# Two subgroups and semi-direct products

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1 Two subgroups and semi-direct products 1 First remarks Throughout, we shall keep the following notation: G is a group, written multiplicatively, and H and K are two subgroups of G. We define the subset HK of G as follows: HK = {hk : h H, k K}. It is not in general a subgroup. Note the following: Proposition 1.1. Suppose that H K = {1}. Then the function F : H K G defined by F (h, k) = hk is injective and its image is HK. In particular, if H and K are both finite and H K = {1}, then #(HK) = #(H) #(K). Proof. Clearly, the image of F is HK by definition. To see that F is injective, suppose that F (h 1, k 1 ) = F (h 2, k 2 ). Then by definition h 1 k 1 = h 2 k 2. Thus h 1 2 h 1 = k 2 k1 1. Since H is a subgroup, h 1 2 h 1 H, and since K is a subgroup, k 2 k1 1 K. Thus h 1 2 h 1 = k 2 k1 1 H K = {1}, and so h 1 2 h 1 = k 2 k1 1 = 1. It follows that h 1 2 h 1 = 1, so that h 1 = h 2, and similarly k 2 k1 1 = 1 so that k 2 = k 1. Thus (h 1, k 1 ) = (h 2, k 2 ) and F is injective. Corollary 1.2. Suppose that G is finite, that H and K are two subgroups of G with H K = {1} and that #(H) #(K) = #(G). Then the function F : H K G defined by F (h, k) = hk is a bijection from H K to G. 1

2 Proof. Proposition 1.1 says that F : H K G is injective, but both H K and G have the same number of elements, so F is surjective as well, hence a bijection. The next proposition provides some examples where the hypothesis H K = {1} is satisfied. Proposition 1.3. Suppose that H and K are finite subgroups of a group G. (i) If the orders #(H) and #(K) are relatively prime, then H K = {1}. (ii) If #(H) = p is a prime number and H is not contained in K, then H K = {1}. For example, if #(H) = #(K) = p and H K, then H K = {1}. Proof. (i) was a homework problem (by Lagrange s theorem, the order of H K must divide both #(H) and #(K) and hence must be 1). To see (ii), note that, since #(H) is prime, it is isomorphic to Z/pZ, and so the only subgroups of H are either H or {1}. But H K is a subgroup of H, not equal to H since otherwise H K. So H K = {1}. We now consider the case where one of the subgroups, say H, is a normal subgroup of G. In this case, we have the following: Proposition 1.4. Suppose that H G and that K G. Then HK is a subgroup of G, not necessarily normal. Moreover, H HK and K HK. Proof. HK is closed under multiplication since (h 1 k 1 )(h 2 k 2 ) = h 1 (k 1 h 2 k 1 1 )k 1k 2, and this is in HK since, as H is normal, k 1 h 2 k 1 1 H. Identity: since 1 H and 1 K, 1 = 1 1 HK. Inverses: (hk) 1 = k 1 h 1 = (k 1 h 1 k)k 1 and this is in HK since, again by normality, k 1 h 1 k H. Clearly H HK and K HK: for example, 1 K and, for all h H, h = h 1 HK. Thus H HK, hence H HK, and the proof that K HK is similar. Since H G, H HK as well. The question as to what the quotient group HK/H looks like is answered by the Second Isomorphism Theorem: Theorem 1.5. Let H and K be two subgroups of G, with H G. Then HK is a subgroup of G, H HK, H K is a normal subgroup of K, and HK/H = K/(H K). 2

3 In particular, if H and K are both finite, then #(HK) = #(H) #(K) #(H K). Proof. We have seen that HK G and that H HK. Also, H K K, and, if x H K, then, for all k K, kxk 1 H since x H and H is normal, and kxk 1 K since x K and K is a subgroup. Thus H K is a normal subgroup of K. To prove the theorem, we shall find a surjective homomorphism f from K to HK/H whose kernel is H K. The First Isomorphism Theorem then says that K/(H K) = HK/H. Define f(k) = kh, the H-coset containing k. Thus f is the composition π i, where i: K HK is the inclusion and π : HK HK/H is the quotient homomorphism. Hence f is a homomorphism since it is a composition of two homomorphisms, and clearly Ker f = Ker π K = H K. We will be done if we show that f is surjective, in other words that every coset xh, with x HK, is equal to some coset of the form kh with k K. But since x HK, x = hk for some h H and k K. Since H is normal, hk = kh for some h H, and clearly kh H = kh. Thus xh = hkh = kh H = kh = f(k), so that f is surjective. The final equality holds since then #(HK)/#(H) = #(K)/#(H K). In case neither of H, K is normal, a somewhat more involved argument along the lines of the proof of Proposition 1.1 shows the following counting formula: Proposition 1.6. Let H and K be two subgroups of G, and define F : H K G as before by F (h, k) = hk. Then the image of F is HK. Moreover, for all x HK, there is a bijection from the inverse image F 1 (x) to the subgroup H K. In particular, if H and K are both finite, then #(HK) = #(H) #(K) #(H K). 2 Direct products and semi-direct products We turn now to a construction which generalizes the direct product of two groups. To set the background, we start with the following: 3

4 Theorem 2.1. Suppose that H and K are two subgroups of a group G such that (i) H K = {1}; (ii) HK = G; (iii) For all h H and k K, hk = kh. Then G = H K. Proof. Define F : H K G by: F (h, k) = hk. We shall show that F is an isomorphism. By Proposition 1.1, the function F is injective, and its image is HK = G by assumption (ii). Thus F is a bijection. To see that F is an isomorphism, note that F ((h 1, k 1 )(h 2, k 2 )) = F (h 1 h 2, k 1 k 2 ) = (h 1 h 2 )(k 1 k 2 ) = h 1 (h 2 k 1 )k 2 = h 1 (k 1 h 2 )k 2 = (h 1 k 1 )(h 2 k 2 ) Thus F is an isomorphism. = F (h 1, k 1 )F (h 2, k 2 ). Example 2.2. The abelian group C contains two subgroups, R >0 and U(1), the subgroup of positive real numbers (under multiplication). Clearly R >0 U(1) = {1}, since a positive real number of absolute value 1 is 1. Also, every nonzero complex number z can be written as z = z z z, where z R >0 and z/ z U(1). Hence C = R >0 U(1). By the theorem, C = R >0 U(1). Let us give an application to finite groups where we use this criterion: Proposition 2.3. Let p be a prime and let G be a group of order p 2. The either G = Z/p 2 Z or G = (Z/pZ) (Z/pZ). Proof. We have seen previously that G is abelian. By Lagrange s theorem, every element of G, not the identity, has order p or p 2. If there exists an element of G whose order is p 2, then G is cyclic of order p 2 and hence G = Z/p 2 Z. Thus we may assume that every element of G, not the identity, has order p. Choose x G, x 1. Then x has order p. In particular, x G. Choose an element y G, y / x. Then y has order p as well. Since x and y are two different subgroups of G, by Proposition 1.3, x y = {1}. By Corollary 1.2, as #( x )#( y ) = p 2 = #(G), x y = G. Thus Conditions (i) and (ii) above are satisfied, and Condition (iii) is automatic since G is abelian. Thus G = x y = (Z/pZ) (Z/pZ). 4

5 In general, the condition above that, for all h H and k K, hk = kh (i.e. the condition that every element of H commutes with every element of K), or equivalently the condition that G = H K, is too restrictive, and we want to relax it a bit and as a by-product describe a new class of interesting groups. To motivate the construction, consider the following set of assumptions on a pair of subgroups H and K of a group G: Assumption. Let G be a group and let H and K be subgroups of G such that: (i) H K = {1}; (ii) HK = G; (iii) H is a normal subgroup of G. Note: We do not require that K is also a normal subgroup of G. In fact, if K is also a normal subgroup of G, then, as we have seen in the homework, for all h H and k K, hk = kh (by considering the product hkh 1 k 1 which lies in both H and K), so we are back in the situation where G is isomorphic to the direct (Cartesian) product H K of H and K. Using the data above, let us see how to understand the group structure on H. By Proposition 1.1 and the assumptions (i) and (ii) above, we see that every element of G can be written uniquely as hk for h H and k K, i.e. the function F : H K G defined by F (h, k) = hk is a bijection. To understand the group law on G, or equivalently on H K, note that (h 1 k 1 )(h 2 k 2 ) = (h 1 (k 1 h 2 k 1 1 ))(k 1k 2 ), which is of the form h k for some (in fact unique) h H, k K since H is normal. To rewrite this operation in another way, define a function f : K Aut H by f(k) = i k, where i k denotes conjugation by k, i.e. f(k) is the automorphism of H defined by f(k)(h) = khk 1. Then we see that multiplication in G is given by the rule (h 1 k 1 )(h 2 k 2 ) = (h 1 f(k 1 )(h 2 ))(k 1 k 2 ). We can ask if it is possible to reverse this process. Concretely, given two abstract groups H and K and a homomorphism f : K Aut H, define a binary operation on the Cartesian product H K by the rule (h 1, k 1 ) (h 2, k 2 ) = (h 1 f(k 1 )(h 2 ), k 1 k 2 ). 5

6 Note that this operation depends on the choice of f. In particular, is the usual product operation in H K f is the trivial homomorphism from K to Aut H, i.e. f(k) = Id for all k K. Proposition 2.4. Let H and K be two groups and let f : K Aut H be a homomorphism. Define a binary operation on H K by the formula (h 1, k 1 ) (h 2, k 2 ) = (h 1 f(k 1 )(h 2 ), k 1 k 2 ). Then H K, together with this operation, is a group H K, the semi-direct product of H and K. Under this operation, H {1} is a normal subgroup of H K isomorphic to H, {1} K is a subgroup of H K isomorphic to K, and the conjugation action of K = {1} K on H {1} = H can be identified with the action given by the automorphism f. The notation is meant to suggest both the product and the fact that H corresponds to a normal subgroup of H K. Of course, the definition of the group operation depends not just on the groups H and K but also on the homomorphism f : K Aut H. To make this dependence explicit, we sometimes write H f K for the semi-direct product. Proof of Proposition 2.4. As usual, the hardest part to check is associativity: ((h 1, k 1 ) (h 2, k 2 )) (h 3, k 3 ) = (h 1 f(k 1 )(h 2 ), k 1 k 2 ) (h 3, k 3 ) = (h 1 f(k 1 )(h 2 ) f(k 1 k 2 )(h 3 ), k 1 k 2 k 3 ) = (h 1 f(k 1 )(h 2 ) f(k 1 )(f(k 2 )(h 3 )), k 1 k 2 k 3 ), where we have freely used associativity in G and the fact that f is a homomorphism. On the other hand, (h 1, k 1 ) ((h 2, k 2 ) (h 3, k 3 )) = (h 1, k 1 ) (h 2 f(k 2 )(h 3 ), k 2 k 3 ) = (h 1 f(k 1 )(h 2 f(k 2 )(h 3 )), k 1 k 2 k 3 ) = (h 1 f(k 1 )(h 2 ) f(k 1 )(f(k 2 )(h 3 )), k 1 k 2 k 3 ), where we have also used the fact that f(k 1 ) is an automorphism of H. Hence is associative. Clearly (1, 1) (h, k) = (1 f(1)(h), 1 k) = (Id(h), k) = (h, k), and (h, k) (1, 1) = (h f(k)(1), k 1) = (h, k), 6

7 since f(k) is an automorphism and hence f(k)(1) = 1. Finally, it is straightforward to check that the inverse of (h, k) is (f(k) 1 (h 1 ), k 1 ), where the element f(k) 1 (h 1 ) is the inverse function of the automorphism f(k) (which is f(k 1 ) since f is a homomorphism) applied to the element k 1 of K: whereas (h, k) (f(k) 1 (h 1 ), k 1 ) = (h f(k)(f(k) 1 (h 1 )), kk 1 ) = (hh 1, kk 1 ) = (1, 1), (f(k) 1 (h 1 ), k 1 ) (h, k) = (f(k) 1 (h 1 ) f(k 1 )(h), k 1 k) = (f(k) 1 (h 1 h), 1) = ((f(k) 1 (1), 1) = (1, 1), where we have again used the fact that f(k 1 ) is an automorphism, as well as f(k 1 ) = f(k) 1. Note that, given h 1 and h 2 in H, (h 1, 1) (h 2, 1) = (h 1 f(1)(h 2 ), 1 1) = (h 1 Id(h 2 ), 1) = (h 1 h 2, 1). Hence the function g 1 : H H K defined by g 1 (h) = (h, 1) is a homomorphism from H to H K whose image is clearly H {1}, showing that H {1} is a subgroup of H K. Since g 1 is clearly injective, it induces an isomorphism from H to H {1}. Likewise, since, given k 1, k 2 K, (1, k 1 ) (1, k 2 ) = (1 f(k 1 )(1), k 1 k 2 ) = (1, k 1 k 2 ), the function g 2 : K H K is an injective homomorphism whose image is {1} K, so that {1} K is a subgroup of H K isomorphic to K. Note that (h, 1) (1, k) = (h f(1)(1), 1 k) = (h, k), so that every element of H K can be (uniquely) written as a product of an element in H {1} times an element in {1} K. The same statement is true in the other order, but it is more complicated since (1, k) (h, 1) = (f(k)(h), k). We then have the formula (1, k) (h, 1) (1, k) 1 = (f(k)(h), k) (1, k 1 ) = (f(k)(h) f(k)(1), kk 1 ) = (f(k)(h), 1). This says that conjugation by an element of {1} K takes H {1} to itself, and since H {1} is a subgroup, conjugation by an element of H {1} also takes H {1} to itself. Since every element of H K is a product of 7

8 an element in H {1} and an element in {1} K, H {1} is a normal subgroup of H K. Moreover, identifying H {1} with H via g1 1 and identifying {1} K with K via g2 1, the conjugation action of {1} K on H {1} is identified with the action of K on H via the homomorphism f. This concludes the proof of Proposition 2.4. Then the remarks before the statement of Proposition 2.4 show the following: Theorem 2.5. Let G be a group and let H and K be two subgroups of G such that (i) H K = {1}; (ii) HK = G; (iii) H is a normal subgroup of G. Then there is a homomorphism from K to Aut H defined by conjugation. Using this homomorphism to define H K, there is an isomorphism F : G H K such that F (H) is the normal subgroup H {1} and F (K) is the subgroup {1} K. Remark 2.6. If G is isomorphic to a semi-direct product of H and K, then identifying H with the corresponding subgroup of G, H is a normal subgroup of G and G/H = K by the second isomorphism theorem, Theorem 1.5 (or directly). This says that, in the case where G contains a normal subgroup H such that G/H = K, and moreover G is isomorphic to the semi-direct product H K, then we can reconstruct G from the smaller building blocks H and K, together with the knowledge of the homomorphism f : K Aut H. However, while the construction of a semi-direct product is important in group theory (and we shall give many examples below), one should not be too optimistic for its applicability. To say that a group G is a semi-direct product, one needs a normal subgroup H and another subgroup K such that the quotient homomorphism π induces an isomorphism from K to G/H. This is often expressed as follows: we can lift the quotient group G/H to a subgroup of G. This usually turns out to be impossible, even in very simple cases. For example, the quotient of Z/4Z by the subgroup H = 2 = Z/2Z is isomorphic to Z/2Z, but there is no subgroup K of Z/4Z which maps isomorphically onto the quotient Z/2Z, and Z/4Z is not isomorphic to a product, semi-direct or otherwise, of Z/2Z with Z/2Z. There is a gadget 8

9 (group cohomology) which classifies (in a certain sense) all groups G which have a normal subgroup H and such that the quotient is isomorphic to K, in terms of H and K and a homomorphism f : K Aut H, provided that H is abelian, but there is no mechanism for handling the general case. Example 2.7. (1) Every direct product H K is a semi-direct product with f : K Aut H the trivial homomorphism (f(k) = Id for all k K). (2) The dihedral group D n of order 2n is a semi-direct product. Here H = ρ is the cyclic subgroup of order n consisting of the rotations, and K = τ is the order 2 subgroup generated by a reflection. Via the discussion of HW 6, Problem 2, H = A 2π/n and K = R. Since H has index 2, it is normal, and τ ρ τ 1 = τ ρ τ = ρ 1. In terms of the matrices A 2kπ/n and R, since R has order 2 and hence R 1 = R, we have RA 2kπ/n R 1 = RA 2kπ/n R = A 2kπ/n. Thus D n = Z/nZ Z/2Z, where the homomorphism f : Z/2Z Aut(Z/nZ) is given by f(0) = Id and f(1) = Id. (3) Similarly, the group O 2 contains ( the ) normal subgroup SO 2 and the 1 0 subgroup R, where R = B 0 =, in the notation of Problem 2 of 0 1 HW 3. Since R has order 2 and, and, for all A SO 2, RAR 1 = A 1, we see that O 2 is a semi-direct product SO 2 {±1} = U(1) {±1}. Here, the corresponding homomorphism f : {±1} Aut U(1) is defined by f(1) = Id and f( 1) is the automorphism of U(1) sending z to z 1. (4) Let G be the group of linear-affine functions l: R R under function composition. Here a linear affine function is a function of the form l(x) = ax+b with a 0. There are two very natural subgroups of G: the translation subgroup H consisting of all functions of the form t b (x) = x + b, and the similitudes K, consisting of all functions of the form m a (x) = ax, a 0. Note that H = R and K = R. Clearly H K = {Id} and, if l(x) = ax + b, then l = t b m a. Finally, a calculation shows that m a t b m 1 a = t ab. Hence H is normal, and the action of K = R on H = R is by sending m a to the automorphism of the (additive) group R given by multiplication by a. Thus G = R R using the action described above. A very similar argument works for the group G of affine-linear transformations of R n ; here L: R n R n is affine-linear if it is of the form L(x) = Ax + b, where A GL n (R) and b R n, with L 1 (x) = A 1 x A 1 b. Then G has a subgroup isomorphic to R n (the translation subgroup) and a subgroup isomorphic to GL n (R). In this case, if T p is translation by p, 9

10 so that T p (x) = x + p, and L(x) = Ax + b is as above, then a calculation shows that (L T p L 1 )(x) = x + Ap = T Ap (x). Hence the translation subgroup is a normal subgroup isomorphic to R n, GL n (R) is a (non-normal) subgroup of G, and G = R n GL n (R), where the action of GL n (R) on R n is the usual one (matrix multiplication). Finally, one often (especially in physics, and for n = 3) considers the Euclidean group, the subgroup E of G, where A SO n, so an element L of E is of the form L(x) = Ax + b, A SO n. Thus A is a rigid motion of R n fixing the origin followed by a translation. In case n = 4 replacing SO n by the Lorentz group gives a group of symmetries well-adapted to electromagnetism and special relativity. (5) As in the homework, let B be the group of upper triangular 2 2 matrices: {( ) } a b B = : a, b, d R, ad 0. 0 d Inside B there are two natural subgroups: the subgroup T of strictly upper triangular matrices, {( ) } 1 b T = : b R, 0 1 and the subgroup D of diagonal matrices D: {( ) } a 0 D = : a, d R, ad 0. 0 d Clearly T D = {Id}, and since ( ) ( ) 1 bd 1 a 0 = d ( ) a b, 0 d we see that T D = B. We have seen in the homework that T B. Hence B = T D. In this case, the action of D = (R ) 2 on T = R is given by: (a, d) (R ) 2 is sent to the automorphism of R given by multiplication by ad 1. (6) Let G be a group of order pq, where p and q are distinct primes and p < q. If P is a p-sylow subgroup and Q is the q-sylow subgroup, then we have seen in class that Q G and P Q = {1} since their orders are relatively prime. Finally, QP = G since #(G) = #(Q)#(P ), by Corollary 1.2. Hence G = Q P. 10

11 The additional piece of information is the homomorphism f : P = Z/pZ Aut Q = Aut(Z/qZ). Clearly f is trivial G is abelian P is normal there is a unique p-sylow subgroup. Otherwise, f corresponds to a non-trivial homomorphism from Z/pZ to Aut(Z/qZ). As we have seen, Aut(Z/qZ) = (Z/qZ) is a group of order q 1, and since Z/pZ is simple, f is non-trivial f is injective. This is only possible if p divides q 1, by Lagrange s theorem, since the image of f is then a subgroup of (Z/qZ) of order p. Of course, by part (3) of the Sylow theorem, we know that the number of p-sylow subgroups is a divisor of pq and 1 mod p, so if this number is not 1, it must be q and we must have q 1 mod p, i.e. p divides q 1. In fact, using the fact that (Z/qZ) is cyclic, one can show that every two nonabelian groups of order pq are isomorphic. 11

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