Math 371 Lecture #32 x7.6 (Ed.2), 8.2 (Ed.2): Normal Subgroups, Part II Recall that definition of a normal subgroup: a subgroup N of G is normal if Na = aN for all a 2 G. Recall that to make the set of right cosets of a subgroup N of a group G into a group requires the normality of N, i.e., to ensure that the \product" (Na)(Nc) = N(ac) is well-defined. We formalize this now. Theorem 8.1. Suppose N is a normal subgroup of a group G. If a ≡ b (mod N) and c ≡ d (mod N), then ac ≡ bd (mod N). Proof. Suppose that a ≡ b (mod N) and c ≡ d (mod N). Then there are elements m; n 2 N such that ab−1 = m and cd−1 = n. Since we are after ac ≡ bd (mod N), we want to show that (ac)(bd)−1 2 N. This product is (ac)(bd)−1 = a(cd−1)b−1 = anb−1: By the normality of N as a subgroup, we have aN = Na, so there exists n2 2 N such an = n2a. Hence −1 −1 anb = n2ab = n2m: Since n2; m 2 N, then n2m 2 N. −1 Therefore (ac)(bd) = n2m 2 N, so that ac ≡ bd (mod N). We will discuss in depth the group of right cosets of a normal subgroup next lecture. For now we develop some tests to determine whether a subgroup of a group is normal or not. We will need some notation for these tests. Notation. For a subgroup N of a group G and an element a 2 G, we set a−1Na = fa−1na : n 2 Ng and aNa−1 = fana−1 : n 2 Ng: What kind of subsets are these in G? Theorem. For a subgroup N of a group G, and for each a 2 G, the subsets aNa−1 and a−1Na are subgroups of G isomorphic to N. Proof. WLOG, we consider for each a 2 G, the mapping fa : G ! G defined by −1 fa(g) = a ga, which is an inner automorphism of G. We show that that subset a−1Na of G is a subgroup of G. −1 −1 −1 −1 −1 −1 −1 −1 For a n1a; a n2a 2 a Na, we have (a n1a)(a n2a) = a n1n2 a. −1 −1 −1 −1 −1 Since N is a subgroup, we have n1n2 2 N, so that a n1n2 a 2 a Na, and so a Na is a subgroup of N. −1 The restriction of fa to N gives a isomorphism from N to a Na (the restriction of fa to N is a homomorphism and a bijection, and hence an isomorphism). Theorem 8.11. The following conditions on a subgroup N of a group G are equivalent. (1) N is a normal subgroup of G. (2) a−1Na ⊆ N for all a 2 G. (3) aNa−1 ⊆ N for all a 2 G. (4) a−1Na = N for all a 2 G. (5) aNa−1 = N for all a 2 G. Proof. (1) ) (2). Suppose N is normal, and for any a 2 G consider a−1na 2 a−1Na. The goal is to show that a−1na 2 N. Now the product na belongs to Na, and since N is normal, we have Na = aN, so that na 2 aN. Thus there is n1 2 N such that na = an1. −1 −1 −1 This means that a na = a an1 = n1 2 N, and hence that a Na ⊆ N. (2) , (3). If a−1Na ⊆ N for all a 2 G, then it holds for all a−1 2 G, namely that aNa−1 = (a−1)−1Na−1 ⊆ N: On the other hand, if aNa−1 ⊆ N for all a 2 G, then it holds for all a−1 2 G, namely that a−1Na = a−1N(a−1)−1 ⊆ N: (3) ) (4). Suppose that aNa−1 ⊆ N for all a 2 G. Then as (3) ) (2), we have that a−1Na ⊆ N for all a 2 G. It remains to show the opposite inclusion, N ⊆ a−1Na for all a 2 G. Let n 2 N. Then we have for any a 2 G that n = a−1(ana−1)a. −1 −1 −1 −1 Since ana 2 aNa and aNa ⊆ N by (3), there is n2 2 N such that ana = n2. −1 −1 −1 −1 −1 Thus n = a (ana )a = a n2a 2 a Na, so that a Na = N for all a 2 G. (4) , (5). If a−1Na = N holds for all a 2 G, then it holds with a−1 in place of a to give aNa−1 = N for all a 2 G. On the other hand, if aNa−1 = N holds for all a 2 G, then it holds with a−1 in place of a to give a−1Na = N for all a 2 G. (5) ) (1). Suppose that aNa−1 = N for all a 2 G. The goal is to show that N is normal, that Na = aN for all a 2 G, which we do by showing the two containments Na ⊆ aN and aN ⊆ Na. Let an 2 aN. −1 −1 −1 Then ana 2 aNa = N, so that there is n3 2 N such that ana = n3. −1 Multiplying ana on the right by a gives an = n3a, showing that an 2 Na This shows that aN ⊆ Na for all a 2 G. Conversely, let na 2 Na. Then a−1na 2 a−1Na, and since (5) implies (4), that a−1na 2 a−1Na = N. −1 Hence there is n4 2 N such that a na = n4. Multiplying on the left by a gives na = an4. This says that na 2 aN, and so Na ⊆ aN for all a 2 G. Examples. (a) Recall that we showed that the subgroup N = f(1); (123); (132)g of S3 is normal by showing Na = aN for all a 2 S3. With Theorem 8.11 (Ed.3) and knowledge about the order of subgroups of S3, we have a simpler argument for the normality of N. The subgroup N is the only subgroup of order 3 of S3 while every other proper subgroup of S3 has order 2 by Lagrange's Theorem. −1 −1 We know for each a 2 S3 that a Na is a subgroup of S3 because the map n 7! a na is an inner automorphism of S3 for each a 2 S3. Since an automorphism is a bijection, we have ja−1Naj = 3. −1 Since a Na is a subgroup of S3 of order 3, there is only one subgroup of S3 that can contain it, namely N, so that a−1Na ⊆ N. By Theorem 8.11 (Ed.3), we conclude that N is a normal subgroup of S3. (b) For any integer n ≥ 2, is the subgroup SL(n; R) of the group GL(n; R) a normal subgroup? The condition for B 2 SL(n; R) is that det(B) = 1. Because det(A−1BA) = det(A−1)det(B)det(A) = det(B) = 1, we have that A−1BA 2 SL(n; R), and so SL(n; R) is normal. Normal subgroups also appear in groups homomorphisms. Theorem. For groups G and H, if f : G ! H is a homomorphism, then ker f is a normal subgroup of G, and Im f is normal subgroup of H. You will prove the two conclusions of this theorem as homework problems. Example. For an integer n ≥ 2, the map f : GL(n; R) ! R∗ given by f(A) = det(A) is a homomorphism because f(AB) = det(AB) = det(A)det(B) = f(A)f(B) for all A; B 2 GL(n; R): What is the kernel of this homomorphism? It is precisely SL(n; R), and hence it is a normal subgroup of GL(n; R)..