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
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 2, 2022 at 1:27 | vote | accept | Jackson Walters | ||
| Dec 1, 2022 at 15:45 | answer | added | Tom De Medts | timeline score: 3 | |
| Nov 30, 2022 at 21:46 | comment | added | Jackson Walters | @TomDeMedts Good. Was going to check $p=31$ next: github.com/jacksonwalters/fano-plane-symmetries, where I start to compute the groups involved. sage's is_isomorphism isn't working for $n=3$, $p=7$, but I'm still curious about this map; it seems to be more than a bijection of sets between $\mathbb{F}_{2^n} \leftrightarrow \mathbb{F}_p$; feels geometric. In Brown+Loehr, the Möbius transformations have a $c^2$in the denominator, and it's used as the quadratic residues mod $7$. are squares. I'd love to see a non-linearity proof. I'd like to manipulate HDD bits using Möbius transformations. | |
| Nov 30, 2022 at 13:06 | comment | added | Tom De Medts | @JacksonWalters Sorry, but I still disagree. Any map from a vector space to another vector space maps basis elements to a linear combination of basis elements, even if it is non-linear. In fact, looking at the reference by Brown and Loehr you provide, they do quite a bit of work to show linearity, so you can't expect a one-line proof. Even more: I checked this for $p=31$, with $g(x) = x^5 + x^2 + 1$, and it turns out to be false: $T_r$ is not linear. (If you want, I can write down more details in an answer.) | |
| Nov 29, 2022 at 13:01 | comment | added | Jackson Walters | @TomDeMedts Choose a basis $\{x^i\}_{0 \le i < n}$ for $\mathbb{F}_{2^n}$. $T_r(x^i)=x^{r(\infty)}+x^{r(i)}=x^{-1/\infty}+x^{-1/i}=1+x^{n-1/i}$, so $T_r$ maps a basis element to a linear combinations of basis elements. | |
| Nov 29, 2022 at 12:02 | comment | added | Geoff Robinson | I am saying (in the second comment particularly) that when $p>7$ is a Mersenne prime there is no faithful group homomorphism from ${\rm PSL}(2,p)$ into ${\rm PGL}(n,2)$ where $p = 2^{n} - 1$. | |
| Nov 29, 2022 at 9:14 | comment | added | Tom De Medts | @JacksonWalters I don't see why $T_f$ is linear (and I certainly don't understand your short argument). For instance, why is $T_r$ linear? Could you elaborate? | |
| Nov 29, 2022 at 1:46 | comment | added | Jackson Walters | I’m confused. So you’re saying the map $T: PSL(2,p) \rightarrow PGL(n,2)$ is not a homomorphism of groups? Where does it fail? $PSL(2,p)$ is generated by reflections $r: k \mapsto -1/k$ and translations $t: k \mapsto k+1$, so surely $T_r$ and $T_t$ are permutations of $\mathbb{F}_{2^n}$. $T_{f\circ h}=T_f \circ T_h$ because we’re composing functions in the exponent. $T_f$ is linear because it maps a basis element $x^i$ to a linear combination of basis elements. Are you saying $im(T)=\langle T_r, T_t \rangle$ is trivial, or that $im(T)=PGL(2,n)$? Like, are you saying the inclusion isn’t proper? | |
| Nov 28, 2022 at 23:49 | comment | added | Geoff Robinson | What I mean is "for a Mersenne prime $p >7$ there is no such homomorphism", ie there is no embedding of ${\rm PSL}(2,p)$ as a subgroup of ${\rm PGL}(n,2)$. | |
| Nov 28, 2022 at 23:39 | comment | added | Geoff Robinson | I think this rarely happens. An element of order $p$ in ${\rm PGL}(n,2)$ is only conjugate to $n$ of its powers in this situation, while (for $p>3$) an element of order $p$ in the simple group ${\rm PSL}(2,p)$ is conjugate to $\frac{p-1}{2}$ of its powers. But note that in this situation $n = \log_{2}(p+1)$ is smaller than $\frac{p-1}{2}$ for Mersenne primes $p > 7.$ | |
| Nov 28, 2022 at 21:59 | history | edited | LSpice | CC BY-SA 4.0 |
Name of reference
|
| Nov 28, 2022 at 21:38 | history | asked | Jackson Walters | CC BY-SA 4.0 |