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

  • $\begingroup$ 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 ;) $\endgroup$ – Martin Brandenburg Apr 11 '13 at 18:38
  • $\begingroup$ I'm impressed! I don't suppose many of us were knowledgeable enough to make that mistake! $\endgroup$ – Derek Holt Apr 12 '13 at 8:22

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.)

  • $\begingroup$ 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. $\endgroup$ – BHZ Apr 11 '13 at 9:04
  • $\begingroup$ 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. $\endgroup$ – BHZ Apr 11 '13 at 9:27
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Apr 11 '13 at 11:09
  • $\begingroup$ I am really thankful for your help and your very useful comments. $\endgroup$ – BHZ Apr 11 '13 at 11:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.