# generao

 23 Reputation 139 views

## Registered User

 Name generao Member for 4 months Seen May 14 at 12:48 Website Location Age 27
 Feb14 comment Metacyclic groups in $AGL(4,3)$@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? Jan29 comment Metacyclic groups in $AGL(4,3)$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? Jan29 comment Metacyclic groups in $AGL(4,3)$Thank you for both of you answering my question. Both answers are quite close to my desire. I would like to tick Peter's answer however Stefan's computation is also nice. Jan29 awarded ● Scholar Jan27 awarded ● Student Jan27 awarded ● Editor Jan27 revised Metacyclic groups in $AGL(4,3)$improve expression Jan26 asked Metacyclic groups in $AGL(4,3)$ Jan14 comment Is $SL(2,5)$ irreducible?You are right. I should have it isomorphic to a $p$-subgroup of $\mathcal{Z}_p{:}\mathcal{Z}_{q-1}$. Thanks a lot Jan13 comment Is $SL(2,5)$ irreducible?I agree that SL$(2,5)$ can be embedded into a unipotent group and hence soluble, which leads to a contradiction. But I suspect it would be isomorphic to $\mathcal{Z}_p:\mathcal{Z}_{q-1}$ but not just a $p$-group. Jan13 answered Proving a determinant = 0 Jan13 asked Is $SL(2,5)$ irreducible?