Quotient Groups §7.8 (Ed.2), 8.4 (Ed.2): Quotient Groups and Homomorphisms, Part I

Math 371 Lecture #33 §7.7 (Ed.2), 8.3 (Ed.2): Quotient Groups §7.8 (Ed.2), 8.4 (Ed.2): Quotient Groups and Homomorphisms, Part I Recall that to make the set of right cosets of a subgroup K in a group G into a group under the binary operation (or product) (Ka)(Kb) = K(ab) requires that K be a normal subgroup of G. For a normal subgroup N of a group G, we denote the set of right cosets of N in G by G/N (read G mod N). Theorem 8.12. For a normal subgroup N of a group G, the product (Na)(Nb) = N(ab) on G/N is well-defined. We saw the proof of this before. Theorem 8.13. For a normal subgroup N of a group G, we have (1) G/N is a group under the product (Na)(Nb) = N(ab), (2) when |G| < ∞ that |G/N| = |G|/|N|, and (3) when G is abelian, so is G/N. Proof. (1) We know that the product is well-defined. The identity element of G/N is Ne because for all a ∈ G we have (Ne)(Na) = N(ae) = Na, (Na)(Ne) = N(ae) = Na. The inverse of Na is Na−1 because (Na)(Na−1) = N(aa−1) = Ne, (Na−1)(Na) = N(a−1a) = Ne. Associativity of the product in G/N follows from the associativity in G: [(Na)(Nb)](Nc) = (N(ab))(Nc) = N((ab)c) = N(a(bc)) = (N(a))(N(bc)) = (N(a))[(N(b))(N(c))]. This establishes G/N as a group. (2) Suppose that |G| < ∞. By Lagrange’s Theorem we know that |G| = |N| [G : N] where the index [G : N] is the number of distinct right cosets of N in G, which is the number of elements in the group G/N. Dividing by |N| ≥ 1 gives |G/N| = [G : N] = |G|/|N|. (3) Suppose that G is abelian. For any Na, Nb ∈ G/N, we have (Na)(Nb) = N(ab) = N(ba) = (Nb)(Na). Thus G/N is abelian. ꢀDefinition. For a normal subgroup N of a group G, the group G/N is called the quotient group or factor group of G by N. Examples. (a) The subgroup {(1), (123), (132)} of S3 is normal. What is S3/N? ∼Since |S | = 6 and |N| = 3, then |G/N| = 2, so that S /N Z . =<ul style="display: flex;"><li style="flex:1">3</li><li style="flex:1">3</li><li style="flex:1">2</li></ul>Does S3 contain a subgroup isomorphic to Z2? Yes, it contains 3 subgroups isomorphic to Z2, namely {(1), (12)}, {(1), (13)}, and {(1), (23)}. Is S3 isomorphic to Z3 × Z2? No, because S3 is nonabelian while Z3 × Z2 is abelian. (b) Let G = Z4 × Z4 and N = h(3, 2)i = {(0, 0), (3, 2), (2, 0), (1, 2)}. Because G is abelian, the subgroup N is normal. What is G/N? Since |G| = 16 and |N| = 4, then |G/N| = 16/4 = 4. <ul style="display: flex;"><li style="flex:1">∼</li><li style="flex:1">∼</li></ul>So either G/N Z or G/N Z × Z . <ul style="display: flex;"><li style="flex:1">=</li><li style="flex:1">=</li></ul><ul style="display: flex;"><li style="flex:1">4</li><li style="flex:1">2</li><li style="flex:1">2</li></ul>To decide which one we look at the four distinct cosets of N in G, which are N + (0, 0) = {(0, 0), (3, 2), (2, 0), (1, 2)}, N + (1, 0) = {(1, 0), (0, 2), (3, 0), (2, 2)}, N + (1, 1) = {(1, 1), (0, 3), (3, 1), (2, 3)}, N + (2, 1) = {(2, 1), (1, 3), (0, 1), (3, 3)}. We check the order of one of these elements of G/N: (N + (1, 1)) + (N + (1, 1)) = N + (2, 2) = N + (1, 0), (N + (1, 1)) + (N + (1, 0)) = N + (2, 1), (N + (1, 1)) + (N + (2, 1)) = N + (3, 2) = N + (0, 0). ∼This says that |N + (1, 1)| = 4, and so G/N Z . =4Does G contain a subgroup isomorphic to Z4? Yes, it does. Is G isomorphic to N × G/N? Yes it is. We are seeing in these examples that there is some kind of relationship between G, a normal subgroup N of G, and the quotient group G/N. Theorem 8.14. Let N be a normal subgroup of a group G. Then G/N is abelian if and only if aba−1b−1 ∈ N for all a, b ∈ G. Proof. The quotient group G/N is a abelian if and only if Nab = Nba for all a, b ∈ G. We have Nab = Nba for all a, b ∈ G if and only if (ab)(ba)−1 ∈ N for all a, b ∈ G. Since (ab)(ba)−1 = aba−1b−1, then G/N is abelian if and only if aba−1b−1 ∈ N for all a, b ∈ G. ꢀRecall that Z(G) = {c ∈ G : cg = gc for all g ∈ G} is a normal subgroup of G. Theorem 8.15. For a group G, we have G/Z(G) is cyclic if and only if G is abelian. Proof. If G is abelian, then Z(G) = G so that G/Z(G) = G/G consists of only one coset, namely Ge which is trivially cyclic, i.e., G/G = hGei. Now suppose that G/Z(G) is cyclic. Then there is a coset Z(G)d for some d ∈ G which generators G/Z(G), so that every element of G/Z(G) is of the form (Z(G)d)k = Z(G)dk for some k ∈ Z. For a, b ∈ G we want to show that ab = ba. Since a = ea is in the coset Z(G)a and since Z(G)a = Z(G)di for some i ∈ Z, we have that a = c1di for some c ∈ Z(G). Similarly we have b = c2dj for some c2 ∈ Z(G) and j ∈ Z. Since c1, c2 commute with every element of G, we have ab = c1dic2dj = c1c2didj = c2c1djdi = c2djc1di = ba. This holds for all a, b ∈ G, so that G is abelian. ꢀAnother normal subgroup that plays a role in the classification of groups and in the theory of group homomorphisms is kernel of a homomorphism f : G → H, which is K = {a ∈ G : f(a) = eH} (see Theorem 8.16 Ed.3). Theorem 8.17. Let f : G → H be a homomorphism of groups with kernel K. Then f is injective if and only if K = {eG}. Proof. If K = {eG} then for f(a) = f(b) we have f(ab−1) = f(a)f(b−1) = f(a)[f(b)]−1 =f(a)[f(a)]−1 = eH, so that ab−1 ∈ K, implying that ab−1 = eG or a = b. If f is injective, then for any c ∈ K we have f(c) = eH which together with f(eG) = eH implies that c = eG. ꢀTheorem 8.18. Let N be a normal subgroup of a group G. Then the map π : G → G/N given by a → Na is a surjective homomorphism with kernel N. Proof. The map π is surjective because for any element Na of G/N we have π(a) = Na. Since Ne is the identity element of G/N, the kernel of π is {a ∈ G : π(a) = Ne} = {a ∈ G : Na = Ne} = {a ∈ G : ae−1 ∈ N}, which says that the kernel of π is N. ꢀ

Quotient Groups §7.8 (Ed.2), 8.4 (Ed.2): Quotient Groups and Homomorphisms, Part I

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