Let $G$ be a finite abelian $p$-group. What is known about the minimal number of generators of a $p$-sylow of $Aut(G)$? is it bounded in terms of $d(G)$ the minimal number of generators of $G$ (and perhaps $p$)?
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1$\begingroup$ Are there examples showing that the minimal number of generators of $Aut(G)$ can be $\ge 5$? $\endgroup$– user6976Commented Aug 6, 2013 at 3:16
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$\begingroup$ I'm sure that $Aut(G)$ can have arbitrary rank (that is $Aut(G)$ may contain subgroups with arbitrary minimal number of generators). I will be pleased if you can explain why exactely 5? $\endgroup$– Yassine GuerboussaCommented Aug 6, 2013 at 20:13
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$\begingroup$ I did not notice that you are asking about the Sylow $p$ subgroup, not about the group $Aut(G)$ itself. My comment is about the whole $Aut(G)$. $\endgroup$– user6976Commented Aug 6, 2013 at 20:28
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$\begingroup$ Yes, I see. I find the question in your comment interesting. $\endgroup$– Yassine GuerboussaCommented Aug 6, 2013 at 20:35
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1 Answer
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For the special case that $G=(\mathbb{Z}/p\mathbb{Z})^n$, this is true. Then we have $d(G)=n$ and $Aut (G)=GL(n,p)$. In the article of A. Patterson, "The minimal number of generators for $p$-subgroups of $GL(n, p)$" of $1974$ it is shown that any $p$-subgroup of $\text{GL}(n,p)$, where $p$ is an odd prime, can be generated by ${\textstyle\frac 1{4}}n^2$ elements. So the bound is $\frac{1}{4}d(G)^2$ in this case.
answered Aug 6, 2013 at 14:11
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$\begingroup$ Without resorting to a reference, the exact number of minimal generators in this case clearly is $d(G)-1$. $\endgroup$– GuntramCommented Aug 6, 2013 at 17:02
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$\begingroup$ @Guntram: no, not any $p$-subgroup of $GL(n,p)=Aut(G)$ can be generated by exactly $n-1=d(G)-1$ generators. $\endgroup$ Commented Aug 6, 2013 at 18:08
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$\begingroup$ Thank you so much dear Professors. Now I think the bound on the rank $5/4 d(G)^2$ works for arbitrary abelian $p$-groups. $\endgroup$ Commented Aug 6, 2013 at 20:21