CHAPTER 3

Finite Groups; Subgroups

Definition (Order of a Group). The number of elements of a group (ﬁnite or inﬁnite) is called its order. We denote the order of G by G . | | Definition (Order of an Element). The order of an element g in a group G is the smallest positive integer n such that gn = e (ng = 0 in additive notation). If no such integer exists, we say g has inﬁnite order. The order of g is denoted by g . | | Example. U(18) = 1, 5, 7, 11, 13, 17 , so U(18) = 6. Also, { } | | 51 = 5, 52 = 7, 53 = 17, 54 = 13, 55 = 11, 56 = 1, so 5 = 6. | | Example. Z12 = 12 under addition modulo n. | | 1 4 = 4, 2 4 = 8, 3 4 = 0, · · · so 4 = 3. | |

36 3. FINITE GROUPS; SUBGROUPS 37 Problem (Page 69 # 20). Let G be a group, x G. If x2 = e and x6 = e, prove (1) x4 = e and (2) x5 = e. What can we say 2about x ? 6 Proof. 6 6 | | (1) Suppose x4 = e. Then x8 = e = x2x6 = e = x2e = e = x2 = e, ) ) ) a contradiction, so x4 = e. 6 (2) Suppose x5 = e. Then x10 = e = x4x6 = e = x4e = e = x4 = e, ) ) ) a contradiction, so x5 = e. 6 Therefore, x = 3 or x = 6. | | | | ⇤

Definition (Subgroup). If a subset H of a group G is itself a group under the operation of G, we say that H is a subgroup of G, denoted H G. If H is a proper subset of G, then H is a proper subgroup of G. e is the trivial subgroup of G. If H = e and H G, H is called nontrivial. { } 6 { } Theorem (3.1 — One-Step Subgroup Test). Let G be a group and = 1 ; 6 H G. Then H is a subgroup of G if a, b H = ab H (for additive notation,✓ a b H). 2 ) 2 Proof. 2 Associativity of H derives from that of G. 1 [To show e H.] H = , so x H. Then xx = e H by hypothesis. 2 6 ; 9 2 2 1 1 1 [To show x H.] Also, ex = x H by hypothesis. 2 2 1 [To show H is closed.] Suppose x, y H. Then y H and, by hypothesis, 1 1 2 2 xy = x(y ) H. 2 Thus H G. ⇤ 38 3. FINITE GROUPS; SUBGROUPS Problem (Page 74 # 73). Let 2 2 H = a + bi a, b R, a + b = 1 . { | 2 } This is the set of points on the unit circle in the complex plane. Then H C⇤ under complex multiplication. Proof. 1 = 1 + 0i H since 12 + 02 = 1. 2 Suppose a + bi, c + di H. Then a2 + b2 = 1 and c2 + d2 = 1. 2 1 a + bi (a + bi)(c di) (a + bi)(c + di) = = = c + di (c + di)(c di) ac + bd + bci adi = (ac + bd) + (bc ad)i c2 + d2 and (ac + bd)2 + (bc ad)2 = a2c2 + 2abcd + b2d2 + b2c2 2abcd + a2d2 = a2c2 + a2d2 + b2c2 + b2d2 = a2(c2 + d2) + b2(c2 + d2) = (a2 + b2)(c2 + d2) = 1 + 1 = 1, 1 so (a + bi)(c + di) H. Thus H C⇤. 2 ⇤ Theorem (3.2 — Two-Step Subgroup Test). Let G be a group and = H G. Then H G if ; 6 ✓ (1) a, b H = ab H (closed under multiplication); 2 ) 2 1 (2) a H = a H (closed under inverses). 2 ) 2 Proof. 1 1 Suppose a, b H. Then b H by (2). By (1), ab H, so H G by 2 2 2 Theorem 3.1. ⇤ 3. FINITE GROUPS; SUBGROUPS 39

Example. R360/n Dn. Proof. h i

R360/n = R0, R360 , R2 360 , , R(n 1)360 h i n ·n · · · n where Ri + Rj = R(i+j) mod 360n . o a 360 b 360 Let R , R R and suppose i = · and j = · . i j 2 h 360/ni n n Now a + b = qn + r, 0 r < n, so (a + b)360 (qn + r)360 r 360 i + j = = = q 360 + · . n n · n Then r 360 (i + j) mod 360 = · , 0 r < n, n so R mod 360 R . i+j 2 h 360/ni a 360 Now suppose R R , i = · . i 2 h 360/ni n a 360 n 360 a 360 (n a)360 360 i = 360 · = · · = , n n n so R360 i R360/n . Since 2 h i Ri + R360 i = R360 mod 360 = R0 and R360 i + Ri = R360 mod 360 = R0, 1 R360 i = Ri . Thus, by the Two-Step Subgroup Test, R360/n Dn. ⇤ h i 40 3. FINITE GROUPS; SUBGROUPS Theorem (3.3 — Finite Subgroup Test). Let H be a ﬁnite nonempty subset of a group G. If H is closed under the operation of G, then H is a subgroup of G. Proof. 1 In view of Theorem 3.2, we need only show that a H whenever a H. 2 2 1 If a = e, then a = a, and we are done. So suppose a = e. 6 Consider the sequence a, a2, . . . . By closure, each of these elements are in H. Since H is ﬁnite, not all of these elements are distinct. Suppose ai = aj, i > j. i j Then a = e, and since a = e, i j > 1. Thus, 6 i j 1 i j i j 1 1 aa = a = e, so a = a . i j 1 But i j 1 1, so a H and we are done. 2 ⇤ Suppose H G. To show H G, show any of the following: ✓ 6 (1) The identity is not in H. (2) an element of H whose inverse is not in H. 9 (3) two elements of H whose product is not in H. 9 Notation. Let G be a group, a G, and let 2 n a = a n Z . h i { | 2 } Note a0 = e. Theorem (3.4 — a is a Subgroup). h i Let G be a group and a G. Then a G. Proof. 2 h i Since a a , a = . Let an, am a . Then 2 h i h i 6 ; 2 h i n m 1 n m n m a (a ) = a a = a a , 2 h i so by Theorem 3.1, a G. h i ⇤ 3. FINITE GROUPS; SUBGROUPS 41 Example. In U(14), ﬁnd 9 . h i 91 = 9, 92 = 81 = 11 mod 14, 93 = 99 = 1 mod 14. 1 1 Thus 9 = 1, 9, 11 with 9 = 11 and 11 = 9. h i { } Example. In D , R = R , R , R , R D . 4 h 270i { 0 90 180 270} 4 R270 + R270 = R180, R180 + R270 = R90, R90 + R270 = R0.

Example. In Z15, 3 = 3, 6, 9, 12, 0 Z15. h i { } Recall here that an is na since the operation is addition modulo n.

Definition (Center of a Group). The center, Z(G), lof a group G is the subset of elements of G that commute with every element of G: Z(G) = a G ax = xa x G . { 2 | 8 2 } Theorem (3.5 — Center is a Subgroup). Z(G) G. Proof. e Z(G), so Z(G) = . Suppose a, b Z(G). Then 2 6 ; 2 (ab)x = a(bx) = a(xb) = (ax)b = (xa)b = x(ab) x G, 8 2 so ab Z(G). 2 Now suppose a Z(G). Then ax = xa x G, so 2 8 2 1 1 1 1 1 1 1 1 a (ax)a = a (xa)a = (a a)xa = a x(aa ) = ) ) 1 1 1 1 exa = a xe = xa = a x. ) 1 Thus a Z(G), and so Z(G) G by Theorem 3.2. 2 ⇤ 42 3. FINITE GROUPS; SUBGROUPS Problem (Page 75 # 79c). Let G = GL(2, R). Find Z(G). Solution. a b Suppose Z(G). Then c d 2 a b e f e f a b e f = GL(2, R) = c d g h g h c d 8 g h 2 ) ae + bg af + bh ae + cf be + df = e, f, g, h R eh fg = 0. ce + dg cf + dh ag + ch bg + dh 8 2 3 6 Now ae + bg = ae + cf = bg = cf. ) For e = f = g = 1 and h = 0, eh fg = 1 = b = c. ) For e = g = h = 1 and f = 0, eh fg = 1 = b = 0 = c = 0. ) ) Also, for e = h = 0 and f = g = 1, af + bh = be + df = af = df = a = d. ) ) a 0 Thus Z(G) = a = 0 . 0 a 6 ⇤ ⇢ Definition (Centralizer of a in G). Let a G, G a group. The centralizer of a in G is the set of all elements in G that comm2 ute with a: C(a) = g G ga = ag . { 2 | } Example. In D4, C(R ) = R , R , R , R = C(R ). 90 { 0 90 180 270} 270 C(R0) = D4 = C(R180). C(H) = R , H, R , V = C(V ). { 0 180 } C(D) = R , D, R , D0 = C(D0). { 0 180 } Thus Z(D ) = R , R . 4 { 0 180} 3. FINITE GROUPS; SUBGROUPS 43 Theorem (3.6 — C(a) is a Subgroup). For each a in a group G, C(a) G. Proof. Will be assigned as homework (Page 72 # 41). Similar to the proof of Theorem 3.5. ⇤ 1 1 Problem (Page 75 # 79a). Let G = GL(2, ). Find C . R 1 0 Solution. ✓ ◆ a b 1 1 Let C . Then c d 2 1 0 ✓ ◆ 1 1 a b a b 1 1 a + c b + d a + b a = = = = 1 0 c d c d 1 0 ) a b c + d c ) b = c and a = b + d = d = a b. ) Thus

1 1 a b 2 2 C = a ab b = 0, a, b R . 1 0 b a b 6 2 ✓ ◆ ⇢ ⇤ 44 3. FINITE GROUPS; SUBGROUPS Problem (Page 71 #33). Let G be a group. Show that Z(G) = C(a). a G \2 [This means the intersection of all subgroups of the form C(a)]. Proof.

[We show Z(G) = a G C(a) by showing mutual set inclusion.] 2 Let x Z(G). Then x commutes with all a G = x C(a) a G. 2 T 2 ) 2 8 2 Thus Z(G) a G C(a). ✓ 2 Now suppose x a G C(a). Then x commutes with every a G, T2 2 2 so x C(a) a G. Thus a G C(a) Z(G), and so 2 8 2T 2 ✓ Z(G) = a G C(a) by mutual set inclusion. ⇤ 2 T T 3. FINITE GROUPS; SUBGROUPS 45 Problem (Page 71 # 37). Suppose G is the group deﬁned by the following Cayley table: 1 2 3 4 5 6 7 8 1 1 2 3 4 5 6 7 8 2 2 1 8 7 6 5 4 3 3 3 4 5 6 7 8 1 2 4 4 3 2 1 8 7 6 5 5 5 6 7 8 1 2 3 4 6 6 5 4 3 2 1 8 7 7 7 8 1 2 3 4 5 6 8 8 7 6 5 4 3 2 1 (a) Find the centralizer of each member of G. Solution. C(1) = G, C(2) = 1, 2, 5, 6 , C(3) = 1, 3, 5, 7 , C(4) = 1, 4, 5, 8 { } { } { } C(5) = G, C(6) = 1, 2, 5, 6 , C(7) = 1, 3, 5, 7 , C(8) = 1, 4, 5, 8 { } { } { } ⇤ (b) Find Z(G). Solution. 8 Z(G) = C(i) = 1, 5 . { } i=1 \ ⇤ (c) Find the order of each element of G. How are these orders arithmetically related to the order of the group? Solution. 1 = 1, 22 = 1 = 2 = 2, 32 = 5, 33 = 5 3 = 7, 34 = 7 3 = 1 = 3 = 4, | | ) | | · · ) | | 42 = 1 = 4 = 2, 52 = 1 = 5 = 2, 62 = 1 = 6 = 2, ) | | ) | | ) | | 72 = 5, 73 = 5 7 = 3, 74 = 3 7 = 1 = 7 = 4. 82 = 2 = 8 = 2. · · ) | | ) | | We have that for all g G, g G . 2 | | | | ⇤