Timeline for How can we conclude that $2p\nmid s_{2p}$?
Current License: CC BY-SA 3.0
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| when toggle format | what | by | license | comment | |
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| Jan 2, 2015 at 12:30 | answer | added | Geoff Robinson | timeline score: 13 | |
| Jan 2, 2015 at 9:54 | comment | added | Padraig Ó Catháin | You could look at "On a Theorem of Frobenius: solutions of $x^n = 1$ in finite groups" by Isaacs and Robinson. On the way to proving Frobenius Theorem, they show that $\phi(n) \mid s_{n}$ for all $n$, but some of the methods and references might be of use. I assume you want to exclude the case that there are no elements of order $2p$ in $G$, in which case $2p \mid 0$? (E.g. elements of order 6 in $A_{4}$) | |
| Jan 2, 2015 at 8:14 | comment | added | Shukran | In fact, I want to know how we can conclude $2p\nmid\sum_{o(x)=2p}|x^G|$? | |
| Jan 2, 2015 at 8:11 | comment | added | Shukran | sorry I forgot to say that $p$ is odd prime. | |
| Jan 2, 2015 at 8:02 | comment | added | S. Carnahan♦ | If $p=2$ and $G = C_2 \times C_4$, then $s_4 = 4$. | |
| Jan 2, 2015 at 7:33 | review | First posts | |||
| Jan 2, 2015 at 7:49 | |||||
| Jan 2, 2015 at 7:29 | history | asked | Shukran | CC BY-SA 3.0 |