Let $G$ be a finite $2$-group such that $\mid Inn(G)\mid=4$ and $\Phi(G)\subsetneq Z(G)$ where $\Phi(G)$ is frattini subgroup. Then prove that there exists an $\alpha\in Aut(G)$ such that $\alpha(g)\neq g$ for some $g\in Z(G)$.

Thank you

  • 1
    $\begingroup$ Could you provide a bit more context? That would help give an idea of how advanced ideas might be needed to do this. $\endgroup$ – Tobias Kildetoft Feb 4 '14 at 8:18
  • $\begingroup$ @Babgen Not only can we assume it. It follows from the assumptions. $\endgroup$ – Tobias Kildetoft Feb 4 '14 at 8:38
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    $\begingroup$ So $Z(G)\Phi(G)/\Phi(G)$ is a nontrivial direct factor of the elementary abelian group $G/\Phi(G)$. Then you can define $\alpha$ to induce the identity on $G/\Phi(G)$ and on $\Phi(G)$ with $\alpha(g) = gh$, for some $g \in Z(G) \setminus \Phi(G)$, where $h$ is an element of order $2$ in $\Phi(G)$. $\endgroup$ – Derek Holt Feb 4 '14 at 8:58

Let me give some explaination for the Derek's answer: Write $F=\Phi(G)$. Consider $\bar{G}=G/F$. Then $\bar{Z}=<\bar{x}_1, \cdots, \bar{x}_k>$ for some $x_1, \cdots, x_k\in Z(G) - F$. Then $Z=Z(G)=<x_1,\cdots, x_k, F>$. Now define a homomorphism induced by $\alpha_1: x_1\rightarrow x_1h, \cdots, x_k\rightarrow x_k, f\rightarrow f$ for $f\in F$, where $h\in F$ such that $|h|=2$. Cleary $\alpha_1$ is an isomorphism for $<\alpha_1({x_1}), \cdots, \alpha_1(x_k), \alpha_1(F)>=Z$. Now let $\bar{G}=<\bar{x}_1, \cdots, \bar{x}_k> \times <\bar{y}_1, \cdots>$. Extending $\alpha_1$ to $\alpha$ by $y_1\rightarrow y_1, \cdots $, we can get the isomorphism descripted by Derek.

  • $\begingroup$ In fact we are not told that $|Z/F|=4$, only that $|Z/F| > 1$, so $Z = Z(G) = \langle x_1,x_2,\ldots, x_n, F \rangle$ for some $n \ge 1$. But apart from that technicality, the argument is fine! $\endgroup$ – Derek Holt Feb 4 '14 at 11:06
  • $\begingroup$ Yes, you are right. I will correct it. Thanks. $\endgroup$ – Wei Zhou Feb 4 '14 at 11:29
  • $\begingroup$ Derek Holt and Wei Zhou: Thank you for your explainations. $\endgroup$ – maryam Feb 4 '14 at 14:43

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