Timeline for Testing whether two elements of $\text{SL}(2, \mathbb{F}_{2^n})$ generate the entire group
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 26, 2017 at 18:35 | vote | accept | DoomMuffins | ||
| Oct 25, 2017 at 21:58 | answer | added | Luc Guyot | timeline score: 13 | |
| Oct 25, 2017 at 17:50 | answer | added | Derek Holt | timeline score: 15 | |
| Oct 25, 2017 at 17:22 | comment | added | Derek Holt | It is worth mentioning that there is a fast Monte Carlo algorithm for deciding whenther an input subgroup $G$ of ${\rm GL}(n,q)$ contains ${\rm SL}(n,q)$. This takes a maximum error probability as input. A positive answer is guaranteed to be correct, but if $G$ does contain ${\rm SL}(n,q)$, then there is a probability of at most $\epsilon$ that it will incorrrectly answer no. | |
| Oct 25, 2017 at 16:43 | comment | added | Derek Holt | If it not equal to ${\rm SL}_2(2^n)$ then it is reducible, imprimitiive, semilinear, or defined over a proper subfield modulo scalars. Those are all easily tested for. In the imprimitive and semilinear cases, there is an abelian subgroup of index $2$. | |
| Oct 25, 2017 at 16:43 | comment | added | YCor | We also have to exclude the case of $A$, $B$ preserving a 1+1 decomposition. This should be easy, since then one of the following subgroups of index $\le 2$ is abelian: $\langle A,B^2,BAB\rangle$, $\langle A^2,B,ABA\rangle$, $\langle A^2,B^2AB,\rangle$. If I understand correctly the classification of maximal subgroups of $SL_2(q)$, this should be enough. | |
| Oct 25, 2017 at 16:34 | comment | added | YCor | I'm writing as a comment as I'm not sure. First, you can check quickly whether they generate matrices as an algebra: namely this holds iff at least one of the families $(I,A,B,AB)$, $(I,A,B,BA)$, $(I,A,B,A^2)$, $(I,A,B,B^2)$ is linearly free. Second, you need to check that it's not contained in $SL_2$ of a smaller field. If this holds, then the traces $Tr(A)$, $Tr(B)$, $Tr(AB)$, $Tr(AB^{-1})$. I think that the converse is known (assuming absolute irreducibility) but I'm not 100% sure, namely that if these 4 traces belong to a subfield then so do all traces. Maybe somebody can confirm... | |
| Oct 25, 2017 at 16:19 | review | First posts | |||
| Oct 25, 2017 at 16:55 | |||||
| Oct 25, 2017 at 16:19 | history | asked | DoomMuffins | CC BY-SA 3.0 |