SEMIDIRECT PRODUCTS OF GROUPS KEITH CONRAD 1. Introduction For two groups H and K, the most basic construction of a group that contains copies of H and K as subgroups is the direct product H ×K, where the group law is componentwise: (h; k)(h0; k0) = (hh0; kk0). Since (h; 1)(h0; 1) = (hh0; 1) and (1; k)(1; k0) = (1; kk0), we can embed H and K into H × K \on the axes" by h 7! (h; 1) and k 7! (1; k) for h 2 H and k 2 K. This lets us think of H × K as a group generated by subgroups isomorphic to H and K. The significance of direct products is that some groups not initially constructed as direct products might decompose into a direct product of smaller groups, so we get a kind of factorization of the group. For example, every finite abelian group is isomorphic to the direct product of its Sylow subgroups. Other groups are not isomorphic to a direct product of smaller groups, such as Sn for n ≥ 3. There is a group construction using two groups H and K that is more subtle (more cunning?) than H × K, called a semidirect product. Interesting features include: (i) it may be nonabelian even if H and K are abelian (note H × K is abelian if and only if H and K are abelian) and (ii) there can be multiple nonisomorphic semidirect products using the same two groups. 2. Recognizing direct products When H and K are embedded into H × K in the standard way, here are three properties of the images of H and K inside H × K: • they generate H × K:(h; k) = (h; 1)(1; k), • they intersect trivially: (h; 1) = (1; k) =) h = 1; k = 1, • they commute elementwise: (h; 1)(1; k) = (1; k)(h; 1). This can be turned into the following \recognition theorem" for a group G to look like a direct product of two subgroups H and K: we have H × K =∼ G by the specific mapping (h; k) 7! hk precisely when the above conditions hold. Theorem 2.1. Let G be a group with subgroups H and K where (1) G = HK; that is, every element of G has the form hk for some h 2 H and k 2 K, (2) H \ K = f1g in G, (3) hk = kh for all h 2 H and k 2 K. Then the map H × K ! G by (h; k) 7! hk is an isomorphism. Proof. Let f : H × K ! G by f(h; k) = hk. This is a homomorphism: f((h; k)(h0; k0)) = f(hh0; kk0) = hh0kk0 and f(h; k)f(h0; k0) = (hk)(h0k0) = h(kh0)k0 = h(h0k)k0 = hh0kk0; where we use (3) to know that h0 and k commute. 1 2 KEITH CONRAD To show f is injective, we check its kernel is trivial: if f(h; k) = 1 then hk = 1 in G, so h = k−1 2 H \ K. By (2), H \ K = f1g, so h = 1 and k = 1. The function f is surjective by (1). There are many groups that decompose into a product of two subgroups that fit the first and second conditions of Theorem 2.1 but not the third one. Example 2.2. In G = Aff(R), let 1 y x 0 H = : y 2 R =∼ R;K = : x 2 R× =∼ R×: 0 1 0 1 Since x y 1 y x 0 = 0 1 0 1 0 1 | {z } | {z } 2H 2K we have G = HK, and clearly H \ K is trivial, but matrices in H and in K often do not commute with each other. You can find your own such matrices (nearly any random choice will work), but also observe that if elements of H and of K always commute with one another then G =∼ H × K by Theorem 2.1, but G =6∼ H × K since H × K is abelian (H and K are abelian) while G is nonabelian. x 0 × Example 2.3. In G = GL2(R), let H = SL2(R) and K = f( 0 1 ): x 2 R g. For each ∆ 0 g 2 G, the number ∆ = det g is nonzero and the matrix has determinant ∆, so 0 1 ∆ 0−1 ∆ 0 g = g · : 0 1 0 1 | {z } | {z } 2H 2K ∼ × Thus GL2(R) = HK, and easily H \ K = fI2g. It turns out that GL2(R) =6 SL2(R) × R . For example, GL2(R) has infinitely many elements of order 2, such as reflections across any 2 line in R through the origin, but SL2(R) has just one element of order 2 (−I2; check it is × × the only one!) and R obviously does as well, so SL2(R) × R has 3 elements of order 2. Example 2.4. In G = Sn, for n ≥ 3, let H = An and K = h(12)i = f(1); (12)g. We have G = HK: for σ 2 Sn, if σ 2 An then σ = σ · (1), while if σ2 = An then σ = σ(12) · (12) and ∼ ∼ σ(12) 2 An. Clearly H \ K = f(1)g. However, Sn =6 H × K = An × Z=(2) for n > 3 by computing the center: for n ≥ 3, Sn has a trivial center while An × Z=(2) has a nontrivial center since Z=(2) has a nontrivial center. Example 2.5. In G = S4, let H be a 2-Sylow subgroup subgroup and K be a 3-Sylow ∼ subgroup (so H = D4 and K is cyclic of order 3). In the Sylow theorems for S4, n2 = 3 and n3 = 4, so H and K are not normal in S4. The set HK can be written as H-cosets Hk and as K-cosets hK, so jHKj is divisible by jHj = 8 and by jKj = 3, so jHKj = 24. Therefore ∼ ∼ S4 = HK. The subgroups H and K intersect trivially. We have S4 =6 H × K = D4 × Z=(3) since D4 × Z=(3) has a nontrivial center (D4 and Z=(3) both have nontrivial center) while S4 has a trivial center. A difference between the last example and the previous ones is that H and K are not normal in G. In the earlier examples, H C G. Because of that difference, the semidirect product we define below will include Examples 2.2, 2.3, 2.4 as special cases, but will not include Example 2.5. SEMIDIRECT PRODUCTS OF GROUPS 3 3. Semidirect products When H and K are subgroups of a group G, the set-product HK might not be a subgroup. For example, in S3 if H = h(12)i and K = h(13)i then HK = f(1); (12); (13); (132)g has size 4 and is not a subgroup of S3. However, if H or K is normal in G then HK is a subgroup. Taking H C G, for instance, (3.1) (hk)(h0k0) = (hkh0k−1)(kk0) 2 HK; (hk)−1 = k−1h−1 = (k−1h−1k)k−1 2 HK since kh0k−1 2 H and k−1hk 2 H. This includes Examples 2.2, 2.3, and 2.4. (Note that in H × K, the standard copies of H and K are both normal subgroups of H × K, e.g., H and K are the kernels of the projection homomorphisms from H ×K to K and H, respectively.) The formulas in (3.1) are going to be the motivation for our definition of the group law in a semidirect product of two groups. Specifically, the product in (3.1) involves kh0k−1, so K is acting on H by conjugation, which is an action by automorphisms of H. Consider now two arbitrary groups H and K, not initially inside a common group, and suppose we have an action of K on H by automorphisms: this means we are given a homomorphism ': K ! Aut(H). Write the automorphism on H associated to k as 'k, 0 0 0 so 'k : H ! H is a bijection and 'k(hh ) = 'k(h)'k(h ) for all h; h 2 H. That ' is a homomorphism from K to Aut(H) means 'k1 ◦ 'k2 = 'k1k2 and '1 = idH for all k1; k2 2 K. In particular, 'k ◦ 'k−1 = '1 = idH , so −1 (3.2) 'k = 'k−1 : That is, the inverse of 'k 2 Aut(H) is 'k−1 . As an example, we could let K act trivially on H: 'k(h) = h for all k 2 K and h 2 H. We're more interested in nontrivial actions of K on H by automorphisms, but sometimes the only choice is the trivial action, for instance if H and K are finite with jKj and j Aut(H)j being relatively prime. In Examples 2.2, 2.3, and 2.4, K can act on H by conjugation since H is a normal subgroup of G. Definition 3.1. For two groups H and K and an action ': K ! Aut(H) of K on H by automorphisms, the corresponding semidirect product H o' K is defined as follows: as a set it is H × K = f(h; k): h 2 H; k 2 Kg. The group law on H o' K is 0 0 0 0 (h; k)(h ; k ) = (h'k(h ); kk ): 0 This group operation is inspired by the first formula in (3.1), with 'k(h ) being an abstracted version of kh0k−1, where the latter notation makes no sense when H and K are not initially inside a common group. The notation o in H o' K has a small C: think of the slanted lines in the small C as \pointing" to the normal subgroup H: we'll see in Theorem 3.5 that f(h; 1) : h 2 Hg in H o' K is isomorphic to H and is a normal subgroup of the semidirect product.