Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
Is this homework? –  Doug Spoonwood Jan 6 '13 at 15:04
    
Show that $\Phi(G) < Z(G)$. –  user641 Jan 6 '13 at 15:21
2  
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
2  
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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