$\operatorname{PSL}(2,\mathbb{F}_p) $ does not embed in $\mathfrak{S}_p$ for $p>11$ 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?
 A: Since Jordan himself knew the finite subgroups of ${\rm SL}(2,\mathbb{C})$, and he knew that the existence of a subgroup of index $p$ for ${\rm PSL}(2,p)$ was equivalent to the existence of a subgroup of index $p$ for $G = {\rm SL}(2,p)$, it might be worth remarking that the cases for which ${\rm SL}(2,p)$ has a subgroup of index $p$ can be understood in terms of the finite subgroups of ${\rm SL}(2,\mathbb{C})$ (I am not suggesting that Jordan would have thought in these terms at that time).
If $H$ is a subgroup of index $p$ in $G = {\rm SL}(2,p)$, then $|H| = (p-1)(p+1)$ is prime to $p$, so that a unimodular two-dimensional representation of $H$ over $\mathbb{F}_{p}$ may be lifted to a unimodular characteristic zero representation of $H$. Hence $H$ is isomorphic to a subgroup of ${\rm SL}(2,\mathbb{C}).$
Note that $H$ is non-Abelian, since $G = {\rm SL}(2,p)$ has a (generalized) quaternion Sylow $2$-subgroup $S$, and $S$ is isomorphic to a Sylow $2$-subgroup of $H$. Hence $H$ is isomorphic to an irreducible subgroup of ${\rm SL}(2,\mathbb{C}).$
Suppose now that $p > 5.$ Then $G$ has a unique conjugacy class of subgroups of order $\frac{p-1}{2}$, and the centralizer of any such subgroup is cyclic of order $p-1$. In particular, any cyclic subgroup of $G$ of order divisible by $\frac{p-1}{2}$ has order dividing $p-1$.
Suppose that $H$ is (isomorphic to) an imprimitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $H$ has an Abelian normal subgroup $A$ of index $2$, which is cyclic by unimodularity. Hence $|A| = \frac{p^{2}-1}{2}$, contrary to the fact that $G$ has no subgroup of that order, as remarked above.
Hence $H$ is a primitive subgroup of ${\rm SL}(2,\mathbb{C}).$ Then $|Z(H)| = 2$ and
$H/Z(H)$ is isomorphic to one of $A_{4},S_{4}$ or $A_{5}.$ Thus $p^{2}-1 = |H| \leq 120$ and $p \leq 11.$
(It would be possible to use more modern methods, or results of Frobenius and Blichfeldt on the eigenvalues of non-central elements in primitive finite complex linear groups, to avoid simply quoting the classification of finite primitive subgroups of ${\rm SL}(2,\mathbb{C})$, but for ease of exposition I don't give the details here.) Notice also that the three possibilities for $H$ as a primitive complex linear group occur as Hall $p^{\prime}$-subgroups of  ${\rm SL}(2,5)$, ${\rm SL}(2,7)$ and ${\rm SL}(2,11)$ for $p = 5,7$ and $11$ respectively.
Later edit: To digress on a remark above, I point out that a theorem of Blichfeldt (actually left as an exercise in his book), can be used to show that if a finite primitive subgroup $H$ of ${\rm SL}(2,\mathbb{C})$ contains an element $x$ of order $p+1$ for some prime $p$, then $p \leq 7.$ For certainly $x \not \in Z(H)$, so Blichfeldt's result shows that the eigenvalues of $x$ can't all lie on an arc of length less than $\frac{2 \pi}{5}$ on the unit circle $S^{1}.$ Replacing $x$ by a suitable generator of $\langle x \rangle$ if necessary, we may suppose that the eigenvalues of $x$ are both on an arc of length $\frac{4\pi}{p+1}.$ Hence
$\frac{p+1}{4} \leq \frac{5}{2}$ and $p \leq 9$ (so $p \leq 7$  as $p$ is prime). We remark that the binary octahedral group of order $48$ contains an element of order $8$, and is isomorphic to a subgroup of both ${\rm SL}(2,7)$ and ${\rm SL}(2,\mathbb{C})$ (primitive in the latter case). Similarly for $H = {\rm SL}(2,3)$, which contains an element of order $6$, and is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and to a subgroup of ${\rm SL}(2,5).$ However, ${\rm SL}(2,5),$ (which is isomorphic to a primitive subgroup of ${\rm SL}(2,\mathbb{C})$ and a subgroup of ${\rm SL}(2,11)$) indeed contains no element of order $12$.
That same result of Blichfeldt also can be used to prove (in the manner alluded to above) that a finite primitive subgroup of ${\rm SL}(2,\mathbb{C})$ has order $24,48$ or $120.$
A: 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. 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.
A: $\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\PGL{PGL}$We can use geometry of algebraic curves to get this bound. As mentioned in the other answer, we reduce to showing that $\PSL_2(\mathbb{F}_p)$ does not have a subgroup of index $p$. If it did, the order of that group would be $\tfrac{(p+1)(p-1)}{2}$. We then show the following:
Lemma Let $k$ be a field of characteristic $p$ (possibly $0$) and let $G$ be a finite subgroup of $\PGL_2(\mathbb{F}_p)$ with $\lvert G\rvert \not\equiv 0 \bmod p$. Then either
(1) $\lvert G\rvert \leq 60$ or
(2) $G$ is contained in the normalizer of a torus (which may or may not be split).
Proof Without loss of generality, we pass to the algebraic closure of $k$; this isn't essential, but it makes the algebraic geometry language simpler. The group $G$ acts on $\mathbb{P}^1$, and thus $\mathbb{P}^1/G$ is an algebraic curve, which is isomorphic to $\mathbb{P}^1$. Thus, we get a $G$-covering $\mathbb{P}^1 \to \mathbb{P}^1$.
Since $\lvert G\rvert $ is not $0 \bmod p$, this is a separable map; let it be ramified over $x_1, x_2, \dotsc, x_r$ with ramification of order $e_i$ over $x_i$.
Since $p$ does not divide $\lvert G\rvert$, we can apply the Riemann–Hurwitz formula and compute
$$2 \lvert G\rvert = 2 + \sum_i \tfrac{\lvert G\rvert}{e_i} (e_i-1)$$
or, in other words,
$$\lvert G\rvert = \frac{2}{2 - \sum_{i=1}^r (1-e_i^{-1})}.$$
Since $2 \leq \lvert G\rvert < \infty$, we see that $1 \leq \sum_{i=1}^r (1-e_i^{-1}) < 2$.
If $r \geq 4$, this is impossible, since each summand is at least $1/2$. Having $r = 1$ is also impossible, as then $\sum_{i=1}^r (1-e_i^{-1})<1$. So we get down to $r=2$ or $r=3$. In the case $r=3$, at least one of the $e_i$ must be $2$, or else we would have $\sum_{i=1}^r (1-e_i^{-1}) \geq 2$. If exactly one of the $e_i$ is $2$ then the closest we can get to $2$ without going over is $\sum_{i=1}^r (1-e_i^{-1}) = 1/2+2/3+4/5 = 59/30$, which gives $\lvert G\rvert = 60$. So we have now reduced to the case of either $r=2$ or of $r=3$ and $e_1=e_2=2$.
I think there is should be a more elementary way to say this next part, but the prime-to-$p$ fundamental group of $\mathbb{P}^1 \setminus \{ \text{$r$ points}\}$ is the profinite completion of $\langle g_1, g_2, \dotsc, g_r \rangle / (g_1 g_2 \dotsm g_r)$, where the element $g_i$ will act with order $e_i$ in a cover ramified of order $e_i$ over the $i$-th point. In particular, $G$ will be generated by $\{ g_1, g_2, \ldots, g_{r-1} \}$.
Now, if $r=2$, then $G$ is generated by $g_1$. So $G$ is cyclic, and of order prime to $p$, so $G$ is contained in a torus.
If $r=3$ and $e_1 = e_2 = 2$, then $G$ is generated by $g_1$ and $g_2$, each of which have order $2$, so $G$ is dihedral. Thus, $G$ has an index two cyclic subgroup $C$ of order prime to $p$, so $C$ is contained in a torus $T$, and $G$ is contained in $N(T)$. $\square$.
Now, if $\lvert G\rvert \leq 60$, then $\tfrac{p^2-1}{2} \leq 60$ and $p \leq 11$. If $G$ is contained in the normalizer of a torus (and now using that $G$ is in $\PSL_2$, not just $\PGL_2$), then $\lvert G\rvert$ divides either $p-1$ (in the case of a split torus) or $\tfrac{p+1}{2}$ (in the case of a nonsplit torus). Either way, this is incompatible with $\lvert G\rvert = \tfrac{p^2-1}{2}$ for $p>2$.
A: Here is the 1868 paper of Jordan mentioned in the comments: C. Jordan. Note sur les équations modulaires, C. R. Acad. Sci. Paris 66 (1868), 308-312.
Here is Jordan's proof from that paper, as I understand it. Assume $p > 11$. First Jordan notes that it will suffice to show that the group $G = \operatorname{SL}_2(p)$ has no subgroup of index $p$. So suppose $G$ has a subgroup $H$ of index $p$, in which case $|H| = (p+1)(p-1)$. The element $t = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ has order $p$, so $\langle t \rangle \cap H = 1$ and $G = \langle t \rangle H$.
Let $\varepsilon$ be a primitive root mod $p$. Then $s = \begin{pmatrix} \varepsilon & 0 \\ 0 & \varepsilon^{-1} \end{pmatrix}$ can be written in the form $t^kx$ for some $x \in H$, so $t^{-k}s = \begin{pmatrix} \varepsilon & -\varepsilon^{-1}k \\ 0 & \varepsilon^{-1} \end{pmatrix}  \in H$. This matrix is conjugate to $s$ in $G$, so by replacing $H$ with a conjugate we may as well assume that $s \in H$.
Jordan notes that the number of elements in $H$, which are not of one of the following forms:
$$\begin{pmatrix} \alpha & 0 \\ 0 & \delta \end{pmatrix}, \begin{pmatrix}0 & \beta \\ \gamma & 0\end{pmatrix}$$
is a multiple of $\frac{1}{2}(p-1)^2$. In other words, let $S = \langle s \rangle$ and $N = N_G(S)$, then $|H \setminus N|$ is a multiple of $\frac{1}{2}(p-1)^2$.
The reason is the following: if $g \in H \setminus N$, then the double coset $SgS$ is contained in $H \setminus N$; moreover any such double coset has size $\frac{1}{2}(p-1)^2$.
Let $r$ be the number of double cosets $SgS$ in $H \setminus N$. Then $|H| = \frac{r}{2} (p-1)^2 + 2(p-1)$ or $|H| = \frac{r}{2} (p-1)^2 + (p-1)$, according to whether $H$ contains $N$ or not. Since $|H| = (p-1)(p+1)$, the only possibility is that $r = 2$ and $H$ contains $N$.
Let $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $g' = \begin{pmatrix} a' & b' \\ c' & d' \end{pmatrix}$ be representives for the two double cosets in $H \setminus N$. Set $n = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$. Then $$H = S \cup nS \cup SgS \cup Sg'S.$$
For any $\alpha \in \mathbb{F}_p^\times$, the product $$h = g \begin{pmatrix} \alpha  & 0 \\ 0 & \alpha^{-1} \end{pmatrix} g = \begin{pmatrix} \alpha a^2 + \alpha^{-1}bc & \alpha ab + \alpha^{-1}bd \\ \alpha ac + \alpha^{-1}dc & \alpha bc + \alpha^{-1} d^2\end{pmatrix}$$ is contained in $H$. Thus one of the following holds:

*

*$h \in S$, in which case diagonal entries are zero.

*$h \in nS$, in which case anti-diagonal entries are zero.

*$h \in SgS$, in which case the product of diagonal entries is $ad$.

*$h \in Sg'S$, in which case the product of diagonal entries is $a'd'$.

In any case, the following equation holds:
\begin{multline}(\alpha a^2 + \alpha^{-1}bc) \cdot (\alpha ab + \alpha^{-1}bd) \cdot [(\alpha a^2 + \alpha^{-1}bc)(\alpha bc + \alpha^{-1} d^2)-ad] \cdot [(\alpha a^2 + \alpha^{-1}bc)(\alpha bc + \alpha^{-1} d^2)-a'd'] = 0\end{multline}
If $d = 0$, this gives us a polynomial equation of degree $6$, if $d \neq 0$ we get a polynomial equation of degree $12$. In any case there are at most $12$ roots, and any $\alpha \in \mathbb{F}_p^\times$ is a root, so $p \leq 13$. It remains to consider the case where $d \neq 0$, $p = 13$, and the equation has $12$ distinct roots.
In particular there are two roots for $\alpha a^2 + \alpha^{-1} bc = 0$, and they correspond to the case where $h \in S$; so for these roots we also have $\alpha bc + \alpha^{-1}d^2 = 0$. This implies $$\frac{a^2}{bc} = \frac{bc}{d^2},$$ so $(bc)^2 = (ad)^2$.
On the other hand $ad - bc = 1$, so it follows that $1 - 2ad = 0$. By the same arguments we have $1-2a'd' = 0$, so $ad = a'd'$. But then the last two factors in the equation are the same, so we have repeated roots, a contradiction.
