Is there a finite non-abelian $2$-group $G$ without non-trivial elementary abelian direct factor and of order $2^9$ satisfying the following condition: $$Z(G) \cap Z(\Phi(G))= \langle \prod_{i=1}^{2^d} a^{x_i} \;|\; a\in Z(\Phi(G)) \rangle,$$ where $\{x_1,\dots,x_{2^d} \}$ is a right transversal of $\Phi(G)$ in $G$?

This means that $Z(\Phi(G))$ is a cohomologically trivial $G/\Phi(G)$-module.

Here $\Phi(G)$ is the Frattini subgroup of $G$ and $Z(H)$ denotes the center of a group $H$.

We know that such groups of order $2^8$ exist and exactly 10 groups of order $2^8$ with the above property exist.


The answer is no. I did a computer search through the $10494213$ groups of order $512$ (which you could have done yourself) and found that the only group that satisfies the condition is the elementary abelian group.

I did successfully confirm that there are $10$ examples of order $2^8$, in addition to the elementary abelian group.

  • $\begingroup$ Many thanks. I could not carry all groups of order 512, but my colleague Maria Guedri was running a program under GAP to do it. She has ruled out about 1/3 of such groups or more. $\endgroup$ – Alireza Abdollahi Apr 14 '16 at 12:13

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.