Timeline for Existence (or the number) of generating triple of involutions of $\operatorname{PGL}_2(p)$ with some conditions
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 16, 2020 at 14:34 | vote | accept | Groups | ||
| Jun 11, 2020 at 11:54 | comment | added | John Shareshian | 2) As you say, $PGL_2(p)$ contains dihedral subgroups of order $2(p+1)$ and $2(p-1)$. Therefore, there exist elements $r$ and $s$ of respective orders $p+1$ and $p-1$ and involutions $c,d,f,h$ such that $cd=r$ and $fh=s$. By comment (1) above, we may assume that $d$ is fixed point free and so is $h$. Now find $u,v$ such that $u^{-1}du=z$ and $v^{-1}hv=z$. Set $a=u^{-1}cu$ and $b=v^{-1}fv$. | |
| Jun 11, 2020 at 11:46 | comment | added | John Shareshian | 1) You are correct that fixed-point-free involutions will act as even permutations when $p \equiv 3 \bmod 4$. This does not matter. Elements of one of the two classes will act as even permutations and elements of the other will act as odd permutations. Which does which depends on $p \bmod 4$. The key point is that elements of order $p-1$ or $p+1$ will act as odd permutations and therefore, if one of these elements is the product of two involutions, then these two involutions are not in the same class. So, one of the two involutions is fixed point free and the other is not. | |
| Jun 10, 2020 at 5:38 | comment | added | Groups | Many thanks for your answer but I have some questions on your solution. Firstly I think the involutions without any fixed points could also be mapped to an even permutation when $p\equiv 3\mod 4$. And could you give more details on the existence of involutions $a,b$ making $|az|=p+1$ and $|bz|=p-1$? | |
| Jun 8, 2020 at 15:34 | history | edited | John Shareshian | CC BY-SA 4.0 |
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| Jun 8, 2020 at 13:07 | history | answered | John Shareshian | CC BY-SA 4.0 |