CHAPTER 9

Normal Subgroups and Factor Groups

Normal Subgroups If H G, we have seen situations where aH = Ha a G. Definition (Normal Subgroup). 6 8 2 A subgroup H of a group G is a normal subgroup of G if aH = Ha a G. 8 2 We denote this by H C G.

Note. This means that if H G, given a G and h H, h0, h00 H C 2 2 9 2 ah = h0a and ah00 = ha. and conversely. It does not mean ah = ha for all h3 H. 2 Recall (Part 8 of Lemma on Properties of Cosets). 1 aH = Ha H = aHa . () Theorem (9.1 — Normal Subgroup Test). 1 If H G, H G xHx H for all x G.  C () ✓ 2 Proof.

(= ) H C G = x G and h H, h0 H xh = h0x = ) 1 ) 8 2 1 8 2 9 2 3 ) xhx = h0 H. Thus xHx H. 2 ✓ 1 1 ( =) Suppose xHx H x G. Let x = a. Then aHa H = ( ✓ 1 8 2 1 1 1 1 ✓ ) aH Ha. Now let x = a . Then a H(a ) = a Ha H = Ha aH.✓ ✓ ) ✓ By mutual inclusion, Ha = aH = H G. ) C ⇤

117 118 9. NORMAL SUBGROUPS AND FACTOR GROUPS Example. (1) Every subgroup of an Abelian group is normal since ah = ha for all a G and for all h H. 2 2 (2) The center Z(G) of a group is always normal since ah = ha for all a G and for all h Z(G). 2 2 Theorem (4). If H G and [G : H] = 2, then H G.  C Proof. If a H, then H = aH = Ha. If a H, aH is a left coset distinct from H and Ha 2is a right coset distinct from H. Since62 [G : H] = 2, G = H aH = H Ha and H aH = = H Ha = aH = Ha. Thus H G. [ [ \ ; \ ) C ⇤ Example. An C Sn since [Sn : An] = 2. Note, for example, that for (1 2) S and (1 2 3) A , 2 n 2 n (1 2)(1 2 3) = (1 2 3)(1 2), 6 but (1 2)(1 2 3) = (1 3 2)(1 2) and (1 3 2) A . 2 n Example. R D since [D : R ] = 2. h 360/ni C n n 360/n Example. SL(2, R) C GL(2, R). Proof. 1 1 1 Let x GL(2, R). Recall det(x ) = = (det(x)) . Then, for all 2 det(x) h SL(2, R), 2 1 1 det(xhx ) = (det(x))(det(h))(det(x)) = 1 (det(x))(det(x)) (det(h)) = 1 1 = 1, · 1 1 so xhx SL(2, R) = x SL(2, R)x SL(2, R). Thus SL(2, R) GL(2, R). 2 ) ✓ C ⇤ 9. NORMAL SUBGROUPS AND FACTOR GROUPS 119 Example. Consider A4, with group table from page 111 shown below:

Let H = ↵ , ↵ , ↵ , ↵ A , A = ↵ , ↵ , ↵ , ↵ A , { 1 2 3 4}  4 { 5 6 7 8} ✓ 4 1 1 and B = ↵ , ↵ , ↵ , ↵ . a A, a B, and b B, b A. { 9 10 11 12} 8 2 2 8 2 2 Let x A : 2 4 1 1 Case 1: x H. Then xH H. Since x H, xHx H. 2 ✓ 2 ✓ 1 Case 2: x A. Then xH A = xHx H. 2 ✓ ) ✓ 1 Case 3: x B. Then xH B = xHx H. 2 ✓ ) ✓ 1 Thus, x A , xHx H = H A by Theorem 9.1. 8 2 4 ✓ ) C 4 Now let K = ↵ , ↵ , ↵ A . Now ↵ K, but { 1 5 9}  4 5 2 1 ↵ ↵ ↵ = ↵ ↵ ↵ = ↵ ↵ = ↵ K, 2 5 2 2 5 2 2 8 7 62 so K A . 6C 4 120 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 G is a group under the operation (aH)(bH) = abH. This group is{ cal|led2the} factor group or quotient group of G by H.

Proof. 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 = a0H and bH = b0H. Then a0 = ah1 and b0 = bh2 for some h , h H = a0b0H = ah bh H = ah bH = ah Hb = aHb = abH by 2 2 2 ) 1 2 1 1 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). 1 The identity is eH = H, and the inverse of aH is a 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 Z, k = 3q + r where 0 r < 3. Thus 2  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 Z3. ⇡ In general, for n > 0, nZ = 0, n, 2n, 3n, . . . , and Z/nZ Zn. { ± ± ± } ⇡ 122 9. NORMAL SUBGROUPS AND FACTOR GROUPS Example. Consider the multiplication table for A4 below, where i repre- sents 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 in{to coset }blocks. Since { } H C A4, when we replace the various boxes by their coset names, we get the Cayley table below for A4/H.

The factor group collapses all the elements of a coset to a single group element of A4/H. When H C G, one can always arrange a Cayley table so this happens. When H G, one cannot. 6C 9. NORMAL SUBGROUPS AND FACTOR GROUPS 123 Example. Is U(30)/U5(30) isomorphic to Z2 Z2 (the Klein 4-group) or Z4? Solution.

U(30) = 1, 7, 11, 13, 17, 19, 23, 29 , U5(30) = 1, 11 . 7U5(30) = 7, 17 , 13U (30) ={ 13, 23 , 19U (30) = 19} , 29 . { } { } 5 { } 5 { } Thus U(30/U (30) = U (30), 7U (30), 13U (30), 19U (30) . 5 { 5 5 5 5 } (7U (30))2 = 19U (30), so 7U ((30) = 2 = 7U ((30) = 4 = 5 5 | 5 | 6 ) | 5 | ) U(30)/U5(30) Z4. ⇡ ⇤ 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, 7U5(30) = 17U5(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 { } Z8, Z4 Z2, or Z2 Z2 Z2 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 . { } { } { } { } { } 2 (3H) = 9H = H, do 3H has order at least 4, ruling out Z2 Z2 Z2. 6 (3H)4 = 17H = H, so 3H = 4 = 3H = 8. 6 | | 6 ) | | Thus G/H Z8. ⇡ ⇤ Applications of Factor Groups Why are factor groups important? When G is finite and H = e , G/H is smaller than G, yet simulates G in many ways. One can often deduce6 { }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. Proof. Let gZ(G) be a generator of G/Z(G), and let a, b G. Then i, j Z 2 9 2 3 aZ(G) = (gZ(G))i = giZ(G) and bZ(G) = (gZ(G))j = gjZ(G). Thus a = gix and b = gjy for some x, y Z(G). Then 2 ab = (gix)(gjy) = gi(xgj)y = gi(gjx)y = (gigj)(xy) = (gjgi)(yx) = (gjy)(gix) = ba. Thus G is Abelian. ⇤ 9. NORMAL SUBGROUPS AND FACTOR GROUPS 125 Note. (1) From the proof above, we have a stronger e↵ect:

Theorem (9.30). Let G be a group and H Z(G). If G/H is cyclic, then G is Abelian. 

(2) Contrapositive:

Theorem (9.300). 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) = p and Z(G) = q, so Z(G) = 1 and Z(G) = e . | | 6 | | 6 | | { } (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). ⇡

Proof. Consider T : G/Z(G) INN(G) defined by T (gZ(G)) = . ! g 1 [To show T is a well-defined function.] Suppose gZ(G) = hZ(G) = h g Z(G). Then, x G, ) 2 8 2 1 1 1 1 1 h gx = xh g = gx = hxh g = gxg = hxh = (x) = (x). ) ) ) g h Thus g = h. Reversing the above argument shows T is 1–1. T is clearly onto. For operation preservation, suppose gZ(G), hZ(G) G/Z(G). Then 2 T (gZ(G) hZ(G)) = T (ghZ(G)) = = = T (gZ(G)) T (hZ(G)). · gh g h · 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 ) = R , R for n even and n { 0 180} Z(Dn) = R0 for n odd. Thus, for n odd, Dn/Z(Dn) = Dn Inn(Dn) by Theorem 9.4,{ }Inn(D ) = D = 2n for n odd. ⇡ | n | | n| Now suppose n is even. Then D 2n D /Z(D ) = [D : Z(D )] = | n| = = n. | n n | n n Z(D ) 2 | n | Then, since D /Z(D ) Inn(D ) by Theorem 9.4, n n ⇡ n Inn(D ) = D /Z(D ) = n. | n | | n n | Further, n = 2p, p a prime. Then, by Theorem 7.3, Inn(Dn) Zn or ⇡ Inn(Dn) Dp. If Inn(Dn) were cyclic, Dn/Z(Dn) would also be by Theo- rem 9.4 =⇡ D is Abelian by Theorem 9.3, an impossibility. ) n Thus Inn(D ) D . 2p ⇡ 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. | | Proof. 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 |. [T| o 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 xn = q. Let x be an element of|prime| order q. If q = p, we are finished, so assume| | q = p. 6 Since every subgroup of an Abelian group is normal, we may construct G = G G/ x . Then G is Abelian and p G , since G = | |. By induction, then, G h i | | | | q has an element – call it y x – oforder p. h i 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. Proof. Suppose gH = n. Suppose g = m. Then (gH)m = gmH = eH = H, so by Corollary| 2 to| Theorem 4.1, n| m| . [We just proved Page 202 # 37.] Then | t Z m = g = nt = gH t 9 2 3 | | | | so, by Theorem 4.2, m m gt = = = n. | | gcd(m, t) t ⇤ Example. Consider, for k Z, k Z. 1 + k Z/ k with 1 + k = 2 h i C h i 2 h i | h i| 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 G, K G, G = HK, and H K = e . ⇥ C C \ { }

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 Z35 where, as we recall, the group operation is addition. 5 Z35 and 7 Z35 since Z35 is Abelian. h i C h i C Since gcd(5, 7) = 1, s, t Z 1 = 5s + 7t. In fact, 1 = ( 4)5 + (3)7. 9 2 3 Thus, for m Z35, m = ( 4m)5 + (3m)7 5 + 7 . So Z35 = 5 + 7 . 2 2 h i h i h i h i Also, 5 7 = 0 , so Z35 = 5 7 . h i \ h i { } h i ⇥ h i We also know 5 Z7 and 7 Z5, so that h i ⇡ h i ⇡ Z35 Z7 Z5 5 7 . ⇡ ⇡ h i h i Example. S = (1), (1 2 3), (1 3 2), (2 3), (1 2), (1 3) . 3 { } Let H = (1 2 3) and K = (1 2) . Then h i h i 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), S3 = HK. Also, H K = (1) . \ h i But S3 H K since S3 H K because S3 is not cyclic and H K is cyclic since6⇡ H⇥ and K are6⇡relatively prime. | | | | What is the problem here? 1 K S since (1 3)(1 2)(1 3) = (1 3)(1 2)(1 3) = (2 3) K. 6C 3 62 Definition (Internal Direct Product H H H ). ! 1 ⇥ 2 ⇥ · · · ⇥ n Let H1, H2, . . . , Hn be a finite collection of normal subgroups of G. We say that G is the internal direct product of H1, H2, . . . , Hn and write G = H H H if 1 ⇥ 2 ⇥ · · · ⇥ n (1) G = H H H = h h h h H , 1 2 · · · n { 1 2 · · · n| i 2 i} (2) (H H H ) H = e for i = 1, 2, . . . , n 1. 1 2 · · · i \ i+1 { } 130 9. NORMAL SUBGROUPS AND FACTOR GROUPS Theorem (9.6 — H H H H H H ). If a group G is 1 2 · · · n ⇡ 1 2 · · · n the internal direct product of a finite number of subgroups H1, H2, . . . , Hn, then G = H H H , i.e., 1 2 · · · n H H H H H H . 1 ⇥ 2 ⇥ · · · ⇥ n ⇡ 1 2 · · · n Proof.

[To show h’s from di↵erent Hi’s commute.] Let hi Hi and hj Hj with i = j. Then, since H G and H G, 2 2 6 i C j C 1 1 1 1 1 (h h h )h H h = H and h (h h h ) h H = H . i j i j 2 j j j i j i j 2 i i i 1 1 Thus h h h h H H = e by Page 200 # 5. [WLOG, suppose i < j. i j i j 2 i \ j { } Then, if h Hi Hj, h H1H2 Hi Hj 1 Hj = e from the definition 2 \ 2 · · · · · · \ { } of internal direct product.] Thus hihj = hjhi. [To show each element of G can be expressed uniquely in the form h h h 1 2 · · · n where hi Hi.] From the definition of internal direct product, there exist h H , 2. . . , h H such that g = h h h for g G. Suppose also 1 2 1 n 2 n 1 2 · · · n 2 g = h10 h20 hn0 where hi0 Hi. Using the commutative property shown above, we can solv· ·e· 2 (?) h h h = h0 h0 h0 1 2 · · · n 1 2 · · · n to get 1 1 1 1 hn0 hn = (h10 ) h1(h20 ) h2 (hn0 1) hn 1. · · · Then 1 1 hn0 hn H1H2 Hn 1 Hn = e = hn0 hn = e = hn0 = hn. 2 · · · \ { } ) ) We can thus cancel hn and hn0 from opposite sides of (?) and repeat the preceding to get hn0 1 = hn 1. continuing, we eventually get hi = hi0 for i = 1, 2, . . . , n. 9. NORMAL SUBGROUPS AND FACTOR GROUPS 131 Now define : G H H H by ! 1 2 · · · n (h h h ) = (h , h , , h ). 1 2 · · · n 1 2 · · · n From the above, is well-=defined. If (h h h ) = (h0 h0 h0 ), 1 2 · · · n 1 2 · · · n (h , h , , h ) = (h0 , h0 , , h0 ) = 1 2 · · · n 1 2 · · · n ) h = h0 for i = 1, . . . , n = h h h = h0 h0 h0 , i i ) 1 2 · · · n 1 2 · · · n so is 1–1. That is onto is clear. [To show operation preservation.]

Now let h1h2 hn, h10 h20 hn0 G. Again, using the commutativity shown above, · · · · · · 2

(h h h )(h0 h0 h0 ) = (h h0 )(h h0 ) (h h0 ) = 1 2 · · · n 1 2 · · · n 1 1 2 2 · · · n n (h h0 , h h0 , , h h0 ) = (h h h ) (h0 h0 h0 ) = ⇥ 1 1 2 2 · · · n⇤ n ⇥ 1 2 · · · n · 1 2 · · ·⇤ n (h h h )(h0 h0 h0 ), 1 2 · · · n 1 2 · · · n so operations are preserved and is an isomorphism. ⇤ Note. If G = H H H , then for 1 2 · · · n H = e H e , i = 1, ..n, i { } i { } G = H H H . 1 ⇥ 2 ⇥ · · · n Clearly, each H H . i ⇡ i 132 9. NORMAL SUBGROUPS AND FACTOR GROUPS Theorem (9.7 – Classification of Groups of Order p2). Every group of 2 order p , where p is a prime, is isomorphic to Zp2 or Zp Zp. Proof. Let G be a group of order p2, p a prime. If G has an element of order p2, then G Zp2. Otherwise, by Corollary 2 of Lagrange’s Theorem, we may assume every⇡ non-identity element of G has order p. [To show that a G, a is normal in G.] Suppose this is not the case. Then 8 1 2 h i 1 b G bab a . Then a and bab are distinct subgroups of order 9 2 3 62 1h i h i h i 1 p. Since a bab is a subgroup of both a and bab , h i \ h i h i h i 1 1 a bab = e . Thus the distinct left cosets of bab are h i \ h i { } h i 1 1 2 1 p 1 1 bab , a bab , a bab , . . . , a bab . h i h i h i h i 1 Since b must lie in one of these cosets, 1 i 1 j i j 1 b = a (bab ) = a ba b 1 for some i and j. Cancelling the b terms, we get i j i j e = a ba = b = a a , ) 2 h i a contradiction. Thus, a G, a is normal in G. 8 2 h i Now let x be any non-identity element in G and y any element of G not in x . Then, by comparing orders and from Theorem 9.6, h i

G = x y Zp Zp. h i ⇥ h i ⇡ ⇤ Corollary. If G is a group of order p2, 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 n n where gcd(n , n ) = 1 for i = j, then 1 2 · · · k i j 6 U(m) = U (m) U (m) U (m) U(n ) U(n ) U(n ). m/n1 ⇥ m/n2 ⇥· · ·⇥ m/nk ⇡ 1 2 · · · 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 (105) U (105) · 15 ⇥ 7 = 1, 16, 31, 46, 61, 76 1, 8, 22, 29, 43, 64, 71, 92 { } ⇥ { } U(7) U(15) ⇡ U(105) = U(5 21) = U (105) U (105) · 5 ⇥ 21 = 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 (105) U (105) U (105) · · 35 ⇥ 21 ⇥ 15 = 1, 71 1, 22, 43, 64 1, 16, 31, 46, 61, 76 { } ⇥ { } ⇥ { } U(3) U(5) U(7) ⇡