This is an extension of this MSE question, in which I asked whether there was a counterexample to the following statement,

Conjecture. If a finite group $G$ contains a $\lbrace p,q \rbrace$-Hall subgroup for every pair of primes $p$ and $q$ dividing $|G|$, then $G$ is solvable.

which is a refinement of the converse to Hall's theorem,

Theorem (Hall). Denote by $\pi(G)$ the set of prime divisors of $|G|$. Then $G$ is solvable if and only if $G$ contains a $\pi$-Hall subgroup for every subset $\pi$ of the prime divisors of $|G|$.

Edit: As requested, we call $H\leqslant G$ a $\pi$-Hall subgroup if $|H|$ and $[G:H]$ are coprime and $p$ divides $|H|$ for every $p\in \pi$. So, Hall subgroups are a generalization of Sylow subgroups for multiple primes.

I received a great answer from Geoff Robinson, who thinks there probably isn't counterexample, and proposed to check it case-by-case using the classification of finite simple groups. I am still digesting his answer, however this led me to wonder whether (assuming the statement is true) there is a proof that does not rely on the classification theorem.

My original idea, before I posted the question on MSE, was to was to show that this implies the hypothesis to the to the original converse - that is, $G$ contains a $\pi$-Hall subgroup for every $\pi\subseteq \pi(G)$ iff it contains a $\{ p,q \}$-Hall subgroup for each $p,q\in \pi(G)$ - but I haven't found any way to make this work yet. It has also been suggested to me to try to mimic the original inductive proof, which (for $|\pi(G)|\geq 3$) makes use of this lemma, but again I do not see how to put it together.

So, my question is, is this conjecture provable without using the classification of finite simple groups?

If it is false, is there a counterexample of a non-solvable group with $\lbrace p,q\rbrace$-Hall subgroups for every pair of primes (which is thus missing some other Hall subgroup, e.g. a $\lbrace p,q,r \rbrace$-Hall subgroup)?

  • 2
    $\begingroup$ @Alexander, could you add a definition of a $p,q$-Hall subgroup into the question? $\endgroup$
    – Nick Gill
    Jan 18, 2013 at 10:00
  • $\begingroup$ @NickGill Done. I understand the confusion - I am not sure why the curly braces are not showing up around $p,q$. $\endgroup$ Jan 18, 2013 at 17:48
  • $\begingroup$ @Alexander: Curly braces are tricky in LaTeX, since in math mode they are sometimes needed but when unneeded are just ignored. Typically you have to type \{ and \} as when specifying a set by some condition. $\endgroup$ Jan 19, 2013 at 0:15
  • 2
    $\begingroup$ As an outsider I find the discussion interesting, but I'm always a bit concerned about the existence of multiple mathematicians having the frequent British/American surname Hall while working in overlapping areas of group theory (the American M. Hall has a son J. Hall who also works with finite groups, though fortunately the British P. Hall was a "lifelong bachelor" as the British prefer to discreetly phrase it in obituaries). Is there are clearer way to talk about "Hall theorems"? $\endgroup$ Jan 19, 2013 at 16:24
  • $\begingroup$ That conjecture is due to Philip Hall. I believe that for its proof it suffices to check it for simple groups. $\endgroup$
    – yakov
    Jun 23, 2016 at 6:47

1 Answer 1


For the record, I believe that P. Hall proved that if $|G|$ has $n$ prime divisors, then $G$ is solvable if and only if $G$ has $n$ Sylow subgroups $P_{1},P_{2}, \ldots ,P_{n},$ one for each prime divisor, such that $P_{i}P_{j} = P_{j}P_{i}$ for $1 \leq i,j \leq n.$ You are asking whether the pairwise permutability condition can be dropped. The proof of the more difficult direction Hall's Theorem is something like the following, given Burnside's $p^{a}q^{b}$-theorem. I have forgotten Hall's proof, so have had to concoct a proof which is largely the same as Hall's except that I need to invoke Glauberman's $ZJ$-theorem, which Hall did not require. For suppose that $G$ has such a set of permutable Sylow subgroups, and we wish to prove that $G$ is solvable. Then we may suppose that $n \geq 3,$ otherwise Burnside's $p^{a}q^{b}$-theorem yields the desired result. By induction, for $1 \leq i \leq n,$ $G$ has a solvable subgroup $H_{i}$ with $G = H_{i}P_{i} = P_{i}H_{i}$ and $H_{i} \cap P_{i} = 1$ (we may take $H_{i} = \prod_{j \neq i} P_{j}$ which is a group by the permutability condition, and has order $[G:P_{i}]).$ We may also suppose that $p_{1} \geq 5,$ and we do. If $P_{1}$ normalizes a non-trivial subgroup $N_{1}$ of $H_{1},$ then we have $\cap _{g \in G} H_{1}^{g}$ = $\cap_{x \in P_{1}} H_{1}^{x} \geq N_{1},$ so $G$ has a non-trivial solvable normal subgroup $K,$ and an induction argument shows that $G/K$ is solvable. Hence for $2 \leq j \leq n,$ we have $O_{p_{j}}(P_{1}P_{j}) = 1.$ Since $P_{1}P_{j}$ is solvable, and $p_{1} \geq 5,$ we have $ZJ(P_{1}) \lhd P_{1}P_{j}$ for each such $j,$ by Glauberman's $ZJ$-theorem. But then $ZJ(P_{1}) \lhd G,$ since it is normalized by each of $P_{1},P_{2}, \ldots ,P_{n}.$ Again, and induction argument shows that $G/ZJ(P_{1})$ is solvable. But I emphasize that this proof requires pairwise permutable Sylow subgroups, and the hypotheses of this question do not require that.

  • 1
    $\begingroup$ I don't believe this answer deserved the bonus to be honest. I do think that the MSE answer did contain enough information to settle the matter, allowing for the use of CFSG. I am not sure how easy it would (or more likely, would not) be to resolve the question without CFSG $\endgroup$ Feb 1, 2013 at 20:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .