# Question about Hall Theorem in group theory

I am reading about Hall Theorem on Wikipedia, which essentially says that if $G$ is finite and solvable and $\pi$ is a set of primes. Then $G$ has a Hall $\pi$-subgroup. I then look at the Sylow system - a set of Sylow $p$-subgroups $S_p$ for each prime $p$ such that $S_p S_q = S_q S_p$ for all $p$ and $q$.

If we have a Sylow system, then the subgroup generated by the groups $S_p$ for $p$ in $\pi$ is a Hall $\pi$-subgroup. A more precise version of Hall's theorem says that any solvable group has a Sylow system.

I am stuck on justifying the bold paragraph. Why does the subgroup generated by the groups $S_p$ for $p$ in $\pi$ is a Hall $\pi$-subgroup and why does Hall's theorem imply that a solvable group has a Sylow system?

• For the first question, in general if $H$ and $K$ are subgroups of a group $G$, then $HK$ is a subgroup if and only if $HK=KH$, so it follows from $S_pS_q=S_qS_p$. For the second question, I think you would need to read the proof of Hall's theorem. – Derek Holt Dec 3 '17 at 19:20

If you look at Hall's original paper "On the Sylow systems of a soluble group", he effectively proves the following:

Suppose that a finite group $G$ has a Sylow $p$-complement for every prime $p$. Then $G$ has a Sylow system, that is, it has a permutable family of subgroups forming a Boolean algebra, containing one Sylow $p$-subgroup for each $p$.

The proof is to just take a set $\mathcal{C}$ of Sylow $p$-complements, one for each $p$ (chosen independently of each other), and show that the set $\mathcal{S}$ of intersections of subsets of $\mathcal{C}$ forms a Sylow system. No solubility or other ingenuity is required at this stage.

You can also easily show that in any finite group, a Hall $p'$-subgroup is a Sylow $p$-complement and vice versa. So a finite group has a Sylow system if and only if it has a Hall $p'$-subgroup for every prime $p$.

Both statements above rest on the equation

$$|KL| = \frac{|K||L|}{|K \cap L|};$$

it is valid for any two finite subgroups $K$ and $L$ of a group, without assuming they are permutable.

Where solubility does come into the proof of Hall's theorem is to show Sylow $p$-complements exist, and to show that any two choices of the set $\mathcal{C}$ of $p$-complements are in the same $G$-orbit (where $G$ acts on the collection of sets of subgroups by conjugation in the obvious way).