Timeline for Existence of 'maximal' finite permutation groups?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 8, 2022 at 2:40 | history | bounty ended | Jonas Anderson | ||
| Jun 7, 2022 at 23:44 | vote | accept | Jonas Anderson | ||
| Jun 7, 2022 at 5:56 | comment | added | Peter McNamara | Inside C^{n+1} with the usual action of S_{n+1} and basis e_1,e_2,..., define vectors b_i by b_i=a(e_1+...+e_n)+be_{n+1}+ce_i where a,b,c are constants such that na+b+c=0 and these vectors b_i are orthogonal (which is another equation to solve for a,b,c, and has solutions). These b_i span the standard irrep of S_{n+1} and if you compute the action with respect to this basis, you'll get the usual permuataion representation for S_n and can work out the extra generator by working out how it acts on this basis. | |
| Jun 7, 2022 at 2:57 | comment | added | Jonas Anderson | This being true I get that there's a larger (finite) group in $U(n)$ given by the elements of $S_{n+1}$. Can you talk me through the proof that this additional generator is monomial for $n\ge 4$? I know it's beyond the scope of my question, but if you could outline how to calculate the `extra' generator in a small example I'd be very grateful. | |
| Jun 7, 2022 at 2:54 | comment | added | Jonas Anderson | Thanks Peter! I'm a physicist and apologize in advance if I'm missing something obvious, but I do not fully understand your construction. I understand that there's always the standard irrep of $S_{n+1}$ ($n\times n$ complex unitary matrices). There's clearly a subgroup isomorphic to $S_{n}$ there. Can I always choose the elements in $S_{n}$ to be represented by the natural (or defining) permutation representation of $S_{n}$ ($n\times n$ monomial matrices)? | |
| Jun 6, 2022 at 1:38 | history | answered | Peter McNamara | CC BY-SA 4.0 |