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 '13 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 '13 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 '13 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 '13 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 '16 at 6:47

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 '13 at 20:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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