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22 Quotient Groups Arkansas Tech University MATH 4033: Elementary Modern Algebra Dr. Marcel B. Finan 22 Quotient Groups Let’s look closely on the construction of the group (ZZn; ©): We know that the elements in ZZn are equivalence classes of the equivalence relation defined on ZZ by a » b if and only if nj(a ¡ b): Also, an element of ZZn is a right coset. Indeed, if 0 · k < n then < n > +k = [k] = fnq + k : q 2 ZZg: The operation © is defined by [a] © [b] = [a + b] or using right cosets (< n > +a) © (< n > +b) =< n > +(a + b): Note that the cyclic group < n > is a normal subgroup of ZZsince ZZis Abelian (See Example 21.4). We are going to use the above ideas to construct new groups where G is a group replacing ZZand N is a normal subgroup of G playing the role of < n > : More precisely, we have the following theorem. Theorem 22.1 Let N be a normal subgroup of a group G and let G=N be the set of all right cosets of N in G: Define the operation on G=N £ G=N by (Na)(Nb) = N(ab): Then (G=N; ¢) is a group, called the quotient group of G by N: Proof. ¢ is a well-defined operation We must show that if (Na1; Nb1) = (Na2; Nb2) then N(a1b1) = N(a2b2): To see this, since (Na1; Nb1) = (Na2; Nb2) then Na1 = Na2 and Nb1 = Nb2: Since a1 = ea1 2 Na1 then a1 2 Na2 so that a1 = na2 for some n 2 N: Sim- 0 0 0 ilarly, b1 = n b2 for some n 2 N: Therefore, a1b1 = na2n b2: Since N/G 0 ¡1 0 ¡1 00 0 00 then a2n a2 2 N; say a2n a2 = n 2 N: Hence, a2n = n a2 so that 00 00 a1b1 = nn a2b2: But nn 2 N so that a1b1 2 N(a2b2): Since N(a2b2) is an equivalence class and a1b1 2 N(a2b2) then N(a1b1) = N(a2b2) (Theorem 9.2). ¢ is associative Let a; b; c 2 G: Then Na(NbNc) = Na(Nbc) = N(a(bc)) = N((ab)c) = N(ab)Nc = (NaNb)(Nc) 1 where we used the fact that multiplication in G is associative. Ne = N is the identity element If a 2 G then (Na)(Ne) = N(ae) = Na and (Ne)(Na) = N(ea) = Na: Every element of G=N is invertible If a 2 G then (Na)(Na¡1) = N(aa¡1) = Ne and (Na¡1)(Na) = N(a¡1a) = Ne so that (Na)¡1 = Na¡1: Remark 22.1 Note that for a finite group G, the number of elements of G=N is just the index of N in G, i.e. [G : N]: That is, jG=Nj = [G : N]: By Lagrange’s theorem, jGj jGj jGj = [G : N]jNj so that [G : N] = jNj : Hence, jG=Nj = jNj : Remark 22.2 If N is not a normal subgroup of G; then the operation defined in the pre- vious theorem will not be well-defined. To see this, consider the subgroup ¡1 N =< (12) >= f(1); (12)g of S3: Since (123)(12)(123) = (23) 62 N then N is not a normal subgroup of S3: However, N(123) = N(23) = f(123); (23)g and N(132) = N(13) = f(13); (132)g: But N(123)(132) 6= N(23)(13) since N(123)(132) = N and N(23)(13) = N(123): Example 22.1 Let G = S3 and N =< (123) >= f(1); (123); (132)g: One can verify easily that N / G; G=N = fN; N(12)g: Example 22.2 Quotient groups can be used to show that A4 has no subgroup of order 6 and thus showing that the converse of Lagrange’s Theorem is false in general. To see this, assume that N is a subgroup of A4 of order 6. Then [G : N] = 2 and therefore N/A4 (See Exercise 21.7). Hence, A4=N makes sense. Moreover, for each a 2 N; (Na)¡1 = Na so that Na2 = N and hence a2inN: One can show that (123)2 = (132); (132)2 = (123); (124)2 = (142); (142)2 = (124); (134)2 = (143); (143)2 = (134); (234)2 = (243); and (243)2 = (234). This yields more than six different elements of A4 in N: That is, jNj > 6; a contradiction. Thus, A4 has no subgroup of order 6. By Theorem 21.1(ii), the kernel of a homomorphism is a normal subgroup. Conversely, quotient groups enable us to show that every normal subgroup is the kernel of some homomorphism. Theorem 22.2 Let G be a group and let N / G: The mapping ´ : G ¡! G=N defined by ´(a) = Na for all a 2 G is a homomorphism such that Ker ´ = N: We call ´ the natural homomorphism from G onto G=N: 2 Proof. First we show that ´ is well-defined. Indeed, if a = b then Na = Nb; i.e. ´(a) = ´(b): Since ´(ab) = N(ab) = NaNb = ´(a)´(b) then ´ is a homomor- phism. ´ is onto since any member of G=N is of the form Na for some a 2 G: That is, ´(a) = Na: It remains to show that Ker ´ = N: The proof is by double inclusions. If a 2 Ker ´ then ´(a) = Ne: Thus, Na = Ne so that a = ne = n for some n 2 N: Hence, a 2 N and Ker ´ ⊆ N: Conversely, for all n 2 N we have ´(n) = Nn = N = Ne so that N ⊆ Ker ´: The operation of multiplication on the quotient group was defined in terms of right cosets. The following theorem shows that when working with cosets of a normal subgroup N, it is immaterial whether we use Na or aN: Example 22.3 If G = ZZand N =< n > then G=N = ZZn and ´ : ZZ ¡! ZZn is given by ´(a) =< n > +a = [a] and Ker ´ =< n > : Theorem 22.3 Let N be a normal subgroup of a group G. Then aN = Na for all a 2 G: Proof. Let x 2 aN: Then x = an for some n 2 N: Since N is normal then ana¡1 2 N: Thus, ana¡1 = n0 2 N and therefore an = n0a. That is, x 2 Na: Hence, aN ⊆ Na: A similar argument shows that Na ⊆ aN: We close this section with some properties of quotient groups. More proper- ties of quotient groups are discussed in the exercises. Lemma 22.1 If G is Abelian and N/G then G=N is also Abelian. Proof. We have to show that for any Na; Nb 2 G=N we have (Na)(Nb) = (Nb)(Na): Now, since G is Abelian then ab = ba: Thus, (Na)(Nb) = Nab = Nba = (Nb)(Na): That is, G=N is Abelian. Lemma 22.2 If G is a group such that G=Z(G) is cyclic then G is Abelian, where Z(G) = fg 2 G : xg = gx 8x 2 Gg: Proof. First we show that Z(G) / G: Indeed, if g 2 G and x 2 Z(G) then gxg¡1 = gg¡1x = x 2 Z(G): Since G=Z(G) is a cyclic group then G=Z(G) =< Z(G)g > 3 for some g 2 G: If a; b 2 G then Z(G)a and Z(G)b belong to G=Z(G): Thus, Z(G)a = Z(G)gn and Z(G)b = Z(G)gm for some integers n and m: Hence, a = xgn and b = ygm for some x; y 2 Z(G): This implies that ab = (xgn)(ygm) = xygn+m = yxgmgn = ygmxgn = ba: Hence, G is Abelian. Lemma 22.3 Let G be a group of order p2 where p is a prime number. Assuming that the center of G is nontrivial, then G is Abelian. Proof. Let G be a group such that jGj = p2: Since e 2 C(G) then C(G) 6= ;: By Lagrange’s theorem either jC(G)j = p2 or jC(G)j = p: If jC(G)j = p2 then C(G) = G and so G is Abelian. If jC(G)j = p then jG=C(G)j = p and so G=C(G) is cyclic (See Corollary 3). By the previous lemma, G is Abelian. Review Problems Exercise 22.1 Find the order of each of the following quotient groups. (i) ZZ8= < [4] > : (ii) < 2 > = < 8 > : Exercise 22.2 Let G = A4 and H = f(1); (12)(34); (13)(24); (14)(23)g / G: Write out the dis- tinct elements of G=H and make a Cayley table for G=H: Exercise 22.3 Construct the Cayley table for ZZ12= < [4] > : Exercise 22.4 Let G be a cyclic group. Prove that for every subgroup H of G, G=H is cyclic. Exercise 22.5 Assume N / G: (a) Prove that if [G : N] is prime, then G=N is cyclic. (b) Prove or disprove the converse of the statement in part (a). Exercise 22.6 Determine the order of ZZ12 £ ZZ4= < [3]; [2] > : Exercise 22.7 Prove that if N/G and a 2 G then o(Na)jo(a): Exercise 22.8 Prove that G=N is Abelian if and only if aba¡1b¡1 2 N for all a; b 2 G: 4 Exercise 22.9 Let N be a normal subgroup of G and a 2 G: Prove that the order of Na in G=N is the smallest positive integer n such that an 2 N: Exercise 22.10 Let θ : G ! H be a homomorphism and K = Ker θ: Show that Á : G=K ! θ(G) defined by Á(Ka) = θ(a) is an isomorphism.
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