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

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Quotient Groups §7.8 (Ed.2), 8.4 (Ed.2): Quotient Groups and Homomorphisms, Part I Math 371 Lecture #33 x7.7 (Ed.2), 8.3 (Ed.2): Quotient Groups x7.8 (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 jGj < 1 that jG=Nj = jGj=jNj, 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 2 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 jGj < 1. By Lagrange's Theorem we know that jGj = jNj [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 jNj ≥ 1 gives jG=Nj = [G : N] = jGj=jNj. (3) Suppose that G is abelian. For any Na; Nb 2 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 f(1); (123); (132)g of S3 is normal. What is S3=N? ∼ Since jS3j = 6 and jNj = 3, then jG=Nj = 2, so that S3=N = Z2. Does S3 contain a subgroup isomorphic to Z2? Yes, it contains 3 subgroups isomorphic to Z2, namely f(1); (12)g, f(1); (13)g, and f(1); (23)g. 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 = f(0; 0); (3; 2); (2; 0); (1; 2)g: Because G is abelian, the subgroup N is normal. What is G=N? Since jGj = 16 and jNj = 4, then jG=Nj = 16=4 = 4. ∼ ∼ So either G=N = Z4 or G=N = Z2 × Z2. To decide which one we look at the four distinct cosets of N in G, which are N + (0; 0) = f(0; 0); (3; 2); (2; 0); (1; 2)g; N + (1; 0) = f(1; 0); (0; 2); (3; 0); (2; 2)g; N + (1; 1) = f(1; 1); (0; 3); (3; 1); (2; 3)g; N + (2; 1) = f(2; 1); (1; 3); (0; 1); (3; 3)g: 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 jN + (1; 1)j = 4, and so G=N = Z4. Does 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 2 N for all a; b 2 G. Proof. The quotient group G=N is a abelian if and only if Nab = Nba for all a; b 2 G. We have Nab = Nba for all a; b 2 G if and only if (ab)(ba)−1 2 N for all a; b 2 G. Since (ab)(ba)−1 = aba−1b−1, then G=N is abelian if and only if aba−1b−1 2 N for all a; b 2 G. Recall that Z(G) = fc 2 G : cg = gc for all g 2 Gg 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 2 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 2 Z. For a; b 2 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 2 Z, we have i that a = c1d for some c 2 Z(G). j Similarly we have b = c2d for some c2 2 Z(G) and j 2 Z. Since c1; c2 commute with every element of G, we have i j i j j i j i ab = c1d c2d = c1c2d d = c2c1d d = c2d c1d = ba: This holds for all a; b 2 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 = fa 2 G : f(a) = eH g (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 = feGg. −1 −1 −1 Proof. If K = feGg then for f(a) = f(b) we have f(ab ) = f(a)f(b ) = f(a)[f(b)] = −1 −1 −1 f(a)[f(a)] = eH , so that ab 2 K, implying that ab = eG or a = b. If f is injective, then for any c 2 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 7! 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 fa 2 G : π(a) = Neg = fa 2 G : Na = Neg = fa 2 G : ae−1 2 Ng; which says that the kernel of π is N. .
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