## In a finite solvable group, Hall subgroups exist and are conjugate

Let $\pi$ be any set of primes. A subgroup $H$ if a finite group $G$ is called a Hall $\pi$-subgroup if every prime dividing $|H|$ is in $\pi$ and no prime in $\pi$ divides $[G:H]$. (If $\pi = \{p\}$, then Hall $\pi$-subgroups and Sylow $p$-subgroups are the same.) Prove the following generalization of Sylow’s Theorem for solvable groups.

If $G$ is a finite solvable group then for every set $\pi$ of primes, $G$ has a Hall $\pi$-subgroup and moreover any two Hall $\pi$-subgroups are conjugate in $G$.

First we prove a lemma.

Lemma: Let $G$ be a finite solvable group and $H \leq G$ a normal $p$-subgroup, with $p$ a prime. Let $\pi$ be a set of primes with $p \in \pi$. Suppose further that Hall $\pi$-subgroups of $G/H$ exist, and that any two Hall $\pi$-subgroups of $G/H$ are conjugate. Then Hall $\pi$-subgroups of $G$ exist and any two Hall $\pi$-subgroups of $G$ are conjugate. Proof: Let $Q/H \leq G/H$ be a Hall $\pi$-subgroup. Suppose that $q$ is a prime dividing $|Q|$. If $q=p$, then $q \in \pi$. If $q \notin \pi$, then $q$ divides $|Q/H|$, so that $q \in \pi$. Moreover, by the Third Isomorphism Theorem, no prime in $\pi$ divides $[G:Q] = [G/H : Q/H]$. Thus $Q \leq G$ is a Hall $\pi$ subgroup. Suppose now that $R \leq G$ is a Hall $\pi$-subgroup. Using this previous exercise, $RH/H \leq G/H$ is a Hall $\pi$-subgroup. By the induction hypothesis, $RH/H$ is conjugate to $Q/H$. In particular, $|RH| = |Q|$. Since $Q$ and $R$ are Hall $\pi$-subgroups, $|Q| = |R|$. Using $|RH| = |R||H|/|R \cap H|$, we have $|H| = |R \cap H|$; since $G$ is finite, this implies $R \cap H = H$, so that $RH = R$, and in fact $Q$ is conjugate to $R$. $\square$

Now to the main result.

Let $\pi$ be a fixed set of primes. We proceed by induction on the breadth of $G$; that is, the number of prime divisors of $|G|$ including multiplicity.

For the base case, let $G$ be a finite solvable group of breadth 1. Then $G \cong Z_p$ for some prime $p$. If $p \in \pi$, then $G$ is the unique Hall $\pi$-subgroup. If $p \notin \pi$, then $1$ is the unique Hall $\pi$-subgroup.

For the inductive step, suppose now that for some integer $k \geq 1$, every finite solvable group of breadth at most $k$ satisfies the conclusion. Let $G$ be a finite solvable group of breadth $k+1$.

If $G$ is simple, then since $G$ is solvable, we have $G^\prime = 1$. Now $G$ is an abelian simple group, so that $G \cong Z_p$ for some prime $p$. But then the breadth of $G$ is 1, a contradiction.

Suppose now that $G$ is not simple; then nontrivial normal subgroups exist. Let $M \leq G$ be a minimal nontrivial normal subgroup. By this previous exercise, $M$ is an elementary abelian $p$-group for some prime $p$.

If $p \in \pi$, then by induction on $G/M$ and using the lemma, Hall $\pi$-subgroups of $G$ exist and any two Hall $\pi$ subgroups are conjugate.

Suppose now that $p \notin \pi$. $G/M$ is a finite solvable group of breadth at most $k$, so that by the induction hypothesis, there exists a Hall $\pi$-subgroup $R/M \leq G/M$. If $q$ divides $|R|$ and $q \neq p$, then $q$ divides $|R/M|$, and we have $q \in \pi$. Thus $|R| = p^an$, where $n$ is a product of elements in $\pi$. Moreover, we can see that Hall $\pi$-subgroups of $G$ and $G/M$ have the same order, so that $n$ is the order of a Hall $\pi$-subgroup of $G$ and $|M| = p^a$. (Note that $p$ does not divide $|R/M|$.)

If $R \leq G$ is proper, then $R$ is a finite solvable group of breadth at most $k$; by the induction hypothesis, there exists a Hall $\pi$-subgroup $T$ of $R$, and thus one of $G$. Moreover, if $S \leq G$ is a Hall $\pi$-subgroup, then $SM/M \leq G/M$ is a Hall $\pi$-subgroup, so that $SM$ is conjugate to $TM$. Since $S \cap M = T \cap M = 1$, we have $S \times M$ conjugate to $T \times M$. Thus $S$ is conjuage to $T$.

Suppose now that $R \leq G$ is not proper; that is, $R = G$. In this case, $M \leq G$ is in fact a Sylow $p$-subgroup.

Now $G/M$ is a finite solvable group. If $G/M$ is simple, then $(G/M)^\prime = 1$, so that $G/M$ is abelian; thus $G/M \cong Z_q$ for some prime $q$. Now the Sylow $q$-subgroups of $G$ and the Hall $\pi$-subgroups of $G$ coincide; by Sylow’s theorem, Hall $\pi$ subgroups exist and are conjugate. If $G/M$ is not simple, then nontrivial normal subgroups exist. Let $N/M \leq G/M$ be a minimal nontrivial normal subgroup. By this previous exercise, $N/M$ is an elementary abelian $q$-group for some prime $q \in \pi$; say $|N/M| = q^b$. In particular, note that $q \neq p$.

Now let $Q \in \mathsf{Syl}_q(N)$.

If $Q \leq G$ is normal, then by the induction hypothesis and the lemma, Hall $\pi$-subgroups of $G$ exist and are conjugate.

Suppose $Q \leq G$ is not normal. Now $N \leq G$ is normal and $Q \leq N$ is a Sylow subgroup. By Frattini’s Argument, we have $G = N \cdot N_G(Q)$ and that $[G:N]$ divides $|N_G(Q)|$. Note that $G = N \cdot N_G(Q)$ $= MQ \cdot N_G(Q)$ $= M \cdot N_G(Q)$. Now by the remarks after the proof of the Second Isomorphism Theorem, we see that $[G : N_G(Q)] = [M : M \cap N_G(Q)] = p^t$ is a power of $p$. In particular, $n$ divides $|N_G(Q)|$, and moreover every Hall $\pi$-subgroup of $N_G(Q)$ is a Hall $\pi$-subgroup of $G$. Since $N_G(Q)$ is proper in $G$, this subgroup has breadth at most $k$, and by induction a Hall $\pi$-subgroup $T$ of $N_G(Q)$, hence $G$, exists. Now suppose $U \leq G$ is a Hall $\pi$-subgroup. Now $U \leq N_G(Q)$ by Lagrange, so that (again by induction) $U$ is conjugate to $T$.