Timeline for A question on conjugacy classes of central involutions in a finite group
Current License: CC BY-SA 3.0
14 events
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| Aug 11, 2014 at 23:18 | comment | added | Geoff Robinson | In fact, the number of classes of central involutions is the number of $N_{G}(S)$ orbits by conjugation on $\Omega_{1}(Z(S)) \backslash \{1_{G}\},$ where $S$ is a Sylow $2$-subgroup and $\Omega_{1}(Z(S))$ is the subgroup of $Z(S)$ generated by its involutions. | |
| Aug 11, 2014 at 22:55 | history | edited | Anurag | CC BY-SA 3.0 |
Changed "is equal to" to "is at most" in EDIT2, following Martin Isaacs' comment.
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| Aug 11, 2014 at 22:51 | comment | added | Anurag | Thanks for correcting me. It should be that the number of classes of central involutions is at most the number of involutions in the center of a Sylow 2-subgroup. Since given a central involution f, if H is a Sylow 2-subgroup contained in the normalizer of f then f must itself belong to H as otherwise, H and f would generate a subgroup of order 2|H|. | |
| Aug 11, 2014 at 21:37 | comment | added | Marty Isaacs | There is something wrong with "EDIT2" of the question. It is not true in general that the number of classes of central involutions in G is the number of involutions in the center of a Sylow 2-subgroup. Look at the alternating group A_5. It has just one class of involutions but the Sylow center contains three involutions. | |
| Aug 7, 2014 at 11:43 | history | edited | Anurag | CC BY-SA 3.0 |
added 259 characters in body
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| Aug 7, 2014 at 9:20 | history | edited | Anurag | CC BY-SA 3.0 |
added 138 characters in body
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| Aug 7, 2014 at 9:08 | answer | added | Derek Holt | timeline score: 7 | |
| Aug 7, 2014 at 2:50 | comment | added | Colin Reid | The question reduces to fusion in a Sylow $2$-subgroup $S$ of $G$. We want to know, given $a \in S, b \in Z(S)$ distinct involutions that are in a given conjugacy class $C$ of $G$, does $ab$ lie in $C$? In other words, is $C'$ a union of cosets of $C' \cap Z(S)$, where $C' = (C \cup \{1\}) \cap S$? (We certainly need $C' \cap Z(S)$ to be a non-trivial group.) | |
| Aug 7, 2014 at 1:55 | comment | added | Anurag | Kindly note that I am interested in only those conjugacy classes of involutions which contain a central involution. Since conjugate of a central involution is also central, it would mean that the conjugacy class would consist entirely of central involutions. | |
| Aug 7, 2014 at 1:50 | history | edited | Anurag | CC BY-SA 3.0 |
clarified a few ambiguous statements
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| Aug 7, 2014 at 1:43 | comment | added | Anurag | Yes, I didn't notice that it is quite straightforward to show that the result is true if G has a unique conjugacy class of involutions. I am slightly rephrasing the question now. @ColinReid: I was talking about conditions on G and not the part of the statement following "if and only if" | |
| Aug 6, 2014 at 23:49 | comment | added | Geoff Robinson | I don't know if you really said what you mean, but if there is a unique conjugacy class of involutions in $G$, the statement is true. By the way, I think you need to specify that $a$ and $b$ are distinct involutions. | |
| Aug 6, 2014 at 23:49 | comment | added | Colin Reid | The condition is certainly necessary, since the product of two involutions is an involution if and only if they are distinct and commute. | |
| Aug 6, 2014 at 23:33 | history | asked | Anurag | CC BY-SA 3.0 |