7 Hall Subgroups

Deﬁnition 7.1. A Hall subgroup of a ﬁnite group G is a subgroup H so that |H| and |G/H| are relatively prime.

Sylow subgroups are examples of Hall subgroups. There is another way to say the deﬁnition. Suppose that π is a set of primes. Then a π-group is a ﬁnite group whose order is a product of powers of primes in π. For example a 20-group is a group of odd order since 20 denotes the set of odd primes. A Hall subgroup of G is a π-subgroup with π0 index for some π.(π0 is the complementary set of primes.) Since the order of a subgroup divides the order of the group we have the following trivial observation.

Lemma 7.2. Every Hall π-subgroup of G is a maximal π-subgroup of G.

Theorem 7.3. If G is a solvable group of order |G| = ab where a, b are relatively prime then G has a subgroup of order a and any two such subgroups are conjugate.

Remark 7.4. This theorem can be rephrased as follows. Let π be the set of primes that divide a. Then we want to show that any ﬁnite solvable group G contains a Hall π-subgroup and any two Hall π-subgroups of G are conjugate.

Proof. Let H be a minimal normal subgroup of G. Then H is elementary abelian so |H| = pm. Case 1.(p ∈ π, so pm divides a.) By induction, G/H contains a Hall π-subgroup A/H where A is necessarily a Hall π-subgroup of G. If A1 is another Hall π-subgroup of G then H ≤ A1 since otherwise A1H is a larger π-subgroup of G. By induction A1/H is conjugate to A/H. But this implies −1 that A1 is conjugate to A. [If A1/H = g(A/H)g where g = gH then −1 A1 = gAg .] This proves the following lemma:

Lemma 7.5. If H E G and A, A1 are subgroups of G containing H then A is conjugate to A1 iﬀ A/H is conjugate to A1/H. Case 2.(p ∈ π0, so pm|b, and suppose pm < b.) In this case G/H has a Hall π-subgroup B/H where H E B ≤ G of order |B| = apm. Since pm < b, |B| < |G| so B contains a Hall π-subgroup A. Since |A| = a this is also a Hall π-subgroup of G. If A1 is another Hall π-subgroup of G then A1H/H is another Hall π-subgroup of G/H so it is conjugate to B/H. Therefore A1H is conjugate to B (by the lemma) so A1 is conjugate to a subgroup A2 of B. Since |B| < |G|, any two Hall π-subgroups of B are conjugate so A2 is conjugate to A. Case 3.(pm = b) Thus H is a normal p-Sylow subgroup of G. Also we may assume that H is the unique minimal normal subgroup of G since any other minimal normal subgroup would fall under Case 1.

1 Since G/H is solvable its minimal normal subgroup K/H is also elementary abelian of degree, say qn, making K a normal subgroup of G of order pmqn. Let Q ≤ K be a q-Sylow subgroup of K. Then K = QH and Q ∩ H = 1. We claim that N(Q) is the Hall subgroup of G of order a that we are looking for. To see this note that KN(Q) = G by the Frattini argument. But KN(Q) = HQN(Q) = HN(Q). Therefore it suﬃces to show that H ∩ N(Q) = 1. We show this in two steps:

(i) H ∩ N(Q) ≤ Z(K)

(ii) Z(K) = 1

(i) Suppose that z ∈ H ∩ N(Q). Since z ∈ H and H is abelian, z centralizes H. To see that z also commutes with any x ∈ Q note that

[z, x] = (zxz−1)x−1 = z(xz−1x−1) ∈ Q ∩ H = 1

(ii) Since Z(K) is characteristic in K E G it follows that Z(K) E G. If Z(K) were nontrivial it would have to contain a minimal normal subgroup, a.k.a., H. This would imply that Q is normal in K. But a normal Sylow subgroup is characteristic. Therefore, Q would be a normal subgroup of G contradicting the uniqueness of H. It remains to show that all Hall subgroup of G of order a are conjugate. Suppose that A is another subgroup of G of order |A| = a. Then the order of AK (K as above) is divisible by a and pm so AK = G.

G AK A ∩ K

Thus pm = |G : A| = |K : A ∩ K| so |A ∩ K| = qn, i.e., A ∩ K is a q-Sylow subgroup of K. Thus A ∩ K is conjugate to Q. Therefore it normalizer is conjugate to N(Q) and so has index pm. But A normalizes A ∩ K and also has index pm so A = N(A ∩ K) which is conjugate to N(Q) as required.

HW3.ex 05 (5.32(i) in Rotman 4th ed) If G is a ﬁnite solvable group show that any π-subgroup of G is contained in a Hall π-subgroup. (Equivalently, show that any maximal π-subgroup is a Hall π-subgroup.) [Hint: divide into three cases as in the proof of Theorem 7.3.]

HW3.ex 06 (5.32(ii) in Rotman 4th ed) Let π = {3, 5}. Show that S5 has maximal π-subgroups of order 3 and 5. Conclude that S5 is not solvable.