Timeline for Number of squares in a finite group
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 18, 2015 at 11:42 | comment | added | YCor | @GeoffRobinson Well, $S_3$ or even the cyclic group of order 6 being counterexamples, I think nobody conjectured such a fact :) | |
| Oct 18, 2015 at 11:27 | comment | added | Geoff Robinson | It might be worth remarking that it is not generally true that in a finite group $G$ of even order, the number of squares is even (so there is indeed something to do for $S_{n}$): for example, in a group $G$ which has Sylow $2$-subgroups of exponent $2$, only elements of odd order are squares, and the number of elements of odd order in any finite group $G$ is odd ( as all elements of odd order except the identity occur in mutually inverse pairs). | |
| Oct 18, 2015 at 0:44 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed typos
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| Oct 17, 2015 at 22:56 | comment | added | YCor | @GHfromMO Thanks very much. For $p$ odd this is indeed a simple verification, but for $p=2$ I had to modify the argument in a quite unexpected way. | |
| Oct 17, 2015 at 22:55 | history | edited | YCor | CC BY-SA 3.0 |
Fixed a flaw in the argument, detected by "GH from MO".
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| Oct 17, 2015 at 21:21 | comment | added | GH from MO | I think it should be remarked (in the case $p>2$) that an element of $Q^C$ is automatically a square within the centralizer of $C$. This is not obvious, because a square-root of an element from $Q^C$ can be outside the centralizer of $C$, e.g. a $2p$-cycle of the form $(1\tau_1\dots p\tau_p)$ is a square-root of $(1\dots p)(\tau_1\dots\tau_p)$. | |
| Oct 17, 2015 at 20:57 | history | edited | GH from MO | CC BY-SA 3.0 |
edited body
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| Oct 17, 2015 at 20:27 | history | edited | GH from MO | CC BY-SA 3.0 |
fixed two typos
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| Oct 17, 2015 at 20:20 | history | answered | YCor | CC BY-SA 3.0 |