Hölder's theorem (Group Theory). Proof collection and applications In (finite) group theory, the number $6$ seems to be special in the sense that we have the result of Hölder that this is the only natural number $n=6$ such that there exists an outer automorphism of $S_n$. My question is twofolds


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*I would like a collection of possible proofs of this result. For instance, I would like to also know if there is a proof that is purely number theoretic or a non-constructive proof of this result. I don't know if Hölder originally proved this constructively, but most of the proof that I see are constructive (and more or less the same). One starts with a transitive subgroup of $S_6$ of order 120 (this will be isomorphic to $S_5$)

*I would like to know of any (easy?) application of this result leading to some result in group theory where the number $6$ stands out just because of this result. 
I would appreciate any ideas.
 A: Here is a nice direct application: subgroups of small index in $S_n$. Assume that $n\geq 5$, and let $H\leq S_n$ be a subgroup of index at most $n$. Then either $H$ is $A_n, S_n$, or a stabilizer of some element $1\leq i \leq n$ in $S_n$, except if $n=6$. 
If $n=6$, then $S_6$ can be viewed as the group of all permutation of the 6 elements of the projective line over $\mathbb{F}_5$. But the group of projective transformations $PGL(2,5)$ form an index 6 subgroup. 
The connection is: if $H\leq S_n$ with index $k$, then the action of $S_n$ on the left cosets of $H$ by left multiplication defines a homomorphism $\varphi: S_n\rightarrow S_k$. If $k< n$, then $\varphi$ cannot be injective. But as $n\geq 5$, the only possible kernels are $A_n, S_n$, yielding these subgroups of small indices. 
If $k=n$, then $\varphi$ can be injective. However, in that case it has to be surjective. So $\varphi\in Aut(S_n)$. Inner automorphims correspond to the stabilizers. Outer automorphisms correspond to different index $n$ subgroups. (Think about it.) 
This line of thought can be extended to obtain simple proofs to results such as $PGL(2,5)\cong S_5$ and $PSL(2,5)\cong A_5$. 
A: There are at least two fundamentally different ways to get the exceptional outer automorphism of $S_6$.  One is to find the permutation representation of $S_5$ on six letters, as already alluded to.  The other is to notice that $A_6\cong PSL(2,9)$ is an index four normal subgroup of $P\Gamma L(2,9)$.  The quotient is the Klein four group, and the intermediate index two subgroups are of distinct isomorphism types: One is $S_6$, one is $PGL(2,9)$, and one is $M_{10}$ (a point stabilizer in $M_{11}$).  Since they are pairwise non-isomorphic, they are characteristic subgroups, so since $P\Gamma L(2,9)$ has no order $2$ normal subgroup (and since $S_6$ has trivial center) it acts faithfully by conjugation on $S_6$, and thus embeds monomorphically into $Aut(S_6)$.  
So it boils down to whether you like $\Bbb{F}_5$ or $\Bbb{F}_9$ better.
If you are unfamiliar with the notation, $\Gamma L(n,k)$ is the group of semilinear automorphisms of $k^n$, where a semilinear map $f$ satisfies $f(u+v)=f(u)+f(v)$ and $f(av)=\sigma_f(a)f(v)$ for some $\sigma_f\in Aut(k)$.  $P\Gamma L(n,k)$ is just that group mod the scalar matrices.
