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In the following post by DavidLHarden : See Here

He quoted the following claim: "There is a theorem that says that if $p$ is a prime and $|G|=p^n $ , then $|AutG| $ divides $ \Pi_{k=0}^{n-1} (p^{n}-p^{k}) $ " .

I can't find any reference for this theorem ,

Does someone know of any reference for this fact?

Thanks in advance

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I think the theorem in this form may be due to Peter M. Neumann- so try Mathscinet if you have access- assuming the theorem is stated correctly. It is certainly true that ${\rm Aut}(G)$ has a normal $p$-subgroup whose index is bounded by the number you quote (this is a Theorem of W.Burnside). –  Geoff Robinson Nov 23 '12 at 8:28
    
I don't have Khukhro's book "$p$-Automorphisms of Finite $p$-Groups" here, but it would be a good place to try. –  m_t Nov 23 '12 at 10:38
    
Geoff: I would try Mathscinet on Monday or something... I am not sure if this is due to Neumann or not... Thanks anyway! @mt: I couldn't find this theorem in the book you just mentioned... Thanks for the suggestion! –  Jason Mraz Nov 23 '12 at 19:16
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2 Answers

I don't have a reference, but here's the next best thing: a proof.

First, let's fix some notation. Let $P$ be a p-group, $G$ it's group of automorphisms, and $\Phi(P) = P^p[P,P]$ it's Frattini subgroup. Define inductively $\Phi^k(P)$ as $\Phi(\Phi^{k-1}(P)).$

The subgroups $\Phi^k(P)$ form a decreasing chain of characteristic subgroups which exhaust $P.$

Let $$G_k := \ker(G \rightarrow Aut(P/\Phi^k(P)))$$ and $$G_k' := \ker(G \rightarrow Aut(\Phi^k(P)/\Phi^{k+1}(P))).$$

Then the subgroups $G_k$ form an decreasing chain of normal subgroups of $G$ which exhaust $G.$

Let $d_k = \dim_{\mathbb{F}_p}(\Phi^k(P)/\Phi^{k+1}(P)).$ The group $P$ can be generated by $d_0$ elements. Choose a generating set $g_1 ... g_{d_0}$ and consider the map from $G_k \cap G_k'/G_{k+1}$ to $(\Phi^k(P)/\Phi^{k+1}(P))^{d_0}$

given by

$$\sigma \mapsto (\sigma(g_i)g_i^{-1})_{i=1}^{d_0}.$$

This map is an injective group homomorphism.

On the other hand $G_k/(G_{k} \cap G_{k}')$ injects into $Aut(\Phi^k(P)/\Phi^{k+1}(P)) \cong GL_{d_i}( \mathbb{F}_p).$

Note that if $p^n = |P|$ and $r = max\{d_i : d_i \neq 0\},$ then $\displaystyle\sum_{i=0}^r d_i = n.$ It follows that the order $$|G| = \displaystyle\prod_{k=0}^{r}|G_k/G_{k+1}| = \displaystyle\prod_{k=0}^{r} |(G_k \cap G_k')/G_{k+1}||(G_k \cap G_k')/G_{k+1}|$$ divides

$$\displaystyle\prod_{s=0}^{d_0-1} (p^{d_0} - p^s)\displaystyle\prod_{k=1}^{r} p^{d_kd_0}\displaystyle\prod_{s=0}^{d_k-1} (p^{d_k} - p^s)$$

which divides

$$\displaystyle\prod_{k=0}^{n-1} (p^n - p^k).$$

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Note that it is completely pointless to put \displaystyle (on outer bracing level) into displayed equations (they are already set in display style), and even more pointless to put more than one of these in one formula (the first declaration remains in effect). Judging from the usage, you may be confusing \displaystyle with \limits. –  Emil Jeřábek Nov 23 '12 at 13:05
    
Thanks @JSpecter! I'll go over your proof later today... The problem is that a friend of mine needs this fact for a paper he's writing, and I don't think a proof of a known fact is something that he would want to put in his paper (due to copyrighting rights) Thanks anyway ! –  Jason Mraz Nov 23 '12 at 19:30
    
@Mraz, it isn't anything to do with copyright. However, you are right that it's best to cite a published paper if it exists. –  Brendan McKay Nov 24 '12 at 0:27
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@McKay, when $p>2$, the automorphism group of a cyclic group of order $p^3$ contains elements of order $p^2$, but $GL(3,p)$ does not. –  Rene Schoof Nov 24 '12 at 14:56
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Why is a reference considered preferable to a proof? A proof is what you want to find in the reference. –  DavidLHarden Nov 25 '12 at 11:15
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As @DavidLHarden explains in the link that you gave, this theorem is proved by attending to the $p$-part and $p'$-part separately.

For the $p'$-part the result follows from the following theorem of Burnside:

Let $\psi$ be a $p'$-automorphism of the $p$-group $P$ which induces the identity on $P/\Phi(P)$. Then $\psi$ is the identity automorphism of $P$.

This is the result that Geoff refers to in his comment above. It is discussed and proved in Section 5 of Gorenstein's Finite Groups, specifically Theorem 1.4 of that section.

I do not know of a reference for the $p$-part of the proof. You should certainly look at the paper by Neumann that Geoff mentions, however if I understand that proof correctly it only proves your bound for $|Out P|$, rather than $|Aut P|$. On the other hand Neumann is considering a much more general setting than just $p$-groups.

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@Nick Gill: Thanks ! Sorry for my ignorance, but what is the $p'$ part of the result? (I can't understand what is $p'$ in your response) . I also have trouble getting into the link you just gave... It gives me an error... can you please fix it? Thanks ! –  Jason Mraz Nov 23 '12 at 19:26
    
@Jason, I've fixed the link. Sorry about that. You can split this bound into two parts: prove that the biggest power of p dividing Aut G divides the given product, and then prove that the biggest integer coprime to p dividing Aut G divides the given product. When I write the p'-part I mean the latter. –  Nick Gill Nov 23 '12 at 21:31
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