7 Hall subgroups Definition 7.1. A Hall subgroup of a finite group G is a subgroup H so that jHj and jG=Hj are relatively prime. Sylow subgroups are examples of Hall subgroups. There is another way to say the definition. Suppose that ¼ is a set of primes. Then a ¼-group is a finite group whose order is a product of powers of primes in ¼. 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 ¼-subgroup with ¼0 index for some ¼.(¼0 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 ¼-subgroup of G is a maximal ¼-subgroup of G. Theorem 7.3. If G is a solvable group of order jGj = 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 ¼ be the set of primes that divide a. Then we want to show that any finite solvable group G contains a Hall ¼-subgroup and any two Hall ¼-subgroups of G are conjugate. Proof. Let H be a minimal normal subgroup of G. Then H is elementary abelian so jHj = pm. Case 1.(p 2 ¼, so pm divides a.) By induction, G=H contains a Hall ¼-subgroup A=H where A is necessarily a Hall ¼-subgroup of G. If A1 is another Hall ¼-subgroup of G then H · A1 since otherwise A1H is a larger ¼-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 2 ¼0, so pmjb, and suppose pm < b.) In this case G=H has a Hall ¼-subgroup B=H where H E B · G of order jBj = apm. Since pm < b, jBj < jGj so B contains a Hall ¼-subgroup A. Since jAj = a this is also a Hall ¼-subgroup of G. If A1 is another Hall ¼-subgroup of G then A1H=H is another Hall ¼-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 jBj < jGj, any two Hall ¼-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 elemen- tary 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 2 H \ N(Q). Since z 2 H and H is abelian, z centralizes H. To see that z also commutes with any x 2 Q note that [z; x] = (zxz¡1)x¡1 = z(xz¡1x¡1) 2 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 jAj = a. Then the order of AK (K as above) is divisible by a and pm so AK = G. G AK A \ K Thus pm = jG : Aj = jK : A \ Kj so jA \ Kj = 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 ¼-subgroup of G is contained in a Hall ¼-subgroup. (Equivalently, show that any maximal ¼-subgroup is a Hall ¼-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 ¼ = f3; 5g. Show that S5 has maximal ¼-subgroups of order 3 and 5. Conclude that S5 is not solvable. 2.