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Let $G$ be a finite 2-group such that $Z(G)$ is elementary abelian 2-group ($\mid Z(G)\mid\geq 4$) and $Inn(G)$ is of order 4. Then prove that there exists an $\alpha\in Aut(G)$ such that $\alpha(g)\neq g$ for some $g\in Z(G)$. Thank you

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Is this homework? –  Doug Spoonwood Jan 6 '13 at 15:04
Show that $\Phi(G) < Z(G)$. –  user641 Jan 6 '13 at 15:21
Why are you asking people to prove a false statement? –  Derek Holt Jan 6 '13 at 17:24
But you never answer any questions. Why do you think that there is no such group? –  Derek Holt Jan 6 '13 at 20:03
I am puzzled about this because it unlikely to be a homework problem, because it is wrong. But I am not offering any more help until she provides some more background. –  Derek Holt Jan 6 '13 at 22:06

1 Answer 1

up vote 3 down vote accepted

Let $G = \mathbb{Z}_4 \rtimes \mathbb{Z}_4 = \langle a \rangle \rtimes \langle b \rangle$ where $b$ acts on $\langle a \rangle$ by inversion. We can write $G$ with the polycyclic presentation $$G=\langle a,b,c,d | a^2=c,b^2=d,a^b=ac\rangle,$$ from which it is clear that $Z(G)=\langle c,d\rangle\cong \mathbb{Z}_2\times \mathbb{Z}_2$.

$\text{Aut}(G)$ is isomorphic to the subgroup of the upper triangular unipotent matrix group $U(4,2)$ consisting of matrices of the form $$\left(\begin{array}{cccc}1&\star&\star&\star\\0&1&\star&\star\\0&0&1&0\\0&0&0&1\end{array}\right)$$ where the $\star$'s are $0$'s or $1$'s. You can see the implied isomorphism from the polycyclic presentation; $\text{Aut}(G)$ is generated by the automorphisms $$a\mapsto ab,\hspace{10pt} a\mapsto ac,\hspace{10pt} a\mapsto ad, \hspace{10pt}b\mapsto bc, \hspace{6pt}\text{and}\hspace{6pt}b\mapsto bd.$$But each one of these fixes $c$ and $d$, and thus fixes the center. So your claim is false.

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I thought you said you weren't answering any more of Maryam's questions. –  Derek Holt Jan 7 '13 at 9:06
@DerekHolt She raised her acceptance rate, which I figure earned her one. If she continues not answering our questions or providing background, I will stop again. –  Alexander Gruber Jan 7 '13 at 14:54
@Alexander Gruber: Thank you for eample –  maryam Jan 7 '13 at 15:09
@maryam: OK, thanks for the explanation! –  Derek Holt Jan 7 '13 at 15:27

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