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$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}$I am looking to classify the homomorphisms from the group $\PSL(2,p)=\Aut(\mathbb{P}^1\mathbb{F}_p)$ to $\PGL(n,2)=\GL(n,2)=\Aut(\mathbb{F}_{2^n})$ when $p$ is a Mersenne prime, i.e. $2^n=p+1$.

There is a bijection $\mathbb{P}^1\mathbb{F}_p \rightarrow \mathbb{F}_{2^n} \cong \mathbb{F}_2[x]/g(x)$ given by $k \mapsto x^\infty+x^k$ where $x^\infty=0$. Given a Möbius transformation $f$, define the homomorphism $T$ by $T_f(x^\infty+x^k)=x^{f(\infty)}+x^{f(k)}$.

When $n=3$, $p=7$, $T$ is an isomorphism (see Brown and Loehr - Why is $\PSL(2, 7) \cong \GL(3, 2)$?). What is $\operatorname{im}(T)\subset \PGL(n,2)$?

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    $\begingroup$ 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 $p > 7.$ $\endgroup$
    – Geoff Robinson
    Commented Nov 28, 2022 at 23:39
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    $\begingroup$ 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 ${\rm PGL}(n,2)$. $\endgroup$
    – Geoff Robinson
    Commented Nov 28, 2022 at 23:49
  • $\begingroup$ 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? $\endgroup$
    – Jackson Walters
    Commented Nov 29, 2022 at 1:46
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    $\begingroup$ 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 $p = 2^{n} - 1$. $\endgroup$
    – Geoff Robinson
    Commented Nov 29, 2022 at 12:02
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    $\begingroup$ @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.) $\endgroup$
    – Tom De Medts
    Commented Nov 30, 2022 at 13:06

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The map $T \colon \mathrm{PSL}_2(p) \to \operatorname{Sym}(\mathbb{F}_{2^n}) \colon f \mapsto T_f$ does not have its image in $\mathrm{GL}_n(2)$ for other Mersenne primes $p = 2^n - 1$, unlike the case $p = 7 = 2^3 - 1$.

For instance, let $p = 31$ and consider $\mathbb{F}_{32} \cong \mathbb{F}_2[x] / (g)$ with $g(x) = x^5 + x^2 + 1$, as in the question, and let $f \colon \mathbb{P}_1(\mathbb{F}_p) \to \mathbb{P}_1(\mathbb{F}_p) \colon a \mapsto -a^{-1}$ (so $f \in \mathrm{PSL}_2(p)$). The map $T_f$ is then the permutation of $\mathbb{F}_{32}$ mapping each $x^k$ to $1 + x^{f(k)}$. In particular, \begin{align*} T_f(1) &= 1, \\ T_f(x^2) &= 1 + x^{15}, \\ T_f(x^5) &= 1 + x^{6}. \end{align*} However, $x^5 + x^2 + 1 = 0$ in $\mathbb{F}_{32}$, but $$ T_f(x^5) + T_f(x^2) + T_f(1) = 1 + x^6 + x^{15} = 1 + x^2 + x^4 \neq 0 ,$$ so $T_f$ cannot be linear, i.e., it is not contained in $\mathrm{GL}_n(2)$.

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  • $\begingroup$ 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? $\endgroup$
    – Jackson Walters
    Commented Dec 2, 2022 at 1:27
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    $\begingroup$ @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. $\endgroup$
    – Tom De Medts
    Commented Dec 2, 2022 at 9:20
  • $\begingroup$ 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. $\endgroup$
    – Jackson Walters
    Commented Dec 2, 2022 at 13:30
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    $\begingroup$ 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. $\endgroup$
    – Geoff Robinson
    Commented Dec 4, 2022 at 12:50

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