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Let $p$ be a Merssene prime, i.e. $p=2^a-1$, where $a$ is a prime.

Let $R$ be a 2-group of order $2(p+1)=2^{a+1}$. Also we know that $|Z(R)|=2$ and $R/Z(R)$ is abelian. Can we conclude that $R$ has no automorhism of order $p$?

I know that there is a theorem that says that if $p$ is a prime and $G$ is a $p$ -group with $|G|=p^n$, $|Aut(G)|$ divides $\prod_{k=0}^{n-1} (p^n −p^k)$. But this is not enough for getting this result.

Thanks for your helps

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It took me a while to realize that $2$-group does not refer to $2$-groups in the sense of math.ucr.edu/home/baez/hda5.pdf ;) –  Martin Brandenburg Apr 11 '13 at 18:38
    
I'm impressed! I don't suppose many of us were knowledgeable enough to make that mistake! –  Derek Holt Apr 12 '13 at 8:22

1 Answer 1

up vote 4 down vote accepted

For $a=2$, you can take $R=Q_8$, which does indeed have an automorphism of order 3.

For $a>2$ there are no groups $R$ satisfying your hypothesis. In such a group, the commutator map would be a non-generate alternating (in fact symmetric in this case) bilinear map $G/Z(G) \times G/Z(G) \to Z(G)$, which forces $G/Z(G)$ to be elementary abelian of even dimension over the field of order 2. (So $R$ would be extraspecial, but it is well-knoiwn that such groups have order $2^k$ with $k$ odd.)

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Thank you very much for your kindness and your helps. Could you kindly introduce me some references for these subjects for example why $k$ is odd and why $G/Z(G)$ must be elementary abelian. –  BHZ Apr 11 '13 at 9:04
    
Excuse me, the commutator map is from $G/Z(G)\times G/Z(G)$ to $G/Z(G)$ or $Z(G)$? I am sorry, since I am not fammilar with these subjects. –  BHZ Apr 11 '13 at 9:27
    
In any nilpotent group of class 2, we have $[ab,c]=[a,c][b,c]$ and $[a,b]^{-1} =[b,a]$, so the commutator map is bilinear and alternating, and we get an induced map $G/Z(G) \times G/Z(G) \to Z(G)$. If $|Z(G)|=2$, then $[a^2,b]=1$ for all $a,b$, so $a^2 \in Z(G)$ and $G/Z(G)$ is elementary abelian. So $G$ is extraspecial. –  Derek Holt Apr 11 '13 at 11:09
    
I am really thankful for your help and your very useful comments. –  BHZ Apr 11 '13 at 11:47

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