Here is a statement having a proof that involved the CFSG.


Let $p$ be a prime, and $S$ be a nonabelian finite simple group such that $S$ is isomorphic to a subgroup of $S\_p$ with $p\mid |S|$. Then $\mathrm{Out}(S)$ is a $p'$-group (i.e. $p\nmid |\mathrm{Out}(S)|$).


**Question** Is there a CFSG-free proof for this statement?



Outer automorphism of a finite simple group which is isomorphic to a subgroup of $S_p$
Let $p$ be a prime, and let $G$ be a non-abelian simple group with $p \mid |G|$. Suppose that $G \leq S\_p$ and that $G$ is transitive.


By a theorem of Burnside a non-solvable transitive group of prime degree is $2$-transitive, so $G$ is $2$-transitive.


Let $B \leq G$ be such that $|B| = p$. Then $G = AB$, where $A$ is a point stabilizer. Note that $A \cap B = 1$, so $A$ is a Hall subgroup.


Now it seems to be a theorem of Wielandt that in this case $Aut(G)/G$ is cyclic of order dividing $p-1$. This is according to the following paper of Cameron (I haven't been able to find the paper by Wielandt), see §3 there.



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> Cameron, Peter J. On groups with several doubly-transitive permutation representations. Math. Z. 128 (1972), 1–14.
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The paper by Wielandt is



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> Wielandt, Helmut. On automorphisms of doubly transitive permutation groups. 1967 Proc. Internat Conf. Theory of Groups (Canberra, 1965) pp. 389–393 Gordon and Breach, New York
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You could also use the proofs given by Cameron, which do not need CFSG.


Denote the image of $N\_{S\_p}(G)$ in $Aut(G)$ by $P(G)$, this is called the group of permutation automorphisms. Then Theorem 3 in Cameron shows that $P(G)$ is normal in $Aut(G)$ and $Aut(G)/P(G)$ is an elementary abelian $2$-group.


Next we can show that $P(G)/Inn(G)$ is cyclic of order dividing $p-1$, which gives you the desired result. For this let $R = N\_{S\_p}(G)$. By the Frattini argument $R = GN\_R(B)$. Then $$\frac{P(G)}{Inn(G)} \cong \frac{R}{GC\_{S\_p}(G)} = \frac{GN\_R(B)}{GC\_{S\_p}(G)} \cong \frac{N\_R(B)}{GC\_{S\_p(G)} \cap N\_R(B)}.$$


Because $B \leq GC\_{S\_p(G)} \cap N\_R(B)$, it follows that $P(G)/Inn(G)$ is isomorphic to a quotient of $N\_R(B)/B$. Now $B$ is self-centralizing in $S\_p$, so $N\_R(B)/B$ is isomorphic to a subgroup of $Aut(B)$, hence cyclic of order dividing $p-1$.


