Timeline for Are generalized symmetric groups maximal finite groups (in a certain sense)?
Current License: CC BY-SA 4.0
9 events
when toggle format | what | by | license | comment | |
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May 21, 2022 at 13:31 | vote | accept | Jonas Anderson | ||
May 21, 2022 at 3:42 | comment | added | Jonas Anderson | @PadraigÓCatháin Thanks again! $M(1,4)$ and $M(2,4)$ are subgroups of the 2-qubit Clifford group which is a maximal finite subgroup of $U(4)$. This Clifford group contains $H_2 \otimes H_2$ (and $H_2 \otimes I_2$ and $I_2 \otimes H_2$ for that matter). I'm especially interested in whether the above conjecture is true for $M(1, 8)$, a purely permutation group. | |
May 21, 2022 at 3:34 | comment | added | Jonas Anderson | @PadraigÓCatháin Excellent point! The matrix groups $M(1, 2), M(2, 2)$, and $M(4,2)$ are all subgroups of the 1-qubit Clifford group (which is isomorphic to the octahedral group). The Clifford group is a maximal finite subgroup of $U(2)$ which contains $H_2$. See arxiv.org/abs/math/0001038 for more details. | |
May 20, 2022 at 16:46 | answer | added | Geoff Robinson | timeline score: 5 | |
May 20, 2022 at 14:56 | comment | added | Padraig Ó Catháin | Let $H_2 = 2^{-1/2} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$, then the group generated with $S_{2}$ is finite, as is the group generated by $H_4 = H_2 \otimes H_2$ with $S_{4}$, I think. Computing minimal polynomials of a few random products $H_{16}PH_{16}$ shows that many of these matrices have infinite order, though. | |
May 20, 2022 at 13:52 | comment | added | Jonas Anderson | Yes. That is equivalent to the question I intended to ask. If it is finite for $m=1$, I'm interested in when/if it becomes infinite for larger $m$. | |
May 20, 2022 at 12:17 | comment | added | YCor | The question is equivalent to the question for $m=1$: is it true that for $U'$ unitary and not monomial, $\langle U',S_n\rangle$ is infinite? | |
S May 20, 2022 at 11:02 | review | First questions | |||
May 20, 2022 at 11:06 | |||||
S May 20, 2022 at 11:02 | history | asked | Jonas Anderson | CC BY-SA 4.0 |