generao
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Registered User
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Feb 14 |
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? |
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Jan 29 |
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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? |
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Jan 29 |
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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. |
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Jan 29 |
awarded | ● Scholar |
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Jan 27 |
awarded | ● Student |
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Jan 27 |
awarded | ● Editor |
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Jan 27 |
revised |
Metacyclic groups in $AGL(4,3)$ improve expression |
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Jan 26 |
asked | Metacyclic groups in $AGL(4,3)$ |
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Jan 14 |
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 |
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Jan 13 |
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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. |
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Jan 13 |
answered | Proving a determinant = 0 |
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Jan 13 |
asked | Is $SL(2,5)$ irreducible? |
