If G is a p-group which is finitely generated with order p^n then what is the upper bound of |Aut(G)|.
2 Answers
There is a theorem by P. Hall which states the following. Let $G$ have a minimal generating set consisting of $d$ elements. Then the order of $\mathrm{Aut}(G)$ divides $p^{d(n-d)}|\mathrm{GL}(d,p)|$. This follows from Burnside's basis theorem for $p$-groups.
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$\begingroup$ For which groups |GL(d,p)| is 1 or more precisely for which groups order of Aut(G) is relatively prime with |GL(d,p)|. $\endgroup$– TusharApr 10, 2015 at 3:17
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$\begingroup$ $|\mathrm{GL}(d,p)|=\prod_{i=0}^{d-1} (p^d-p^i)$. $\endgroup$ Apr 10, 2015 at 5:16
An upper bound is $|{\rm GL}(n,p)|$, and if $G$ is not elementary Abelian, the bound is better. This is well-known, but I am not sure of a reference for the first proof in the literature: here is the argument: Let $\{x_{1},x_{2},\ldots, x_{t}\}$ be a minimal generating set for $G$, and let $\phi: G \to G$ be an automorphism. Then $\phi$ is clearly specified by $\{\phi(x_{1}),\ldots, \phi(x_{t}) \}$, and the latter set is also a minimal generating set. Let $X_{1} = \{1_{G} \}$ and $X_{i} = \langle \phi(x_{1}),\ldots,\phi(x_{i-1}) \rangle$ for $i > 1.$ Then $\phi(x_{i}) \in G \backslash X_{i}$ for $1 \leq i \leq t$, and $|X_{i}| \geq p^{i-1}$ for each $i$. Hence $|{\rm Aut}(G)| \leq \prod_{i=1}^{t} (p^{n}-p^{i-1})$, while we certainly have $t \leq n.$
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$\begingroup$ As per Oliver solution the order of Aut(G) divides p^d(n−d)|GL(d,p)|. But For which groups |GL(d,p)| is 1 or more precisely for which groups order of Aut(G) is relatively prime with |GL(d,p)|. $\endgroup$– TusharApr 10, 2015 at 4:36
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1$\begingroup$ I think perhaps you mean to ask when ${\rm Aut}(G)$ is a $p$-group, for it is never the case when $G$ is a finite $p$-group of order greater than $2$ that ${\rm Aut}(G)$ has order coprime to $p(p-1).$ $\endgroup$ Apr 10, 2015 at 5:27