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Math 371 Lecture #32 §7.6 (Ed.2), 8.2 (Ed.2): Normal , Part II

Recall that definition of a normal : a subgroup N of G is normal if Na = aN for all a ∈ G. Recall that to make the set of right of a subgroup N of a 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 ∈ 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 ∈ 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 ∈ N such an = n2a. Hence −1 −1 anb = n2ab = n2m.

Since n2, m ∈ N, then n2m ∈ N. −1 Therefore (ac)(bd) = n2m ∈ 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 ∈ G, we set

a−1Na = {a−1na : n ∈ N} and aNa−1 = {ana−1 : n ∈ N}.

What kind of are these in G? Theorem. For a subgroup N of a group G, and for each a ∈ G, the subsets aNa−1 and a−1Na are subgroups of G isomorphic to N.

Proof. WLOG, we consider for each a ∈ G, the mapping fa : G → G defined by −1 fa(g) = a ga, which is an inner of G. We show that that a−1Na of G is a subgroup of G. −1 −1 −1 −1 −1 −1 −1 −1 For a n1a, a n2a ∈ a Na, we have (a n1a)(a n2a) = a n1n2 a. −1 −1 −1 −1 −1 Since N is a subgroup, we have n1n2 ∈ N, so that a n1n2 a ∈ a Na, and so a Na is a subgroup of N. −1 The restriction of fa to N gives a from N to a Na (the restriction of fa to N is a 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 ∈ G. (3) aNa−1 ⊆ N for all a ∈ G. (4) a−1Na = N for all a ∈ G. (5) aNa−1 = N for all a ∈ G. Proof. (1) ⇒ (2). Suppose N is normal, and for any a ∈ G consider a−1na ∈ a−1Na. The goal is to show that a−1na ∈ N. Now the product na belongs to Na, and since N is normal, we have Na = aN, so that na ∈ aN.

Thus there is n1 ∈ N such that na = an1. −1 −1 −1 This means that a na = a an1 = n1 ∈ N, and hence that a Na ⊆ N. (2) ⇔ (3). If a−1Na ⊆ N for all a ∈ G, then it holds for all a−1 ∈ G, namely that

aNa−1 = (a−1)−1Na−1 ⊆ N.

On the other hand, if aNa−1 ⊆ N for all a ∈ G, then it holds for all a−1 ∈ G, namely that a−1Na = a−1N(a−1)−1 ⊆ N. (3) ⇒ (4). Suppose that aNa−1 ⊆ N for all a ∈ G. Then as (3) ⇒ (2), we have that a−1Na ⊆ N for all a ∈ G. It remains to show the opposite inclusion, N ⊆ a−1Na for all a ∈ G. Let n ∈ N. Then we have for any a ∈ G that n = a−1(ana−1)a. −1 −1 −1 −1 Since ana ∈ aNa and aNa ⊆ N by (3), there is n2 ∈ N such that ana = n2. −1 −1 −1 −1 −1 Thus n = a (ana )a = a n2a ∈ a Na, so that a Na = N for all a ∈ G. (4) ⇔ (5). If a−1Na = N holds for all a ∈ G, then it holds with a−1 in place of a to give aNa−1 = N for all a ∈ G. On the other hand, if aNa−1 = N holds for all a ∈ G, then it holds with a−1 in place of a to give a−1Na = N for all a ∈ G. (5) ⇒ (1). Suppose that aNa−1 = N for all a ∈ G. The goal is to show that N is normal, that Na = aN for all a ∈ G, which we do by showing the two containments Na ⊆ aN and aN ⊆ Na. Let an ∈ aN. −1 −1 −1 Then ana ∈ aNa = N, so that there is n3 ∈ N such that ana = n3. −1 Multiplying ana on the right by a gives an = n3a, showing that an ∈ Na This shows that aN ⊆ Na for all a ∈ G. Conversely, let na ∈ Na. Then a−1na ∈ a−1Na, and since (5) implies (4), that a−1na ∈ a−1Na = N. −1 Hence there is n4 ∈ N such that a na = n4.

Multiplying on the left by a gives na = an4. This says that na ∈ aN, and so Na ⊆ aN for all a ∈ G. 

Examples. (a) Recall that we showed that the subgroup N = {(1), (123), (132)} of S3 is normal by showing Na = aN for all a ∈ S3.

With Theorem 8.11 (Ed.3) and knowledge about the 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 ∈ S3 that a Na is a subgroup of S3 because the map n 7→ a na is an of S3 for each a ∈ S3. Since an automorphism is a bijection, we have |a−1Na| = 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 n ≥ 2, is the subgroup SL(n, R) of the group GL(n, R) a normal subgroup? The condition for B ∈ 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 ∈ SL(n, R), and so SL(n, R) is normal. Normal subgroups also appear in groups . 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 ∈ GL(n, R). What is the of this homomorphism? It is precisely SL(n, R), and hence it is a normal subgroup of GL(n, R).