Timeline for Finite simple groups and conjugacy classes with 2p elements
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 23, 2011 at 19:19 | vote | accept | Leandro Vendramin | ||
| Sep 17, 2011 at 14:03 | vote | accept | Leandro Vendramin | ||
| Dec 23, 2011 at 19:19 | |||||
| Sep 17, 2011 at 14:03 | vote | accept | Leandro Vendramin | ||
| Sep 17, 2011 at 14:03 | |||||
| Sep 17, 2011 at 12:22 | answer | added | Derek Holt | timeline score: 13 | |
| Sep 16, 2011 at 19:20 | comment | added | Steve D | I would guess that this can't happen, because such an element gives a centralizer in a finite simple group $G$ of index $2p$. That is, you have a subgroup of small index with nontrivial center, which I think is not possible, after doing some examples in GAP and looking at the ATLAS. | |
| Sep 16, 2011 at 9:50 | comment | added | Alex B. | Unfortunately, this argument also doesn't immediately generalise to $|C|=2p$. | |
| Sep 16, 2011 at 9:50 | comment | added | Alex B. | Actually, more generally a conjugacy class with $p^n$ elements is impossible for $n>0$. That's because if $C$ is a conjugacy class and $\chi$ is an irreducible char, and if $(\chi(1),|C|)=1$, then any $g\in C$ is either in the centre of $\chi$ or $\chi(C)=0$. If $G$ is simple, then the centres of characters are trivial, so $\chi(C)=0$ for all $\chi$ with $p\nmid \chi(1)$. The column orthogonality equation between 1 and $g\in C$ now gives a contradiction. This is somewhere in Isaacs, but for me, it's easier to link to my notes: math.postech.ac.kr/~bartel/docs/reptheory.pdf, Thm 6.7. | |
| Sep 16, 2011 at 9:24 | comment | added | Tim Dokchitser | I can see why a conjugacy class $C$ with $p$ elements is impossible: $G$ acts on $C$ by conjugation, faithfully as it is simple, so $G\subset S_p$. Under this embedding, the point-stabilizers $G\cap S_{p-1}$ are then the normalizers of elements of $C$. But every element normalizers itself, so the sizes of these subgroups must be a multiple of $p$, contradiction. This argument does not immediately work for $2p$, but it does put restrictions on $G\subset S_{2p}$ (e.g. it should contain the full $p$-Sylow $C_p\times C_p$ of $S_{2p}$). Maybe it can help to prove that this is impossible? | |
| Sep 16, 2011 at 6:17 | comment | added | Alon Amit | Is it known that this can't occur with the alternating groups? Empirically it looks like the sizes of the conjugacy classes of $A_n$ all have at least 3 prime factors once $n>8$. | |
| Sep 16, 2011 at 3:30 | comment | added | Noam D. Elkies | Did you trawl the ATLAS for examples? | |
| Sep 16, 2011 at 1:41 | history | asked | Leandro Vendramin | CC BY-SA 3.0 |