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10 Semidirect Products A semidirect product is a groups having two complementary subgroups one of which is normal. Definition 10.1. Two subgroups H, K ≤ G are called complementary (to each other) if 1. H ∩ K = 1 2. HK = G [HK = {hk | h ∈ H, k ∈ K}]1 For example, if G is a group of order pa q b then the p-Sylow subgroup P and q-Sylow subgroup Q are complementary. The complement of H is not unique, even if both H, K are normal. [E.g., take G = R2 as an additive group and let H, K be any two distinct lines through the origin.] Proposition 10.2. If H, K ≤ G are complementary then each element of G can be written uniquely in the form g = hk where h ∈ H, k ∈ K. Proof. We are assuming that G = HK so each g = hk. So we only need −1 the uniqueness. But g = h1 k1 = h2 k2 implies h−1 2 h1 = k2 k1 . This is in H ∩ K = 1 so h1 = h2 and k1 = k2 . Proposition 10.3. If K E G then any complementary subgroup H is iso≈ morphic to Q = G/K. Furthermore the isomorphism q : H − → Q is the restriction to H of the quotient map G ³ Q. Proof. The kernel of the map φ : H → Q induced by the quotient map is K ∩ H = 1. Thus φ is injective. To show that φ is onto take any element gK of Q = G/K. Then g = hk so gK = hkK = hK = φ(h) ∈ im(φ). Definition 10.4. G is a semidirect product of K by Q if K E G and there exists H ≤ G is a complement of K isomorphic to Q. By Proposition 10.3, Q ∼ =H∼ = G/K. Thus a semidirect product of K by Q is an extension of K by Q. Example 10.5. Sn is a semidirect product of An by Z/2 where the complement of An is given, e.g., by H = h(12)i. Example 10.6. The dihedral group D2n is a semidirect product of Z/n = hsi by Z/2 where H = hti is the complement of hsi. 1 Note that, in general, HK is not a subgroup of G unless either H or K is normal. For example H = h(12)i, K = h(234)i generate all of S4 but HK has only 6 elements. 1 Example 10.7. If n is odd, the cyclic group Z/2n is also a semidirect product of Z/n by Z/2 where the complement of K = 2Z/2n is the 2-Sylow subgroup of Z/2n. HW4.ex05: Prove that Q (the quaternion group of order 8) is not a semidirect product of Z/4 by Z/2. Theorem 10.8. G is a semidirect product of K by Q = G/K iff the quotient map q : G ³ Q splits, i.e., if there exists a homomorphism s : Q → G so that qs = idQ . [s is called a section of q.] Proof. The image of the section s, if it exists, is a complement for K so G is a semidirect product. Conversely, if G is a semidirect product, then K has a complement H and the inverse of the isomorphism φ : H → Q is a section s. Definition 10.9. Suppose that θ : Q → Aut(K) (x 7→ θx ) is a homomorphism. Then we say that a semidirect product G of K by Q realizes θ if the conjugation action of H = s(Q) on K is given by θ, i.e., γs(x) = θx for all x ∈ Q. The basic theorem about semidirect products is that they exist and are uniquely determined (up to isomorphism) by K, Q, θ. I will outline the usual proof and then explain a fancy categorical proof (using holomorphs and pullbacks) which although more complicated has a certain appeal. Definition 10.10. Given K, Q and θ : Q → Aut(K) let K oθ Q be the cartesian product K × Q with the following multiplication rule: (a, x)(b, y) = (aθx (b), xy). We need to check that this is a group operation. 1. (associativity) Use the fact that θ is a homomorphism (θxy = θx θy ): (a) ((a, x)(b, y))(c, z) = (aθx (b), xy)(c, z) = (aθx (b)θxy (c), xyz) (b) (a, x)((b, y)(c, z)) = (a, x)(bθy (c), yz) = (aθx (b)θx θy (c), xyz) 2. (left identity) (1, 1)(a, x) = (1θ1 (a), x) = (a, x) 3. (left inverse) (θz (a)−1 , z)(a, x) = (θz (a)−1 θz (a), zx) = (1, 1) if z = x−1 . Theorem 10.11. K oθ Q is a semidirect product of K by Q realizing θ. 2 Proof. We identify K with the set of all (a, 1) where a ∈ K. This is a normal subgroup of K oθ Q: (∗, x)(c, 1)(∗, x−1 ) = (∗θx (c)∗, xx−1 ) ∈ K H = {(1, x) | x ∈ Q} is a subgroup complementary to K and isomorphic to Q. Thus K oθ Q is a semidirect product of K by Q. It is also clear that this realizes the action θ. [Replace ∗ by 1 in the calculation above.] Theorem 10.12. Any semidirect product G of K by Q realizing θ is isomorphic to K oθ Q. Proof. By definition 10.9 there is a section s : Q → G of the quotient map q : G → Q so that θx = γs(x) . Let ψ : K oθ Q → G be given by ψ(a, x) = as(x). This is a homomorphism since ψ(a, x)ψ(b, y) = as(x)bs(y) = aγs(x) (b)s(x)s(y) = aθx (b)s(xy) = ψ(aθx (b), xy) and ψ is a bijection by Proposition 10.2. There is a “universal” semidirect product of K called the “holomorph” of K. Definition 10.13. For any group K its holomorph Hol(K) is defined to be the group of permutations of K generated by 1. automorphisms ψ : K → K 2. left translations Lg : K → K (x 7→ gx) In other words, ® Hol(K) = Aut(K), K ` where K ` = {Lg | g ∈ K}. Note that Aut(K) normalizes K ` since ψLg ψ −1 = Lψ(g) Consequently, K ` E Hol(K) and Hol(K) = K ` Aut(K). Theorem 10.14. Hol(K) ∼ = K oid Aut(K) where id : Aut(K) → Aut(K) is the identity homomorphism. Proof. The elements of Hol(K) can be written uniquely in the form Lg ψ and multiplication is given by: Lg ψLh φ = Lg Lψ(h) ψφ = Lgψ(h) ψφ which corresponds to the multiplication rule in K oid Aut(K) The semidirect product K oθ Q is now given by the “pull-back” construction: K oθ Q = {(x, f ) | x ∈ Q, f = Lg ψ ∈ Hol(K), θ(x) = ψ} 3