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

Suppose that a finite group $G$ contains a Hall $\{p,q\}$-subgroup for every pair of prime divisors $p,q$ of $|G|$. Does it follow that $G$ is solvable?

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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.

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Thanks. I'm going to digest this a bit and see if I can construct a proof. –  Alexander Gruber Dec 31 '12 at 20:51
There are still some parts of this I don't understand. I am quite familiar with the ZJ theorem and the structure of $H$ lacking elements of order $pq$ - those parts I get fully. Before the edit, you begin with "P does not normalize any nontrivial $p$'-group" - is that a property of groups of lie type in odd characteristic? If so where can I read more about this specifically? In your later edit, why is it a problem that $O_p(H)\not= 1$? I thought Borel subgroups normalized $P$ and were then contained in parabolic subgroups, so isn't that OK? –  Alexander Gruber Jan 17 '13 at 3:55
Yes, that is a property of Lie type groups. Try Solomon Lyons, etc. Later, I was saying that there is likely to be prime $q$ which does not divide the order of any parabolic, and was assuming $q$ was one such. You need to look at the various tori to find such a prime. Eg, in GL(n,p), take a prim $q$ which divides $p^{n}-1$ but not $p^{i}-1$ for $i <n.$ Such a prime $q$ almost always exists, and $q$ does not divide the order of any parabolic. –  Geoff Robinson Jan 17 '13 at 4:16