User generao - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:31:53Z http://mathoverflow.net/feeds/user/30610 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118788/is-sl2-5-irreducible Is $SL(2,5)$ irreducible? generao 2013-01-13T08:38:42Z 2013-02-09T13:06:06Z <p>Maybe this is just a very fundamental problem, but I am not too sure the answer. It is well-known that $SL(2,5)$ is contained in $SL(2,q)$ iff $q$ is odd and $5\mid q(q^2-1)$. My question is whether we can always say $SL(2,5)$ is irreducible on the vector space $GF(q)^2$ ?</p> http://mathoverflow.net/questions/119938/metacyclic-groups-in-agl4-3 Metacyclic groups in $AGL(4,3)$ generao 2013-01-26T14:55:13Z 2013-01-27T22:16:49Z <p>Just wonder if there is any known results on the transitive metacyclic groups of $AGL(4,3)$? </p> <p>Sorry, I should elaborate a bit here. I am working on a graph $\Gamma$ which admits a metacyclic group transitive on its vertices. By some other restricted conditions, I have reduced that $Aut\Gamma$ is contained in $AGL(4,3)$. So it turns out that I need to work on the following problems:</p> <ol> <li>Check whether there exist such metacyclic subgroups in $AGL(4,3)$.</li> <li>If exist, what are they?</li> </ol> <p>One of my attempt was to inspect subgroups of $AGL(4,3)$ and rule out those impossible cases. But it seems tedious and somehow I still have to seek help from machinery search...</p> <p>Alternatively, I am hoping to do it on a more elegant and efficient way. That is, to find the generators of the metacyclic groups. We know that a metacyclic group can be generated by two elements. Hence if we can find all the possible generators, then we can construct the groups and verify if these groups are transitive or not. There are some helpful information we already know: written $AGL(4,3)=\mathbb{Z}_3^4{:}GL(4,3)$, so a transitive metacyclic group $G=\langle(a_1,b_1),(a_2,b_2)|a_i\in\mathbb{Z}_3^4,b_i\in GL(4,3),~\mbox{and}~o(b_1),o(b_2)~\mbox{divisible by}~9\rangle$. Then the next question is which exact generators we should choose?</p> <p>Another thought is that if it is possible to do a pure computer search? as the group is not too big, but I am not too sure how to write up the codes...</p> <p>Any suggestion would be much appreciated.</p> http://mathoverflow.net/questions/118789/proving-a-determinant-0/118791#118791 Answer by generao for Proving a determinant = 0 generao 2013-01-13T09:09:44Z 2013-01-13T09:09:44Z <p>To answer your minor question, it might be referred as a calculation of Laplace expansion of a determinant.</p> http://mathoverflow.net/questions/119938/metacyclic-groups-in-agl4-3/120018#120018 Comment by generao generao 2013-02-14T07:21:01Z 2013-02-14T07:21:01Z @Peter: There was something I didn't reckon at the first time. At your last step, you said one &quot;computes&quot; 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? http://mathoverflow.net/questions/119938/metacyclic-groups-in-agl4-3/120018#120018 Comment by generao generao 2013-01-29T10:17:43Z 2013-01-29T10:17:43Z 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? http://mathoverflow.net/questions/119938/metacyclic-groups-in-agl4-3 Comment by generao generao 2013-01-29T10:13:37Z 2013-01-29T10:13:37Z 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. http://mathoverflow.net/questions/118949/infinitely-differentiable-function/118950#118950 Comment by generao generao 2013-01-15T08:32:07Z 2013-01-15T08:32:07Z which other possibilities you could have if assuming the function is smooth? By Taylor series we have to have $f^{(n)}(0)=0$ for $n&gt;0$. http://mathoverflow.net/questions/118788/is-sl2-5-irreducible Comment by generao generao 2013-01-14T03:17:59Z 2013-01-14T03:17:59Z You are right. I should have it isomorphic to a $p$-subgroup of $\mathcal{Z}_p{:}\mathcal{Z}_{q-1}$. Thanks a lot http://mathoverflow.net/questions/118788/is-sl2-5-irreducible Comment by generao generao 2013-01-13T13:38:20Z 2013-01-13T13:38:20Z 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.