This is an extension of [this MSE](https://math.stackexchange.com/q/268091/12952) question, in which I asked whether there was a counterexample to the following statement,



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



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> **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 $|G|$.
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*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](http://groupprops.subwiki.org/wiki/Three_solvable_subgroups_of_pairwise_coprime_indices_implies_solvable), 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)?



An extension of the converse to Hall's theorem.
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.


