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Timeline for Metacyclic groups in $AGL(4,3)$

Current License: CC BY-SA 3.0

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Feb 14, 2013 at 23:34 comment added Peter Mueller If $c$ is a generator for $C$, and $u$ normalizes $C$, then $u$ induces an automorphism of order $1$ or $3$ on $C$, hence $cu=uc^m$ with $m=1$, $4$, or $7$. Given $c$ in Jordan normal form, this yields a system of linear equations of $u$, and one verifies that $u^3\in C$. I think this can be done by hand, but I believe that I used Sage for the computation.
Feb 14, 2013 at 7:21 comment added Easy @Peter: There was something I didn't reckon at the first time. At your last step, you said one "computes" the exponent is 3. Did you do it by machine or do it manually? I verified that using GAP, but when I tried calculating manually I found that is a huge calculation. Or is there any trick I didn't know?
Jan 29, 2013 at 14:47 comment added Peter Mueller @generao: Do whatever you want, the argument is so simple that I'm not proud of it ... However, as you raised the question, it would probably be scientifically more correct to mention the two answers.
Jan 29, 2013 at 10:17 comment added Easy Peter, I think there is a better way to show that $G=\mathbb{Z}_9.\mathbb{Z}_9$. Since $G$ is transitive on $3^4$, then so is its Sylow $3$-group $P$, and $81\mid|P|$ by orbit stabiliser theorem. On the other hand, the highest order of $3$-elements in $AGL(4,3)$ is 9. So the only possible structure for $P$ is $\mathbb{Z}_9.\mathbb{Z}_9$. Also I am gonna to partially quote ur argument in my paper. Do you need ur name on?
Jan 29, 2013 at 10:03 vote accept Easy
Jan 29, 2013 at 10:03 vote accept Easy
Jan 29, 2013 at 10:03
Jan 29, 2013 at 10:03 vote accept Easy
Jan 29, 2013 at 10:03
Jan 27, 2013 at 22:16 history edited Peter Mueller CC BY-SA 3.0
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Jan 27, 2013 at 21:50 history edited Peter Mueller CC BY-SA 3.0
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Jan 27, 2013 at 12:31 history answered Peter Mueller CC BY-SA 3.0