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.

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