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Fix a prime $p$, and let $M$ be the unique nonabelian group of order $p^3$ and exponent $p$. Let us denote by $E_n$ the elementary abelian group of rank $n$.

Is it true that $\operatorname{Aut}(M \times E_n)$ and $\operatorname{GL}(n+3,p)$ have isomorphic Sylow $p$-subgroups?

For $n=0$ they are isomorphic; and If I'm not mistaken they have the same order for all $n$.

Thanks in Advance.

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  • $\begingroup$ Perhaps I'm not understanding the question, but it seems to me that $M\times E_n\cong E_{n+3}$. Viewing $E_{n}$ as a vector space over $\mathbb{F}_p$, we then have the isomorphism of groups $\mathrm{Aut}(E_n)\cong\mathrm{GL}(n,p)$ for all $n$. $\endgroup$
    – Jared
    Commented Apr 7, 2014 at 17:39
  • $\begingroup$ @Jared: thanks for your comment; You may notice that $M$ is not abelian, so that $M \times E_n$ and $E_{n+3}$ cannot be isomorphic. $\endgroup$
    – Yassine Guerboussa
    Commented Apr 7, 2014 at 19:12
  • $\begingroup$ Ok, It is my mistake that I didn't noticed that $M$ is not abelian. $\endgroup$
    – Yassine Guerboussa
    Commented Apr 7, 2014 at 19:14
  • $\begingroup$ You have probably already checked this, but Magma says that for p=3 and n=1, the groups are isomorphic. $\endgroup$
    – Marty Isaacs
    Commented Apr 7, 2014 at 21:37

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