Hall's Theorem

Definition For $\pi$ a set of primes, a $\pi$-group is a group whose orders is a product of prime powers taken from $\pi$.

Definition A Hall $\pi$-subgroup $H$ is a $\pi$-subgroup of $G$ where the index $|G:H|$ is a product of primes not from $\pi$.

Theorem (Hall - 1928) Let $G$ be a solvable group and $\pi$ a set of primes then

1. $G$ has a Hall $\pi$-subgroup.
2. Any two Hall $\pi$-subgroups are conjugate.
3. Any $\pi$-subgroup of $G$ is contained in a Hall $\pi$-subgroup.

proof: We will show by induction on $|G|$ that:

• (a) $G$ has a $\pi$-subgroup $H$
• (b) Any $\pi$-subgroup $L$ of $G$ is contained in a conjugate of $H$.

 (b) implies (2) and (a,b) together imply (1) since any Sylow $p$-subgroup is a $\pi$-subgroup (when $p \in \pi$) we can ensure that the order $|G:H|$ has no powers of $p$ in it.

The base case is trivial. Let $|G| = mn$ where $m$ is a product of powers of primes from $\pi$ and $n$ contains no primes from $\pi$. The case $m=1$ is trivial. We split the proof into two cases:

(Case A.a) There is a minimal normal subgroup $M$ with order dividing $m$. By earlier results $M$ is elementary abelian and a $p$-group for some $p$. Write $|M| = p^a$ so that $|G/M| = m_1 n$ where $m_1 = \frac{m}{p^a}$. By induction $G/M$ has a subgroup - which by isomorphism theorems, is of the form $H/M$ - with order $m_1$; then $H$ is a subgroup of $G$ with order $m$ this proves part (a).
(Case A.b) Given a $\pi$-subgroup $L$ we have $L M \le G$ (since $L$ is a $\pi$-group and $M$ is normal in $G$ todo: make sure this is right) and by isomorphism theorems $LM/M \simeq L/(L \cap M)$ so $LM/M$ is a $\pi$-subgroup of $G/M$. By induction it lies in some conjugate of $H/M$, say $L M/M \le (H/M)^{Mg} = H^g/M$ and so we have $L \le LM \le H^g$ gving (b).

(Case B) In this case there does not exist a minimal normal subgroup of order dividing $m$. Let $M$ be an minimal normal subgroup then it's elementary abelian and it must be a $q$-group for some prime $q | n$ (i.e. $q$ is a prime not in $\pi$). Write $|M|=q^b$ so $|G/M| = \frac{mn}{q^b} = mn_1$ with $n_1 = \frac{n}{q^b}$ and we split this into two more cases:

(Case Bi.a) Assume $n_1 > 1$, then by induction $G/M$ has a subgroup $K/M$ of order $m$ (because the index $|G/M:K/M|$ is $n_1$), $|K| = mq^b < mn$ so by induction again $K$ has a $\pi$-subgroup $H$ of order $m$ as required for (a)
(Case Bi.b) Given $L$ as before $LM/M$ is a $\pi$-subgroup of $G/M$ so by induction it lies in some conjugate of $K/M$, say $LM/M \le (K/M)^{Mg} = K^g/M$ then $L^{g-1}$ is a $pi$-subgroup of $K$ so by induction it lies in some conjugate of $H$, say $L^{g^{-1}} \le H^k$ whence $L \le H^{k g^{-1}}$ as required for (b).

(Case Bii.) The final case is when $G$ has no minimal normal subgroups of order dividing $m$ and $n_1=1$ (refer back to Case B for definition of $n_1$). Let $N/M$ be a minimal normal subgroup of $G/M$ then we know it is an elementary abelian $p$-group for some $p|m$, say $|N/M| = p^a$ then $N \unlhd G$ and $|N| = p^a q^b$.
Let $P$ be a Sylow $p$-subgroup of $N$ and write $H=N_G(P)$ by the Frattini argument $G=HN$ since $N=PM$ and $P\le H$ so $G=HPM=HM$. Let $J = H \cap M$ then $J \unlhd HM = G$ by minimality of $M$ we must have $J=1$ or $J=M$:
(Case Bii.J=M) This case is easily disposed of. $H \cap M = M$ so $M \le H$ so $G = HM = H = N_G(P)$ so $P \unlhd G$ (by normalizer facts) and some subgroup of $P$ is then a minimal normal subgroup of of $G$ whose order does divide $m$, contradiction.
(Case Bii.J=1.a) In this case $|H \cap M| = 1$ and thus $mq^b = |G| = |H||M| = q^b |H|$ so $|H| = m$ giving (a).
(Case Bii.J=1.b) Given $L$ we have (and we can form this because $M$ is a $q$ group and $L$ is a $p$-group) $LM = LM \cap G = LM \cap HM = (LM \cap H)M$ (by the ABC lemma). Now $LM \cap H$ is a $\pi$-subgroup of $LM$ and $$|LM:LM\cap H| = |(LM\cap H)M|/|LM\cap H| = |M|/|LM \cap H \cap M| = |M|$$ therefore it is in fact a Hall $\pi$-subgroup so if $LM<G$ we are done by induction. It only remains to deal with the case where $LM=G$: in this case since $L \cap M = 1$ by coprime orders, $|G|=|L||M|$ implying $|L| = m$. Since $M \le N$ (since $M \unlhd N$ from the fact the quotient makes sense) $LN = G$ and thus $|L \cap N| = |L||N|/|G| = p^a$ so by Sylow's theorem $L \cap N$ is conjugate to $P$ ($L = P^n$ for some $n \in N$). $L \cap N \unlhd L$ since $N \unlhd G$ so $$L \le N_G(L \cap N) = N_G(P^n) = N_G(P)^n = H^n$$ giving (b).

Thus we have (a) and (b) for all groups, this implies (i) and (ii). For (iii) let $K$ be any Hall $\pi$-subgroup of $G$ by (b) we know that $K \le H^g$ for some $g \in G$, since $|H|=|K|$, $K=H^g$.