of the trivial automorphism only.2012-09-03
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    oh, the deleted part filled the details. You may take a look.2012-09-03
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    And as another point, where did you use the assumption that $n\ge2$? Without it the claim as false. For example the cyclic group of order $p$ has an automorphism group of order $p-1$.2012-09-03
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    Ok! That looks better, but I think that those were still hints, and left enough work for the OP to do :-)2012-09-03
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    The proof will break if $n=1$, since $\mathbb{Z}_{p}^{*}$ has order $p-1$. I did not think seriously on that.2012-09-03
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    @JyrkiLahtonen: Indeed this approach does not show anything for the case $G$ is Abelian. I'm not entirely convinced it reduces the problem to the Abelian case either; in case a (the) Sylow-$p$ subgroup is in the center, I think I can see that it is a direct factor, but it is not _obvious_.2012-09-03
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    @MarcvanLeeuwen: You can take a look on the deleted parts.2012-09-03
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    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.2012-09-03
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    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.2012-09-03
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    @user32240: I did2012-09-03
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    @user32240: Really Thanks for every step you did here. Honestly, I wanted to put myself to a challenge and that is why I really wanted some hints. Anyway, I am reading your neat answer right now and thinking about your approach. Thanks :-)2012-09-03
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    I removed a false claim (p=7 for instance) and replaced it with an easy proof.2012-09-03
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    @JackSchmidt: thanks!2012-09-04
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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.2012-09-03
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    Thanks for your attepmt. Thanks. I am reading yours and above answer right now. :)2012-09-03