Is there a counterexample to this weakened converse of Hall's theorem?

Solution 1:

I think this may be known (using the classification of finite simple groups), but a quick search of Mathscinet and some googling did not really help. The question easily reduces to simple groups. I'll summarize a little bit of evidence to suggest that there is no counterexample. Let $G$ be a simple group of Lie type in odd characteristic $p.$ Then a Sylow $p$-subgroup $P$ of $G$ normalizes no non-trivial $p^{\prime}$ subgroup. Suppose that for each prime $q \neq p,$ there is a Hall $\{p,q\}$-subgroup of $G$. Then for each prime $q \neq p,$ there is a Sylow $q$-subgroup $Q$ of $G$ such that $PQ = QP.$ Now $PQ$ is a solvable group, and we have $O_{q}(PQ) = 1.$ Hence, by Glauberman's theorem, we have $ZJ(P) \lhd PQ.$ Then $ZJ(P) \lhd G$ as $ZJ(P)$ is normalized by a Sylow $r$-subgroup of $G$ for each prime divisor $r$ of $|G|$. This is a contradiction.

Feit and Thompson showed that it is rare for the symmetric group (and, implictly, the alternating group) to have a Hall $\{2,3\}$-subgroup. That leaves sporadic groups and groups of Lie type in characteristic $2$ to consider. The latter should succumb to a more sophisticated version of the argument above- possible using a result of Stellmacher on a $ZJ$-type subgroup for $2$-constrained $S_{4}$-free groups to show that there would be a $2$-local subgroup of $3$-power index. For sporadic groups, it's a question of a smal amount of checking (assuming there really is no counterexample).

Later edit: In view of the bounty (in case it helps others), I make few more remarks.

If our group $G$ is a simple group of Lie type in characteristic $p,$ then every prime $q \neq p$ must divide the order of some parabolic subgroup of $G$ (this seems likely to happen rarely, if at all, for all primes $q$). For if $H$ is a Hall $\{p,q \}$-subgroup of $G,$ then $H$ is solvable. If $O_{p}(H) \neq 1,$ then $H$ is contained in a maximal $p$-local subgroup of $G,$ which is a parabolic. If $O_{p}(H) = 1,$ and $q$ does not divide the order of any parabolic, then there is no element of order $pq$ in $G.$ Then $H$ (and also $G$) has a cyclic or generalized quaternion Sylow $p$-subgroup. As $G$ is simple, $p$ must be odd, and $G$ must be ${\rm PSL}(2,p)$, which never has a Hall $\{2,p\}$-subgroup ( for $p>3$, which the case here), a contradiction.

In general, if our simple group $G$ has a Hall $\{p,q\}$-subgroup $H$, but no element of order $pq,$ then either $G$ ha cyclic Sylow $q$-subgroups or cyclic Sylow $p$-subgroups. For $H$ is solvable, ad we may label so that $O_{q}(H) \neq 1.$ Then $H$ (and so $G$) must have cyclic or generalized quaternion Sylow $p$-subgroups ( so $p$ must odd, and they must be cyclic, as $G$ is simple). This should be useful in eliminating sporadic groups.