Timeline for Counting conjugacy classes with a subgroup of prime index
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 9, 2018 at 12:42 | comment | added | Mark Wildon | Yes. Sorry, I seem to have used $H$ and $N$ interchangeably. | |
| Jun 9, 2018 at 11:34 | history | edited | Siddhartha | CC BY-SA 4.0 |
edited body
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| Jun 9, 2018 at 11:33 | comment | added | Siddhartha | @Mark : I guess you mean $H$ by $N$. | |
| Jun 8, 2018 at 13:54 | comment | added | Mark Wildon | I think some confusion is reasonable: the action is not defined in the article. Let $s'$ be the number of conjugacy classes in $gN$. The result we need is $s' = s$. I think the intended argument is essentially as follows. By Burnside's Counting Lemma for the conjugacy action of $H$ on $gN$, $|H|s' = \sum_{h \in H} |C_{gN}(h)|$. If $h$ is not in an invariant class then $h$ commutes with no element of $gN$. For $h \in n_1^G \cup \ldots \cup n_s^G$ we have $|C_{gN}(h)| = |C_G(h)|/p$. Therefore the sum is $|n_1^G||C_G(n_1)|/p + \cdots + |n_s^G||C_G(n_s)|/p| = |G|s/p$. This is $|H|s$, as required. | |
| Jun 8, 2018 at 7:35 | comment | added | Geoff Robinson | Probably belongs on Maths Stackexchange. | |
| Jun 8, 2018 at 6:07 | history | asked | Siddhartha | CC BY-SA 4.0 |