A famous result of Galois, in his letter to Auguste Chevalier, is that for $p$ prime $>11$ the group $\operatorname{PSL}(2,\mathbb{F}\_p) $ does not embed in the symmetric group $\mathfrak{S}\_p$. The standard proof nowadays goes through the classification of subgroups of $\operatorname{PSL}(2,\mathbb{F}\_p) $ (Dickson's theorem), which is far from trivial. Does anyone know a simpler argument?



$\operatorname{PSL}(2,\mathbb{F}_p) $ does not embed in $\mathfrak{S}_p$ for $p>11$
A few months ago Péter Pál Pálfy has given a talk about this exact topic. The abstract of the talk was the following:



> 
> In his "testamentary letter" Galois claims (without proof) that PSL(2,p) does not have a subgroup of index p whenever p>11, and gives examples that for p = 5, 7, 11 such subgroups exist. The attempt by Betti in 1853 to give a proof does not seem to be complete. Jordan's proof in his 1870 book uses methods certainly not known to Galois. Nowadays we deduce Galois's result from the complete list of subgroups of PSL(2,p) obtained by Gierster in 1881.
> 
> 
> In the talk I will give a proof that might be close to Galois's own thoughts. Last October I exchanged a few e-mails on this topic with Peter M. Neumann. So the talk is in some way a commemoration of him.
> 
> 
> 


The recording of the talk is available [here](https://drive.google.com/file/d/10T_AuJb-J12movAkKJNXbIz87sw1-wM9/view?usp=sharing). The presentation of the proof begins around 32 minutes in.


The proof is elementary, but is itself nontrivial. We easily reduce to showing $G=\operatorname{PSL}(2,\mathbb{F}\_p)$ has no index $p$ subgroup for large $p$. The idea is to study the natural doubly transitive action of $G$ on the projective line and its interrelation between an index $p$ subgroup and the subgroup of affine transformations inside $G$. The bounds on $p$ come out of realizing that all elements of $\mathbb F\_p^\times$ must satisfy quadratic relations coming from that action.


