Timeline for How to classify homomorphisms from $\operatorname{PSL}(2,p)$ to $\operatorname{PGL}(n,2)$ when $2^n=p+1$?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2022 at 12:50 | comment | added | Geoff Robinson | Indeed my comments stand. There is no group homomorphism at all from PSL(2,p) into GL(n ,2) when p>7 is a prime with p+1 =2^n , except for the homomorphism sending everything to the identity. | |
| Dec 2, 2022 at 13:30 | comment | added | Jackson Walters | I think that takes care of the question: $im(T)$ is a non-linear group of permutations of $\mathbb{F}_{2^n}$ when $n\ne 3$. One can compute the linear subgroup $im(T) \cap GL(n,2)$, but why? Really I’m content now that I can perform but manipulations with Möbius transformations, linearly or non-linearly. | |
| Dec 2, 2022 at 9:20 | comment | added | Tom De Medts | @JacksonWalters Of course, $\langle T_r, T_t \rangle$ is still a group, because it's a subgroup of $\operatorname{Sym}(\mathbb{F}_{2^n})$. To answer the question about "how non-linear" the map $T$ is, you could try intersecting the image of $T$ with $\mathrm{GL}_n(2)$ and computing its index in $\operatorname{im}(T)$, but I'm not sure what sort of information you would hope to get out of this. | |
| Dec 2, 2022 at 1:27 | vote | accept | Jackson Walters | ||
| Dec 2, 2022 at 1:27 | comment | added | Jackson Walters | Cool. I realized after you wrote it wasn't linear that I needed to check $T_f(x+y)=T_f(x)+T_f(y)$. It's clear that $T_f(\lambda x)= \lambda T_f(x)$ for $\lambda \in \{0,1\}$ since $T_f(0)=x^{f(\infty)}+x^{f(\infty)}=0$. Your counter-example settles it. I suppose the problem is $x^k+x^j$ cannot be simply rearranged to $x^l$ for some $l$ depending on $g, j, k$. It just so happens to work for $p=7$, $n=3$. I wonder though, how non-linear is this map? Perhaps $\langle T_r, T_t \rangle$ is not a group, but it is some algebraic object. Which one? | |
| Dec 1, 2022 at 15:45 | history | answered | Tom De Medts | CC BY-SA 4.0 |