CHAPTER 9. Normal Subgroups and Factor Groups. Normal Subgroups

Normal Subgroups CHAPTER 9 Normal Subgroups and Factor Groups If H apple G, we have seen situations where ah 6= Ha 8 a 2 G. Definition (Normal Subgroup). A subgroup H of a group G is a normal subgroup of G if ah = Ha 8 a 2 G. We denote this by H C G. Note. This means that if H C G, given a 2 G and h 2 H, 9 h 0, h 00 2 H 3 ah = h 0 a and ah 00 = ha. and conversely. It does not mean ah = ha for all h 2 H. Recall (Part 8 of Lemma on Properties of Cosets). ah = Ha () H = aha 1. Theorem (9.1 Normal Subgroup Test). If H apple G, H C G () xhx 1 H for all x 2 G. (=)) H C G =) 8 x 2 G and 8 h 2 H, 9 h 0 2 H 3 xh = h 0 x =) xhx 1 = h 0 2 H. Thus xhx 1 H. ((=) Suppose xhx 1 H 8 x 2 G. Let x = a. Then aha 1 H =) ah Ha. Now let x = a 1. Then a 1 H(a 1 ) 1 = a 1 Ha H =) Ha ah. By mutual inclusion, Ha = ah =) H C G. 117

118 9. NORMAL SUBGROUPS AND FACTOR GROUPS Example. (1) Every subgroup of an Abelian group is normal since ah = ha for all a 2 G and for all h 2 H. (2) The center Z(G) of a group is always normal since ah = ha for all a 2 G and for all h 2 Z(G). Theorem (4). If H apple G and [G : H] = 2, then H C G. If a 2 H, then H = ah = Ha. If a 62 H, ah is a left coset distinct from H and Ha is a right coset distinct from H. Since [G : H] = 2, G = H [ah = H [Ha and H \ ah = ; = H \ Ha =) ah = Ha. Thus H C G. Example. A n C S n since [S n : A n ] = 2. Note, for example, that for (1 2) 2 S n and (1 2 3) 2 A n, but and (1 3 2) 2 A n. (1 2)(1 2 3) 6= (1 2 3)(1 2), (1 2)(1 2 3) = (1 3 2)(1 2) Example. hr 360/n i C D n since [D n : R 360/n ] = 2. Example. SL(2, R) C GL(2, R). Let x 2 GL(2, R). Recall det(x 1 ) = h 2 SL(2, R), 1 det(x) = (det(x)) 1. Then, for all det(xhx 1 ) = (det(x))(det(h))(det(x)) 1 = (det(x))(det(x)) 1 (det(h)) = 1 1 = 1, so xhx 1 2 SL(2, R) =) x SL(2, R)x 1 SL(2, R). Thus SL(2, R) C GL(2, R).

9. NORMAL SUBGROUPS AND FACTOR GROUPS 119 Example. Consider A 4, with group table from page 111 shown below: Let H = { 1, 2, 3, 4 } apple A 4, A = { 5, 6, 7, 8 } A 4, and B = { 9, 10, 11, 12 }. 8 a 2 A, a 1 2 B, and 8 b 2 B, b 1 2 A. Let x 2 A 4 : Case 1: x 2 H. Then xh H. Since x 1 2 H, xhx 1 H. Case 2: x 2 A. Then xh A =) xhx 1 H. Case 3: x 2 B. Then xh B =) xhx 1 H. Thus, 8 x 2 A 4, xhx 1 H =) H C A 4 by Theorem 9.1. Now let K = { 1, 5, 9 } apple A 4. Now 5 2 K, but so K 6C A 4. 2 5 1 2 = 2 5 2 = 2 8 = 7 62 K,

120 9. NORMAL SUBGROUPS AND FACTOR GROUPS Factor Groups Theorem (9.2 Factor Groups). Let G be a group and H C G. The set G/H = {ah a 2 G} is a group under the operation (ah)(bh) = abh. This group is called the factor group or quotient group of G by H. We first show the operation is well-defined. [Our product is determined by the coset representatives chosen, but is the product uniquely determined by the cosets themselves?] Suppose ah = a 0 H and bh = b 0 H. Then a 0 = ah 1 and b 0 = bh 2 for some h 2, h 2 2 H =) a 0 b 0 H = ah 1 bh 2 H = ah 1 bh = ah 1 Hb = ahb = abh by associativity in G, the Lemma on cosets (page 145), and the fact that H C G. Thus the operation is well-defined. Associativity in G/H follows directly from associativity in G: (ahbh)ch = (abh)ch = (ab)ch = a(bc)h = ah(bch) = ah(bhch). The identity is eh = H, and the inverse of ah is a 1 H. Thus G/H is a group.

9. NORMAL SUBGROUPS AND FACTOR GROUPS 121 Example. Let 3Z = {0, ±3, ±6, ±9,... }. Then 3Z C Z since Z is abelian. Consider the following 3 cosets: 0 + 3Z = 3Z = {0, ±3, ±6, ±9,... }, 1 + 3Z = {1, 4, 7,... ; 2, 5, 8,... }, 2 + 3Z = {2, 5, 8... ; 1, 4, 7,... }. For k 2 Z, k = 3q + r where 0 apple r < 3. Thus k + 3Z = r + 3q + 3Z = r + 3Z. So we have all the cosets. A Cayley table for Z/3Z: 0 + 3Z 1 + 3Z 2 + 3Z 0 + 3Z 0 + 3Z 1 + 3Z 2 + 3Z 1 + 3Z 1 + 3Z 2 + 3Z 0 + 3Z 2 + 3Z 2 + 3Z 0 + 3Z 1 + 3Z Therefore, Z/3Z Z 3. In general, for n > 0, nz = {0, ±n, ±2n, ±3n,... }, and Z/nZ Z n.

122 9. NORMAL SUBGROUPS AND FACTOR GROUPS Example. Consider the multiplication table for A 4 below, where i represents the permutation i on page 117 of these notes. Let H = {1, 2, 3, 4}. The 3 cosets of H are H, 5H = {5, 6, 7, 8}, and 9H = {9, 10, 11, 12}. Notice how the above table is divided into coset blocks. Since H C A 4, when we replace the various boxes by their coset names, we get the Cayley table below for A 4 /H. The factor group collapses all the elements of a coset to a single group element of A 4 /H. When H C G, one can always arrange a Cayley table so this happens. When H 6C G, one cannot.

9. NORMAL SUBGROUPS AND FACTOR GROUPS 123 Example. Is U(30)/U 5 (30) isomorphic to Z 2 Z 4? Solution. Z 2 (the Klein 4-group) or U(30) = {1, 7, 11, 13, 17, 19, 23, 29}, U 5 (30) = {1, 11}. 7U 5 (30) = {7, 17}, 13U 5 (30) = {13, 23}, 19U 5 (30) = {19, 29}. Thus U(30/U 5 (30) = {U 5 (30), 7U 5 (30), 13U 5 (30), 19U 5 (30)}. (7U 5 (30)) 2 = 19U 5 (30), so 7U 5 ((30) 6= 2 =) 7U 5 ((30) = 4 =) U(30)/U 5 (30) Z 4. Note. ah has two possible interpretations: (1) The order of ah as an element of G/H. (2) The size of the set ah. The appropriate interpretation will be clear from the context. Note. When we take a group and factor out by a normal subgroup H, we are essentially defining every element in H to be the identity. In the example above, 7U 5 (30) = 17U 5 (30) since 17 = 7 11 in U(30) and going to the factor group makes 11 the identity.

124 9. NORMAL SUBGROUPS AND FACTOR GROUPS Problem (Page 201 # 25). Let G = U(32) and H = {1, 31}. Which of Z 8, Z 4 Z 2, or Z 2 Z 2 Z 2 is G/H isomorphic to? Solution. G = U(32) = {1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31}. The cosets in G/H are: H = {1, 31}, 3H = {3, 29}, 5H = {5, 27}, 7H = {7, 25}, 9H = {9, 23}, 11H = {11, 21}, 13H = {13, 19}, 15H = {15, 17}. (3H) 2 = 9H 6= H, do 3H has order at least 4, ruling out Z 2 Z 2 Z 2. (3H) 4 = 17H 6= H, so 3H 6= 4 =) 3H = 8. Thus G/H Z 8. Applications of Factor Groups Why are factor groups important? When G is finite and H 6= {e}, G/H is smaller than G, yet simulates G in many ways. One can often deduce properties of G from G/H. Theorem (9.3 G/Z Theorem). Let G be a group and let Z(G) be the center of G. If G/Z(G) is cyclic, then G is Abelian. Let gz(g) be a generator of G/Z(G), and let a, b 2 G. Then 9 i, j 2 Z 3 az(g) = (gz(g)) i = g i Z(G) and bz(g) = (gz(g)) j = g j Z(G). Thus a = g i x and b = g j y for some x, y 2 Z(G). Then ab = (g i x)(g j y) = g i (xg j )y = g i (g j x)y = (g i g j )(xy) = Thus G is Abelian. (g j g i )(yx) = (g j y)(g i x) = ba.

Note. 9. NORMAL SUBGROUPS AND FACTOR GROUPS 125 (1) From the proof above, we have a stronger e ect: Theorem (9.3 0 ). Let G be a group and H apple Z(G). If G/H is cyclic, then G is Abelian. (2) Contrapositive: Theorem (9.3 00 ). If G is non-abelian, then G/Z(G) is not cyclic. Example. Consider a non-abelian group of order pq, where p and q are primes. Then, since G/Z(G) is not cyclic, Z(G) 6= p and Z(G) 6= q, so Z(G) = 1 and Z(G) = {e}. (3) If G/Z(G) is cyclic, it must be trivial (only the identity). Theorem (9.4 G/Z(G) Inn(G)). For any group G, G/Z(G) is isomorphic to Inn(G). Consider T : G/Z(G)! INN(G) defined by T (gz(g)) = g. [To show T is a well-defined function.] Suppose gz(g) = hz(g) =) h 1 g 2 Z(G). Then, 8 x 2 G, h 1 gx = xh 1 g =) gx = hxh 1 g =) gxg 1 = hxh 1 =) g (x) = h (x). Thus g = h. Reversing the above argument shows T is 1 1. T is clearly onto. For operation preservation, suppose gz(g), hz(g) 2 G/Z(G). Then T (gz(g) hz(g)) = T (ghz(g)) = gh = g h = T (gz(g)) T (hz(g)). Thus T is an isomorphism.

126 9. NORMAL SUBGROUPS AND FACTOR GROUPS Problem (Page 204 # 60). Find Inn(D n ). Solution. By Example 14, page 67, Z(D n ) = {R 0, R 180 } for n even and Z(D n ) = {R 0 } for n odd. Thus, for n odd, D n /Z(D n ) = D n Inn(D n ) by Theorem 9.4, Inn(D n ) = D n = 2n for n odd. Now suppose n is even. Then D n /Z(D n ) = [D n : Z(D n )] = D n Z(D n ) = 2n 2 = n. Then, since D n /Z(D n ) Inn(D n ) by Theorem 9.4, Inn(D n ) = D n /Z(D n ) = n. Further, n = 2p, p a prime. Then, by Theorem 7.3, Inn(D n ) Z n or Inn(D n ) D p. If Inn(D n ) were cyclic, D n /Z(D n ) would also be by Theorem 9.4 =) D n is Abelian by Theorem 9.3, an impossibility. Thus Inn(D 2p ) D p.

9. NORMAL SUBGROUPS AND FACTOR GROUPS 127 Theorem (9.5 Cauchy s Theorem for Abelian groups). Let G be a finite Abelian group and let p be a prime such that p G. Then G has an element of order p. Clearly, the theorem is true if G = 2. We use the Second Principle of Induction on G. Assume the staement is true for all Abelian groups with order less than G. [To show, based on the induction assumption, that the statement holds for G also.] Now G must have elements of prime order: if x = m and m = qn, where q is prime, then x n = q. Let x be an element of prime order q. If q = p, we are finished, so assume q 6= p. Since every subgroup of an Abelian group is normal, we may construct G = G/hxi. Then G is Abelian and p G, since G = G. By induction, then, G q has an element call it yhxi of order p. For the conclusion of the proof we use the following Lemma: Lemma (Page 204 # 67). Suppose H C G, G finite. If G/H has an element of order n, G has an element of order n. Suppose gh = n. Suppose g = m. Then (gh) m = g m H = eh = H, so by Corollary 2 to Theorem 4.1, n m. [We just proved Page 202 # 37.] Then so, by Theorem 4.2, 9 t 2 Z 3 m = g = nt = gh t g t = m gcd(m, t) = m t = n. Example. Consider, for k 2 Z, hki C Z. 1 + hki 2 Z/hki with 1 + hki = k, but all elements of Z have infinite order, so the assumption that G must be finite in the Lemma is necessary.

128 9. NORMAL SUBGROUPS AND FACTOR GROUPS Internal Direct Products Our object here is to break a group into a product of smaller groups. Definition. G is the internal direct product of H and K and we write G = H K if H C G, K C G, G = HK, and H \ K = {e}. Note. (1) For an internal direct product, H and K must be di erent normal subgroups of the same group. (2)For external direct products, H and K can be any groups. (3) One forms an internal direct product by starting with G, and then finding two normal subgroups H and K within G such that G is isomorphic to the external direct product of H and K.

9. NORMAL SUBGROUPS AND FACTOR GROUPS 129 Example. Consider Z 35 where, as we recall, the group operation is addition. h5i C Z 35 and h7i C Z 35 since Z 35 is Abelian. Since gcd(5, 7) = 1, 9 s, t 2 Z 3 1 = 5s + 7t. In fact, 1 = ( 4)5 + (3)7. Thus, for m 2 Z 35, m = ( 4m)5 + (3m)7 2 h5i + h7i. So Z 35 = h5i + h7i. Also, h5i \ h7i = {0}, so Z 35 = h5i h7i. We also know h5i Z 7 and h7i Z 5, so that Z 35 Z 7 Z 5 h5i h7i. Example. S 3 = {(1), (1 2 3), (1 3 2), (2 3), (1 2), (1 3)}. Let H = h(1 2 3)i and K = h(1 2)i. Then H = {(1), (1 2 3), (1 3 2)} and K = {(1), (1 2)}. Since (1 2 3)(1 2) = (1 3) and (1 3 2)(1 2) = (2 3), S 3 = HK. Also, H \ K = h(1)i. But S 3 6 H K since S 3 6 H K because S 3 is not cyclic and H K is cyclic since H and K are relatively prime. What is the problem here? K 6C S 3 since (1 3)(1 2)(1 3) 1 = (1 3)(1 2)(1 3) = (2 3) 62 K. Definition (Internal Direct Product H 1 H 2 H n ).! Let H 1, H 2,..., H n be a finite collection of normal subgroups of G. We say that G is the internal direct product of H 1, H 2,..., H n and write G = H 1 H 2 H n if (1) G = H 1 H 2 H n = {h 1 h 2 h n h i 2 H i }, (2) (H 1 H 2 H i ) \ H i+1 = {e} for i = 1, 2,..., n 1.

130 9. NORMAL SUBGROUPS AND FACTOR GROUPS Theorem (9.6 H 1 H 2 H n H 1 H 2 H n ). If a group G is the internal direct product of a finite number of subgroups H 1, H 2,..., H n, then G = H 1 H 2 H n, i.e., H 1 H 2 H n H 1 H 2 H n. [To show h s from di erent H i s commute.] Let h i 2 H i and h j 2 H j with i 6= j. Then, since H i C G and H j C G, (h i h j h 1 i )h 1 j 2 H j h 1 j = H j and h i (h j h 1 i h 1 j ) 2 h i H i = H i. Thus h i h j hi 1 hj 1 2 H i \ H j = {e} by Page 200 # 5. [WLOG, suppose i < j. Then, if h 2 H i \H j, h 2 H 1 H 2 H i H j 1 \H j = {e} from the definition of internal direct product.] Thus h i h j = h j h i. [To show each element of G can be expressed uniquely in the form h 1 h 2 h n where h i 2 H i.] From the definition of internal direct product, there exist h 1 2 H 1,..., h n 2 H n such that g = h 1 h 2 h n for g 2 G. Suppose also g = h 0 1h 0 2 h 0 n where h 0 i 2 H i. Using the commutative property shown above, we can solve (?) h 1 h 2 h n = h 0 1h 0 2 h 0 n to get h 0 nhn 1 = (h 0 1) 1 h 1 (h 0 2) 1 h 2 (h 0 n 1) 1 h n 1. Then h 0 nhn 1 2 H 1 H 2 H n 1 \ H n = {e} =) h 0 nhn 1 = e =) h 0 n = h n. We can thus cancel h n and h 0 n from opposite sides of (?) and repeat the preceding to get h 0 n 1 = h n 1. continuing, we eventually get h i = h 0 i for i = 1, 2,..., n.

9. NORMAL SUBGROUPS AND FACTOR GROUPS 131 Now define : G! H 1 H 2 H n by (h 1 h 2 h n ) = (h 1, h 2,, h n ). From the above, is well-=defined. If (h 1 h 2 h n ) = (h 0 1h 0 2 h 0 n), (h 1, h 2,, h n ) = (h 0 1, h 0 2,, h 0 n) =) so is 1 1. That is onto is clear. [To show operation preservation.] h i = h 0 i for i = 1,..., n =) h 1 h 2 h n = h 0 1h 0 2 h 0 n, Now let h 1 h 2 h n, h 0 1h 0 2 h 0 n 2 G. Again, using the commutativity shown above, (h1 h 2 h n )(h 0 1h 0 2 h 0 n) = (h 1 h 0 1)(h 2 h 0 2) (h n h 0 n) = (h 1 h 0 1, h 2 h 0 2,, h n h 0 n) = (h 1 h 2 h n ) (h 0 1h 0 2 h 0 n) = (h 1 h 2 h n ) (h 0 1h 0 2 h 0 n), so operations are preserved and is an isomorphism. Note. If G = H 1 H 2 H n, then for Clearly, each H i H i. H i = {e} H i {e}, i = 1,..n, G = H 1 H 2 H n.

132 9. NORMAL SUBGROUPS AND FACTOR GROUPS Theorem (9.7 Classification of Groups of Order p 2 ). Every group of order p 2, where p is a prime, is isomorphic to Z p 2 or Z p Z p. Let G be a group of order p 2, p a prime. If G has an element of order p 2, then G Z p 2. Otherwise, by Corollary 2 of Lagrange s Theorem, we may assume every non-identity element of G has order p. [To show that 8 a 2 G, hai is normal in G.] Suppose this is not the case. Then 9 b 2 G 3 bab 1 62 hai. Then hai and hbab 1 i are distinct subgroups of order p. Since hai \ hbab 1 i is a subgroup of both hai and hbab 1 i, hai \ hbab 1 i = {e}. Thus the distinct left cosets of hbab 1 i are hbab 1 i, ahbab 1 i, a 2 hbab 1 i,..., a p 1 hbab 1 i. Since b 1 must lie in one of these cosets, b 1 = a i (bab 1 ) j = a i ba j b 1 for some i and j. Cancelling the b 1 terms, we get e = a i ba j =) b = a i j 2 hai, a contradiction. Thus, 8 a 2 G, hai is normal in G. Now let x be any non-identity element in G and y any element of G not in hxi. Then, by comparing orders and from Theorem 9.6, G = hxi hyi Z p Z p. Corollary. If G is a group of order p 2, where p is a prime, then G is Abelian.

9. NORMAL SUBGROUPS AND FACTOR GROUPS 133 Example. We use Theorem 8.3, its corollary, and Theorem 9.6 for the following. If m = n 1 n 2 n k where gcd(n i, n j ) = 1 for i 6= j, then U(m) = U m/n1 (m) U m/n2 (m) U m/nk (m) U(n 1 ) U(n 2 ) U(n k ). We use the = sign for the internal direct product since the elements are all within U(m). U(105) = U(15) U(7) = U 15 (105) U 7 (105) = {1, 16, 31, 46, 61, 76} {1, 8, 22, 29, 43, 64, 71, 92} U(7) U(15) U(105) = U(5 21) = U 5 (105) U 21 (105) = {1, 11, 16, 26, 31, 41, 46, 61, 71, 76, 86, 101} {1, 22, 43, 64} U(21) U(5) U(105) = U(3 5 7) = U 35 (105) U 21 (105) U 15 (105) = {1, 71} {1, 22, 43, 64} {1, 16, 31, 46, 61, 76} U(3) U(5) U(7)