Question: Is it true that a non-solvable primitive group $G$ of degree $n$ satisfies $|G| \ge 6n$?

It is checked below for $n<2500$. If it is true in general, then the bound is optimal because $A_5$ has a maximal subgroup of index $10$. If moreover we assume $n>10$, then is it true that $|G| \ge 12n$?

IOW: Let $G \neq A_5$ be a non-solvable group and $M$ a core-free maximal subgroup. Is $|M|\ge 12?$

gap> PrimitiveNonSolvableBound(2,2499);
        [ [ 5, 60, 12 ], [ 6, 60, 10 ], [ 10, 60, 6 ], [ 28, 336, 12 ], [ 55, 660, 12 ], [ 91, 1092, 12 ], [ 819, 9828, 12 ], [ 2109, 25308, 12 ] ]  

    local o,i,G,No,T,ord,r;
    T:=[]; r:=12;
    for o in [o1..o2] do
        for i in [1..No] do
            if not IsSolvable(G) and ord/o<=r then
    return T;
  • $\begingroup$ It seems like you could maybe approach this by looking at the various values of $|M|$ less than $12$: a lot of them would imply $M$ was a maximal Sylow subgroup. For example, I think this line of thought would be enough to show the $6n$ bound. $\endgroup$ – Steve D Jan 23 '18 at 23:14

Yes. Let $M$ be a maximal corefree subgroup of a finite group $G$, with $|M|\leq 11$. If $M$ is abelian, one can easily show that $G$ is primitive of affine type and thus is soluble:

Let $M$ be the stabilizer of a point in $G$, assumed to be abelian. If $M=1$, then $G$ is a cyclic group of prime order. Suppose that $M>1$. Let $g\in G\setminus M$. By the maximality of $M$ in $G$, it follows that $\langle M,M^g \rangle=G$. Now $M\cap M^g$ is centralized by $M$ and $M^g$ and hence, by $G$. It follows that $M\cap M^g =1$. We have shown that $M\cap M^g =1$ for every $g\in G\setminus M$, from which it follows that $G$ is a Frobenius group with complement $M$. We have that $G=N\rtimes M$ for $N$ nilpotent (by Thompson, say). In fact, in this case, $N$ must be elementary abelian.

See also on primitive group actions with abelian stabilizers .

This can be generalised, with some effort. For example, Theorem 1 of Deskins "A condition for the solvability of a finite group" (1961) is that a finite group with a maximal nilpotent subgroup of nilpotency class at most $2$ is soluble.

Since $|M|\leq 11$, this only leaves open the case when $M$ is isomorphic to $S_3$ or $D_{10}$.

The $S_3$ case is considered in the proof of Case II of Theorem in Wong "Determination of a Class of Primitive Permutation Groups" (1966). He shows that $G$ is soluble or $G\cong A_5$.

The $D_{10}$ case is follows from the proof of Theorem 3.3 in Quirin, "Primitive Permutation Groups with Small Orbitals" (1971). Again, he shows that $A_5$ is the only nonsoluble example.

Note that all these results predate the classification of finite simple groups.

If you are willing to rely on the CFSG, then you could look at Li and Zhang "The finite primitive groups with soluble stabilizers, and the edge-primitive s-arc transitive graphs" (2011) where all primitive groups with a soluble stabiliser have been classified. You could go through their lists and look for small stabilisers, and you could classify all examples with $|G|\leq 59n$.


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