Timeline for The number in the join of conjugate class and centralizer
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 19, 2012 at 10:16 | comment | added | Wei Zhou | Nick, you are right. The Frobenius groups with abelian Kernel satisfy the equality. So these groups seem more than I thought. | |
| May 19, 2012 at 9:21 | comment | added | Nick Gill | I could be wrong but it seems to me that $G=C_p\rtimes C_{p-1}$ would satisfy the equality for all $g\in G$. In fact Frobenius groups are probably a good source of examples in general - one would just need to be sure that elements in the kernel satisfy the equality. Taking the kernel to be abelian would be enough (as for $G=C_p\rtimes C_{p-1}$) but there may well be other examples... | |
| May 19, 2012 at 0:58 | comment | added | Wei Zhou | Nick, do you know any groups satisfying the equality for every $h \in G$, except Dedekind groups? | |
| May 18, 2012 at 13:16 | comment | added | Nick Gill | Wei Zhou, you are right, and the condition in the edit above confirms this. | |
| May 18, 2012 at 13:14 | history | edited | Nick Gill | CC BY-SA 3.0 |
added 77 characters in body
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| May 18, 2012 at 13:08 | history | edited | Nick Gill | CC BY-SA 3.0 |
Characterization completed
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| May 16, 2012 at 15:08 | comment | added | Wei Zhou | I think that the hypotheses that $C_G(a)=\langle a \rangle$ also implies that equality holds | |
| May 16, 2012 at 14:54 | comment | added | Mark Wildon | Hi Nick. Maybe there is some sort of converse, but it will definitely need some extra hypotheses. For example, take $G = S_6$ and let $a = (123456) \in S_6$. Then $C = \left< a \right>$ and $N$ is the dihedral group of order $12$, so $|N : C| = 2$ and $a^G \cap C = \lbrace a,a^{-1} \rbrace$. But $C$ is not a trivial intersection subgroup of $G$ because both $C^{(23)(56)} = \left< (132465) \right>$ and $C$ contain $(14)(25)(36)$. | |
| May 16, 2012 at 13:01 | history | answered | Nick Gill | CC BY-SA 3.0 |