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



The automorphisms of a 2-group of nilpotency class 2
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.)


