Intuition on Hall subgroups and solvability

Question

I see there are many questions on Hall subgroups, but I can't find one that answers my question.

Let $G$ be a finite group. A subgroup $H<G$ is a Hall $\pi$-subgroup if its order is the product of some set of primes $\pi$ (with multiplicity), and $|H|$ and $[G:H]$ are relatively prime. Any such order of $|H|$ is a Hall factor.

Philip Hall proved that a finite group is solvable if and only if the group has subgroups of every Hall factor order. The "$\Rightarrow$" direction is proved in Hungerford, but "$\Leftarrow$" was deemed beyond the scope of that book.

This result sees quite deep, in that this characterization of solvable group seems to impose a lot less structure than does the definition of solvable group. From the definition of having a derived series that terminates trivially, a solvable group can be viewed as having abelian factors (quotient groups of the derived series) that can be assembled in a nonabelian way (via semidirect products and Frattini extensions). But at least to me, these two characterizations seem miles apart.

Consider another direction. Finite abelian groups are direct products of cyclic groups. They contain subgroups of every order that divides the group, and every subgroup is normal in every containing subgroup. This is very beautiful structure.

Nilpotent groups ("almost abelian") can be characterized as being isomorphic to the direct products of their Sylow subgroups. Now Sylow groups are $p$-groups, and these have very nice structure. So again we get that nilpotent groups have subgroups of every order that divides the group, but not every subgroup is normal (though there are many normality relationships). We still have lots of nice structure.

But in light of Hall's Theorem, we have that solvable groups need only satisfy what seem to be rather weak divisibility conditions. This seems quite a leap. The only intuition I have been able to grasp is that many Hall subgroups are relatively "large", as many of them have order equal to the product of all but 1 or 2 primes. But I am at a loss as to how this connects to the definition of solvability.

So to the actual question: Is there any intuition as to how the definition of solvable group connects to the Hall Theorem characterization of solvable group?

Answer 1

Both directions are proven in Rotman's book "An Introduction to Group Theory". The proof (which only uses the existence of $p$-complements), in broad strokes, proceeds as follows:

1) Proceed by induction on $|G|$ by supposing you have a "minimal criminal": a group $G$ of smallest order that violates the claim (has all such subgroups but is not solvable).

2) Argue that $G$ must be simple.

3) Proceed through a fair amount of technical work to derive a contradiction by constructing a proper normal subgroup. Burnside's Theorem on the solvability of groups with two prime factors makes an appearance.

Obviously, the third part is the big thing, and while Rotman does it all in the span of a single paragraph and about 1/3 of a page, it's reasonably excluded from Hungerford.

Here is the proof of the second part:

If $G$ has a nontrivial normal subgroup $N$, and if $H$ is any Hall $p'$-subgroup of $G$, then checking orders shows that $H\cap N$ is a Hall $p'$-subgroup of $N$ and $HN/N$ is a Hall $p'$-subgroup of $G/N$. So both $N$ and $G/N$ satisfy the hypotheses, and so by minimality of $G$ both $N$ and $G/N$ are solvable, whence $G$ is solvable, a contradiction.

So the "intuition" here, perhaps, is that possession of (all) such subgroups is inherited by normal subgroups and quotients thereof, which eliminates the non-simple case. The third part then says that no simple group can be "separated" well enough to have this property (by a more complicated argument), completing the argument.