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 = fhk : h 2 H; k 2 Kg: It is not in general a subgroup. Note the following: Proposition 1.1. Suppose that H \ K = f1g. 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 = f1g, then #(HK) = #(H) · #(K): Proof. Clearly, the image of F is HK by definition. To see that F is injec- tive, suppose that F (h1; k1) = F (h2; k2): −1 −1 Then by definition h1k1 = h2k2. Thus h2 h1 = k2k1 . Since H is a −1 −1 subgroup, h2 h1 2 H, and since K is a subgroup, k2k1 2 K. Thus −1 −1 −1 −1 h2 h1 = k2k1 2 H \ K = f1g, and so h2 h1 = k2k1 = 1. It follows −1 −1 that h2 h1 = 1, so that h1 = h2, and similarly k2k1 = 1 so that k2 = k1. Thus (h1; k1) = (h2; k2) and F is injective. Corollary 1.2. Suppose that G is finite, that H and K are two subgroups of G with H \ K = f1g 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 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 = f1g 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 = f1g. (ii) If #(H) = p is a prime number and H is not contained in K, then H \ K = f1g. For example, if #(H) = #(K) = p and H 6= K, then H \ K = f1g. 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 f1g. But H \ K is a subgroup of H, not equal to H since otherwise H ⊆ K. So H \ K = f1g. 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 C G and that K ≤ G. Then HK is a subgroup of G, not necessarily normal. Moreover, H C HK and K ≤ HK. Proof. HK is closed under multiplication since −1 (h1k1)(h2k2) = h1(k1h2k1 )k1k2; −1 and this is in HK since, as H is normal, k1h2k1 2 H. Identity: since 1 2 H and 1 2 K, 1 = 1·1 2 HK. Inverses: (hk)−1 = k−1h−1 = (k−1h−1k)k−1 and this is in HK since, again by normality, k−1h−1k 2 H. Clearly H ≤ HK and K ≤ HK: for example, 1 2 K and, for all h 2 H, h = h · 1 2 HK. Thus H ⊆ HK, hence H ≤ HK, and the proof that K ≤ HK is similar. Since H C G, H C 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 C G. Then HK is a subgroup of G, H C HK, H \ K is a normal subgroup of K, and HK=H =∼ K=(H \ K): 2 In particular, if H and K are both finite , then #(H) · #(K) #(HK) = : #(H \ K) Proof. We have seen that HK ≤ G and that H C HK. Also, H \ K ≤ K, and, if x 2 H \ K, then, for all k 2 K, kxk−1 2 H since x 2 H and H is normal, and kxk−1 2 K since x 2 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 2 HK, is equal to some coset of the form kH with k 2 K. But since x 2 HK, x = hk for some h 2 H and k 2 K. Since H is normal, hk = kh0 for some h0 2 H, and clearly kh0H = kH. Thus xH = hkH = kh0H = 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 2 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 #(H) · #(K) #(HK) = : #(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 Theorem 2.1. Suppose that H and K are two subgroups of a group G such that (i) H \ K = f1g; (ii) HK = G; (iii) For all h 2 H and k 2 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 ((h1; k1)(h2; k2)) = F (h1h2; k1k2) = (h1h2)(k1k2) = h1(h2k1)k2 = h1(k1h2)k2 = (h1k1)(h2k2) = F (h1; k1)F (h2; k2): Thus F is an isomorphism. ∗ >0 Example 2.2. The abelian group C contains two subgroups, R and U(1), the subgroup of positive real numbers (under multiplication). Clearly >0 R \ U(1) = f1g, since a positive real number of absolute value 1 is 1. z Also, every nonzero complex number z can be written as z = jzj · , where jzj >0 ∗ >0 jzj 2 R and z=jzj 2 U(1). Hence C = R U(1). By the theorem, ∗ ∼ >0 C = R × 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 p2. The ∼ 2 ∼ either G = Z=p 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 p2. If there exists an element of G whose order is p2, then G is cyclic of order p2 and hence G =∼ 2 Z=p Z. Thus we may assume that every element of G, not the identity, has order p. Choose x 2 G, x 6= 1. Then hxi has order p. In particular, hxi 6= G. Choose an element y 2 G, y2 = hxi. Then y has order p as well. Since hxi and hyi are two different subgroups of G, by Proposition 1.3, hxi\hyi = f1g. By Corollary 1.2, as #(hxi)#(hyi) = p2 = #(G), hxihyi = G. Thus Conditions (i) and (ii) above are satisfied, and Condition (iii) is automatic since G is ∼ ∼ abelian. Thus G = hxi × hyi = (Z=pZ) × (Z=pZ). 4 Another example of an application to finite groups is: Proposition 2.4. Let p and q be distinct primes with p < q, and let G be a finite group such that there is a normal p-Sylow subgroup P and a normal q-Sylow subgroup Q. Then ∼ ∼ G = P × Q = (Z=pZ) × (Z=qZ): Proof. Since p = #(P ) and q = #(Q) are relatively prime, P \ Q = f1g (by a homework problem). Since #(G) = pq = #(P )#(Q), G = PQ. Finally, since P and Q are normal subgroups and P \Q = f1g, gh = hg for all g 2 P and h 2 Q, again by a homework problem. Thus, Theorem 2.1 applies so that G =∼ P × Q. Finally since P is cyclic of order p and Q is cyclic of order ∼ ∼ ∼ q, P = Z=pZ and Q = Z=qZ. Thus P × Q = (Z=pZ) × (Z=qZ). Remark 2.5. By the Sylow theorem, the above condition that both P and Q are normal is satisfied unless q ≡ 1 mod p.