Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 8 down vote accepted

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.