CHAPTER 6 Isomorphisms A while back, we saw that hai with hai = 42 and Z 42 had basically the same structure, and said the groups were isomorphic, in some sense the same. Definition (Group Isomorphism). An isomorphism from a group G to a group G is a 1 1 mapping from G onto G that preserves the group operation. That is, (ab) = (a) (b) 8 a, b 2 G. If there is an isomorphism from G onto G, we say G and G arr isomorphic and write G G. Note. The operations in G and G may be di erent. The operation to the left of = is in G, while that on the right is in G. 70
To establish an isomorphism: (1) Define : G! G. 6. ISOMORPHISMS 71 (2) Show is 1 1: assuming (a) = (b), show a = b. (3) Show is onto: 8 g 2 G. show 9 g 2 G 3 (g) = g. (4) Show (ab) = (a) (b) 8 a, b 2 G. Example. With a = 42, show hai Z 42. Define : hai! Z 42 by (a n ) = n mod 42. [Show 1 1.] Suppose (a n ) = (a m ). Then n = m mod 42 =) 42 n m =) a n = a m by Theorem 4.1. Thus is 1 1. [Show onto.] For all n 2 Z 42, a n 2 hai and (a n ) = n, so is onto. [Show operation preservation.] For all a n, a m 2 hai, (a n a m ) = (a n+m ) = n+m mod 42 = n mod 42+m mod 42 = (a n )+ (a m ). Thus, by definition, hai Z 42. Example. Any finite cyclic group hai with hai = n is isomorphic to Z n under (a k ) = k mod n. The proof of this is identical to that of the previous example.
72 6. ISOMORPHISMS Example. Let G be the positive real numbers with multiplication and G be the group of real numbers with addition. Show G G. Define : G! G by (x) = ln(x). [Show 1 1.] Suppose (x) = (y). Then so is 1 1. ln(x) = ln(y) =) e ln(x) = e ln(y) =) x = y, [Show onto] Now suppose x 2 G. e x > 0 and (e x ) = ln e x = x, so is onto. [Show operation preservation.] Finally, for all x, y 2 G, so G G. (xy) = ln(xy) = ln x + ln y = (x) + (y), [Question: is the only isomorphism?] Example. Any infinite cyclic group is isomorphic to Z with addition. Given hai with a = 1, define (a k ) = k. The map is clearly onto. If (a n ) = (a m ), n = m =) a n = a m, so is 1 1. Also, so hai Z. (a n a m ) = (a n+m ) = n + m = (n) + (m), Consider h2i under addition, the cyclic group of even integers. The h2i Z = h1i with : h2i! Z defined by (2n) = n.
6. ISOMORPHISMS 73 Example. Let G be the group of real numbers under addition and G be the group of positive numbers under multiplication. Define (x) = 2 x 1. is 1 1 and onto but so is not an isomorphism. (1 + 2) = (3) = 4 6= 2 = 1 2 = (1) (2), Can this mapping be adjusted to make it an isomorphism? Use (x) = 2 x. If (x) = (y), 2 x = 2 y =) log 2 2 x = log 2 2 y =) x = y, so is 1 1. Given any positive y, (log 2 y) = 2 log 2 y = y, so is onto. Also, for all x, y 2 R, (x + y) = 2 x+y = 2 x 2 y = Thus is an isomorphism. Example. Are U(8) and U(12) isomorphic? Solution. (x) (y). U(8) = {1, 3, 5, 7} is noncyclic with 3 = 5 = 7 = 2. U(12) = {1, 5, 7, 11} is noncyclic with 5 = 7 = 11 = 2. Define : U(8)! U(12) by 1! (1), 3! 5, 5! 7, 7! 11. is clearly 1 1 and onto. Regarding operation preservation: (3 5) = (7) = 11 and (3) (5) = 5 7 = 11. (3 7) = (5) = 7 and (3) (7) = 5 11 = 7. (5 7) = (3) = 5 and (5) (7) = 7 11 = 5. (1 3) = (3) = 5 and (1) (3) = 1 5 = 5. etc. Since both groups are Abelian, we need only check each pair in a single order. Thus, U(8) U(12).
74 6. ISOMORPHISMS This group of order 4, with all non-identity element of order 2, is called the Klein 4 group. Any other group of order 4 must have an element of order 4, so is cyclic and isomorphic to Z 4. For orders 1, 2, 3, 5, there are only the cyclic groups Z 1, Z 2, Z 3, and Z 5. Problem (Page 140 # 36). Let G = {a + b p apple a 2b 2 a, b 2 Q} and H = a, b 2 Q Show b a that G and H are isomorphic under addition. Prove that G and H are closed under multiplication. Does your isomorphism preserve multiplication as well as addition? (G and H are examples of rings a topic we begin in Chapter 12) Solution. Given a + b p 2, c + d p 2 2 G. (a + b p 2) + (c + d p 2) = (a + c) + (b + d) p 2 2 G and (a + b p 2)(c + d p 2) = (ac + 2bd) + (ad + bc) p 2 2 G since (a + c), (b + d), (ac + 2bd), (ad + bc) 2 Q. apple apple a 2b c 2d Also, if b a, 2 H, d c apple apple apple a 2b c 2d a + c 2(b + d) b a + = and d c b + d a + c apple apple apple a 2b c 2d ac + 2bd 2(ad + bc) = 2 H b a d c ad + bc ac + 2bd since (a + c), (b + d), (ac + 2bd), ad + bc 2 Q. Thus G and H are closed under both addition and multiplication. Also, G and H are groups under addition (you can prove this, if you desire).
Define : G! H by (a + b p apple a 2b 2) = b a. Suppose (a + b p 2) = (c + d p apple apple a 2b c 2d 2). Then b a = d c 6. ISOMORPHISMS 75 a = c and b = d =) a + b p 2 = c + d p 2 =) is 1 1. apple a 2b For b a 2 H, a + bp 2 2 G and (a + b p apple a 2b 2) = b a =), so is onto. Now suppose a + b p 2, c + d p 2 2 G. Then (a + b p 2) + (c + d p 2) = (a + c) + (b + d) p apple a + c 2(b + d) 2 = = b + d a + c apple apple apple a + c 2b + 2d a 2b c 2d b + d a + c = = b a + = (a + b p 2) + (c + d p 2). d c Thus is an isomorphism under addition. Also, (a+b p 2)(c+d p 2) = (ac+2bd)+(ad+bc) p apple ac + 2bd 2(ad + bc) 2 = ad + bc ac + 2bd apple apple apple ac + 2bd 2ad + 2bc a 2b c 2d = = (a + b p 2) (c + d p 2). ad + bc ac + 2bd b a d c So preserves multiplication also. (We will later see that this makes a ring homomorphism.) =
76 6. ISOMORPHISMS Problem (Page 141 # 54). Consider G = hmi and H = hni where m, n 2 Z. These are groups under addition. Show G H. Does this isomorphism also preserve multiplication? Solution. Define : G! H by (x) = n x. Suppose (x) = (y). Then m n m x = n y =) x = y =) is 1 1. m For x 2 H, x = rn where r 2 Z. Then m n x = m rn = rm 2 G. Then n m n x = n m m n x = x, so is onto. Now, suppose x, y 2 G. Then (x + y) = n m (x + y) = n m x + n y = (x) + (y), m so addition is preserved and G H. But, (xy) = n m (xy) 6= n m x n m y = (x) (y), so multiplication is not preserved by.
6. ISOMORPHISMS 77 Theorem (3). is an equivalence relation on the set G of all groups. (1) G G. Define : G! G by (x) = x. This is clear. (2) Suppose G H. If : G! H is the isomrphism and, for g 2 G, (g) = h, then 1 : H! G where 1 (h) = g is 1 1 and onto. Suppose a, b 2 H. Since is onto, 9, 2 G 3 ( ) = a and ( ) = b. Then ( ) = ( ) ( ) = ab, so 1 (ab) = 1 ( ) ( ) = 1 ( ) = = Thus 1 is an isomorphism and H G. 1 ( ) 1 ( ) = 1 (a) 1 (b). (3) Suppose G H and H K with : G! H and : H! K the isomorphisms. Then : G! K is 1 1 and onto. For all a, b 2 G, ( )(ab) = (ab) = (a) (b) = (a) (b) = ( )(a) ( )(b). Thus G K and is an equivalence relation on G. Note. Thus, when two groups are isomorphic, they are in some sense equal. They di er in that their elements are named di erently. Knowing of a computation in one group, the isomorphism allows us to perform the analagous computation in the other group.
78 6. ISOMORPHISMS Example (Conjugation by a). Let G be a group, a 2 G. Define a function a : G! G by a (x) = axa 1. Suppose a(x) = a (y). Then axa 1 = aya 1 =) x = y by left and right cancellation. Thus a is 1 1. Now suppose y 2 G. We need to find x 2 G 3 axa 1 = y. But that will be so if x = a 1 ya, for then so a(x) = a (a 1 ya) = a(a 1 ya)a 1 = (aa 1 )y(aa 1 ) = eye = y, a is onto. Finally, for all x, y 2 G, a(xy) = axya 1 = axeya 1 = axa 1 aya 1 = Thus a is an isomorphism from G onto G. a (x) a (y). Recall U(8) = {1, 3, 5, 7}. Consider 7(1) = 7 1 7 = 1, 7(3) = 7 3 7 = 5 7 = 3, 7(5) = 7 5 7 = 3 7 = 5, 7. Since 7 is its own inverse, 7(7) = 7 7 7 = 1 7 = 7. Conjugation by 7 turns out to be the identity isomorphism, which is not very interesting.
6. ISOMORPHISMS 79 Theorem (6.1 Cayley s Theorem). Every group is isomorphic to a group of permutations. Let G be any group. [We need to find a set A and a permutation on A that forms a group G that is isomorphic to G.] We take the set G and define, for each g 2 G, the function T g : G! G by T g (x) = gx 8x 2 G. Now T g (x) = T g (y) =) gx = gy =) x = y by left cancellation, so T g is 1 1. Given y 2 G, by Theorem 2 (solutions of equations) 9 a unique solution x 2 G 3 gx = y or T g (x) = y. Thus T g is onto and so is a permutation on G. Now define G by {T g g 2 G}. For any g, h 2 G, (T g T h )(x) = T g T h (x) = T g (hx) = g(hx) = (gh)(x) = T gh (x) 8 x 2 G, so G is closed under composition. Then T e is the identity and T 1 g = T g 1. Since composition is associative, G is a group. Now define : G! G by (g) = T g 8 g 2 G. (g) = (h) =) T g = T h =) T g (e) = T h (e) =) ge = he =) g = h, so is 1 1. is clearly onto by construction. Finally, for g, h 2 G, (gh) = T gh = T g T h = (g) (h), so is an isomorphism.
80 6. ISOMORPHISMS Example. U = {1, 3, 5, 7}. apple apple 1 3 5 7 1 3 5 7 T 1 = 1 3 5 7, T 3 = 3 1 7 5, T 5 = Then U(8) = {T 1, T 3, T 5, T 7 }. U(8) 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1 apple 1 3 5 7 5 7 1 3, T 7 = U(8) T 1 T 3 T 5 T 7 T 1 T 1 T 3 T 5 T 7 T 3 T 3 T 1 T 7 T 5 T 5 T 5 T 7 T 1 T 3 T 7 T 7 T 5 T 3 T 1 apple 1 3 5 7 7 5 3 1. Theorem (6.2 Properties of Elements Acting on Elements). Suppose is an isomorphism from a group G onto a group G. Then (1) (e) = e. (e) = (ee) = (e) (e). Also, since (e) 2 G, (e) = e (e) =) e (e) = (e) (e) =) e = (e) by right cancellation. (2) For all n 2 Z and for all a 2 G, (a n ) = (a) n. For n 2 Z, n 0, (a n ) = (a) n from the definition of an isomorphism and induction. For n < 0, n > 0, so e = (e) = (a n a n ) = (a n ) (a n ) = (a n ) (a) n =) (a) n = (a n ). (3) For all a, b 2 G, ab = ba () (a) (b) = (b) (a). ab = ba () (ab) = (ba) () (a) (b) = (b) (a).
(4) G = hai () G = h (a)i. 6. ISOMORPHISMS 81 Let G = hai. By closure, h (a)i G. Since is onto, for any element b 2 G, 9 a k 2 G 3 (a k ) = b. Thus b = (a) k =) b 2 h (a)i. Thus G = h (a)i. Now suppose G = h (a)i. Clearly, hai G. For all b 2 G, (b) 2 h (a)i. Thus 9 k 2 Z 3 (b) = (a) k = (a k ). Since is 1 1, b = a k. Thus hai = G. (5) a = (a) 8 a 2 G (Isomorphisms preserve order). a n = e () (a n ) = (e) () (a) n = e. Thus a has infinite order () (a) has infinite order, and a has finite order n () (a) has infinite order n. (6) For a fixed integer k and a fixed b 2 G, x k = b has the same number of solutions in G as does x k = (b) in G. Suppose a is a solution of x k = b in G, i.e., a k = b. Then so (a) is a solution of x k = (b) in G. (a k ) = (b) =) (a) k = (b), Now suppose y is a solution of x k = (b) in G. Since is onto, 9 a 2 G 3 (a) = y. Then (a) k = (b) () (a k ) = (b) =) a k = b since is 1 1. Thus a is a solution of x k = b in G. Therefore, there exists a 1 1 correspondence between the sets of solutions. (7) If G is finite, then G and G have exactly the same number of elements of every order. Follows directly from property (5).
82 6. ISOMORPHISMS Example. Is C R with multiplication the operation of both groups? Solution. x 2 = 1 has 2 solutions in C, but none in R, so by Theorem 6.2(6) no isomorphism can exist. Theorem (6.3 Properties of Isomorphisms Acting on Groups). Suppose that is an isomorphism from a group G onto a group G. Then (1) 1 is an isomorphism from G onto G. The inverse of is 1 1 since is. Let g 2 G. 1 (g) = g =) is onto. Now let x, y 2 G. Then 1 (xy) = 1 (x) 1 (y) () 1 (xy) = 1 (x) 1 (y) () xy = 1 (x) 1 (y) = xy. Thus operations are preserved and 1 is an isomorphism. (2) G is Abelian () G is Abelian. Follows directly from Theorem 6.2(3) since isomorphisms preserve commutivity. (3) G is cyclic () G is cyclic. Follows directly from Theorem 6.2(4) since isomorphisms preserve order.
6. ISOMORPHISMS 83 (4) If K apple G, then (K) = { (k) k 2 K} apple G. Follows directly from Theorem 6.2(4) since isomorphisms preserve order. (5) If K apple G, then 1 (K) = {g 2 G (g) 2 K} apple G. Follows directly from (1) and (4). (6) Z(G) = Z(G). Follows directly from Theorem 6.2(3) since isomorphisms preserve commutivity. Definition (Automorphism). An isomorphims from a group G onto itself is called an automorphism. Problem (Psge 140 # 35). Show that the mapping (a + bi) = a bi is an automorphism of C under addition. Show that preserves complex multiplication as well. (This means is an automorphims of C as well.) so Solution. is clearly 1 1 and onto. Suppose a + bi, c + di 2 C. (a + bi) + (c + di) = (a + c) + (b + d)i = (a + c) (b + d)i is an automorphism of C under addition. (a bi) + (c di) = (a + bi) + (c + di), Now suppose a + bi, c + di 2 C. (a + bi)(c + di) = (ac bd) + (ad + bc)i = (ac bd) (ad + bc)i = (a bi)(c di) = (a + bi) (c + di), so is an automorphism of C under multiplication.
84 6. ISOMORPHISMS Example (related to Example 10 on Page 135). Consider R 2. Any reflection across a line through the origin or rotation about the origin is an automorphism under componentwise addition. For example, consider the reflection about the line y = 2x. y=2x P (a,b) Consider (a, b) not on y = 2x. The line through (a, b)? y = 2x is y b = 1 x a. To find P (using substitution): 2 2x b = 1 x a =) 5 2 2 x = 1 2 a + b =) x = 1 5 a + 2 5 b =) y = 2 5 a + 4 5 b. Thus the direction vector v = P (a, b) takes (a, b) to P. 4 v = 5 a + 2 5 b, 2 5 a 1 5 b. Define (a, b) = (b, a). is clearly 1 1 and onto.
Finally, (a, b) + (c, d) = (a + c, b + d) = so 6. ISOMORPHISMS 85 (b + d, a + c) = (b, a) + (d, c) = (a, b) + (c, d), is an isomorphism. Definition (Inner Automorphism Induced by a). Let G be a group, a 2 G. The function a defined by a (x) = axa 1 8 x 2 G is called the inner automorphism of G induced by a. Example. The action of the inner automorphism of D 4 induced by R 90 is given in the following table.
86 6. ISOMORPHISMS Problem (Page 145 # 45). Suppose g and h induce the same inner automorphism of a group G. Prove h 1 g 2 Z(G). Let x 2 G. Then g(x) = h (x) =) gxg 1 = hxh 1 =) h 1 gxg 1 h = x =) (h 1 g)x(h 1 g) 1 h = x =) (h 1 g)x = x(h 1 g) =) h 1 g 2 Z(G) since x is arbitrary. Definition. Aut(G) = { is an automorphism of G} and Inn(G) = { is an inner automorphism of G}. Theorem (6.4 Aut(G) and Inn(G) are groups). Aut(G) and Inn(G) are groups under function composition. From the transitive portion of Theorem 3, compositions of isomorphisms are isomorphisms. Thus Aut(G) is closed under composition. We know function composition is associative and that the identity map is the identity automorphism. Suppose 2 Aut(G). 1 is clearly 1 1 and onto. Now 1 (xy) = 1 (x) 1 (y) () ((= by 1 1) 1 (xy) = 1 (x) 1 (y) () xy = 1 (x) 1 (y) () Thus 1 is operation preserving, and Aut(G) is a group. xy = xy.
6. ISOMORPHISMS 87 Now let g, h 2 Inn(G) Aut(G). For all x 2 G, g h(x) = g (hxh 1 ) = ghxh 1 g 1 = (gh)x(gh) 1 = gh (x), so gh = g h 2 Inn(G), and Inn(G) is closed under composition. Also, 1 g = g 1 since g 1 g(x) = g 1(gxg 1 ) = g 1 gxg 1 (g 1 ) 1 = g 1 gxg 1 g = exe = e (x), and so also gg 1 = e. Thus Inn(G) Aut(G) by the two-step test. Example. Find Inn(Z 12 ). Solution. Since Z 12 = {0, 1, 2,..., 11}, Inn(Z 12 ) = { 0, 1, 2,..., 11}, but the second list may have duplicates. That is the case here. For all n 2 Z 12 and for all x 2 Z 12, n(x) = n + x + ( n) = n + ( n) + x = 0 + x = 0 + x + 0 = 0 (x), the identity automorphism. Thus Inn(Z 12 ) = { 0 }.
88 6. ISOMORPHISMS Example. Find Inn(D 3 ). Solution. D 3 = {", (1 2 3), (1 3 2), (2 3), (1 3), (1 2)} as permutations. Then Inn(D 3 ) = { ", (1 2 3), (1 3 2), (2 3), (1 3), (1 2)}, but we need to eliminate repetitions. (1 2 3)(1 2 3) = (1 2 3)(1 2 3)(3 2 1) = (1 2 3). (1 2 3)(1 3 2) = (1 3 2)(1 2 3)(3 2 1) = (1 3 2). (1 2 3)(2 3) = (1 3 2)(2 3)(3 2 1) = (1 3). (1 2 3)(1 3) = (1 3 2)(1 3)(3 2 1) = (1 2). (1 2 3)(1 2) = (1 3 2)(1 2)(3 2 1) = (2 3). Thus (1 2 3) is distinct from ". (1 3 2)(2 3) = (1 3 2)(2 3)(2 3 1) = (1 2). Thus (1 3 2) is distinct. (2 3)(2 3) = (2 3)(2 3)(2 3) = (2 3). (2 3)(1 3) = (2 3)(1 3)(2 3) = (1 2). Thus (2 3) is distinct (1 3)(2 3) = (1 3)(2 3)(1 3) = (1 2). (1 3)(1 3) = (1 3)(1 3)(1 3) = (1 3). (1 3 2)(1 3) = (1 3 2)(1 3)(2 3 1) = (2 3). Thus (1 3) is distinct. (1 2)(2 3) = (1 2)(2 3)(1 2) = (1 3). (1 2)(1 3) = (1 2)(1 3)(1 2) = (2 3). Thus (1 2) is distinct. Therefore, there are no duplicates and Inn(D 3 ) = { ", (1 2 3), (1 3 2), (2 3), (1 3), (1 2)}.
6. ISOMORPHISMS 89 Theorem (6.5 Aut(Z n ) = U(n)). For all n 2 N, Aut(Z n ) = U(n). Let 2 Aut(Z n ). Then (k) = (1 + 1 + {z + 1} ) = ( (1) + (1) + + (1) ) = k (1). {z } k terms k terms Now 1 = n =) (1) = n, so (1) is a generator of Z n since 1 is also a generator. Now consider T : Aut(Z n )! U(n) where T ( ) = (1). Suppose, 2 Aut(Z n ) and T ( ) = T ( ). Then (1) = (1), so 8 k 2 Z n, (k) = k (1) = k (1) = (k). Thus = and T is 1 1. [To show T is onto.] Now suppose r 2 U(n) and consider : Z n! Z n defined by (s) = sr mod n 8 s 2 Z n. [To show 2 Aut(Z n ).] Suppose (x) = (y). Then xr = yr mod n. But r 1 exists modulo n =) xrr 1 = yrr 1 mod n =) x 1 = y 1 mod n =) x = y mod n, so is 1 1. Suppose x 2 Z n. By Theorem2, 9 s 2 Z n 3 (s) = sr = x, so is onto. Now suppose x, y 2 Z n. (x + y) = (x + y)r mod n = (xr + yr) mod n = so 2 Aut(Z n ). Since T ( ) = (1) = r, T is onto. xr mod n + yr mod n = (x) + (y),
90 6. ISOMORPHISMS Finally, let, 2 Aut(Z n ). Then T ( ) = ( )(1) = (1) = (1 + 1 + {z + 1} ) (1) terms (1) + (1) + + (1) = (1) (1) = T ( )T ( ), {z } (1) terms so Aut(Z n ) U(n). Example. Aut(Z 10 ) U(10). The multiplication tables for the two groups are given below: The isomorphism is clear.