$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\PGL{PGL}\DeclareMathOperator\GL{GL}$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 $\PSL(2, 7) \cong \GL(3, 2)$?](https://personal.math.vt.edu/brown/doc/PSL(2,7)_GL(3,2).pdf)). What is $\operatorname{im}(T)\subset \PGL(n,2)$?



How to classify homomorphisms from $\operatorname{PSL}(2,p)$ to $\operatorname{PGL}(n,2)$ when $2^n=p+1$?
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 $\mathrm{GL}\_n(2)$.


