Isomorphic,”|H I| in Some Sense the Same

CHAPTER 6

Isomorphisms

A while back, we saw that a with a = 42 and Z42 had basically the same structure, and said the groupsh i were “isomorphic,”|h i| 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) a, b G. 8 2 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 6. ISOMORPHISMS 71 To establish an isomorphism: (1) Deﬁne : G G. ! (2) Show is 1–1: assuming (a) = (b), show a = b. (3) Show is onto: g G. show g G (g) = g. 8 2 9 2 3 (4) Show (ab) = (a)(b) a, b G. 8 2 Example. With a = 42, show a Z42. | | h i ⇡ Proof. n Deﬁne : a Z42 by (a ) = n mod 42. h i ! [Show 1–1.] Suppose (an) = (am). Then n = m mod 42 = 42 n m = an = am ) | ) by Theorem 4.1. Thus is 1–1. n n [Show onto.] For all n Z42, a a and (a ) = n, so is onto. 2 2 h i [Show operation preservation.] For all an, am a , 2 h i (anam) = (an+m) = n+m mod 42 = n mod 42+m mod 42 = (an)+(am).

Thus, by definition, a Z42. h i ⇡ ⇤ Example. Any finite cyclic group a with a = n is isomorphic to Zn under (ak) = k mod n. The proof ofhthisi is iden|h ticali| 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. ⇡ Proof. Define : G G by (x) = ln(x). ! [Show 1–1.] Suppose (x) = (y). Then ln(x) = ln(y) = eln(x) = eln(y) = x = y, ) ) so is 1–1. [Show onto] Now suppose x G. ex > 0 and (ex) = ln ex = x, so is onto. 2 [Show operation preservation.] Finally, for all x, y G, 2 (xy) = ln(xy) = ln x + ln y = (x) + (y), so G G. ⇡ [Question: is the only isomorphism?] ⇤

Example. Any infinite cyclic group is isomorphic to Z with addition. Given a with a = , define (ak) = k. The map is clearly onto. If (an) = (am), nh =i m =| | an 1= am, so is 1–1. Also, ) (anam) = (an+m) = n + m = (n) + (m), so a Z. h i ⇡ Consider 2 under addition, the cyclic group of even integers. The 2 Z = h i h i ⇡ 1 with : 2 Z defined by (2n) = n. h i h i ! 6. ISOMORPHISMS 73 Example. Let G be the group of real numbers under addition and G be x 1 the group of positive numbers under multiplication. Define (x) = 2 . is 1–1 and onto but (1 + 2) = (3) = 4 = 2 = 1 2 = (1)(2), 6 · so is not an isomorphism. Can this mapping be adjusted to make it an isomorphism? Use (x) = 2x. If (x) = (y), 2x = 2y = log 2x = log 2y = x = y, so is 1–1. ) 2 2 ) log2 y Given any positive y, (log2 y) = 2 = y, so is onto. Also, for all x, y R,(x + y) = 2x+y = 2x2y = (x)(y). 2 Thus is an isomorphism. Example. Are U(8) and U(12) isomorphic? Solution. 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 Z4. For orders 1, 2, 3, 5, there are only the cyclic groups Z1, Z2, Z3, and Z5. Problem (Page 140 # 36). a 2b Let G = a + bp2 a, b Q and H = a, b Q Show { 2 } b a 2 ⇢  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 + bp2, c + dp2 G. 2 (a + bp2) + (c + dp2) = (a + c) + (b + d)p2 G and 2 (a + bp2)(c + dp2) = (ac + 2bd) + (ad + bc)p2 G 2 since (a + c), (b + d), (ac + 2bd), (ad + bc) Q. 2 a 2b c 2d Also, if , H, b a d c 2   a 2b c 2d a + c 2(b + d) + = and b a d c b + d a + c    a 2b c 2d ac + 2bd 2(ad + bc) = H b a d c ad + bc ac + 2bd 2    since (a + c), (b + d), (ac + 2bd), ad + bc Q. 2 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). 6. ISOMORPHISMS 75 a 2b Define : G H by (a + bp2) = . ! b a  a 2b c 2d Suppose (a + bp2) = (c + dp2). Then = = b a d c )   a = c and b = d = a + bp2 = c + dp2 = is 1–1. ) ) a 2b a 2b For H, a + bp2 G and (a + bp2) = , so is onto. b a 2 2 b a   Now suppose a + bp2, c + dp2 G. Then 2 a + c 2(b + d) (a + bp2) + (c + dp2) = (a + c) + (b + d)p2 = = b + d a + c  a + c 2b + 2d a 2b c 2d = + = (a + bp2) + (c + dp2). b + d a + c = b a d c    Thus is an isomorphism under addition. Also, ac + 2bd 2(ad + bc) (a+bp2)(c+dp2) = (ac+2bd)+(ad+bc)p2 = = ad + bc ac + 2bd  ac + 2bd 2ad + 2bc a 2b c 2d = = (a + bp2) (c + dp2) . ad + bc ac + 2bd b a d c    So preserves multiplication also. (We will later see that thismakes a ring homomorphism.) ⇤ 76 6. ISOMORPHISMS Problem (Page 141 # 54). Consider G = m and H = n where m, n h i h i 2 Z. These are groups under addition. Show G H. Does this isomorphism also preserve multiplication? ⇡ Solution. n Define : G H by (x) = x. Suppose (x) = (y). Then ! m n n x = y = x = y = is 1–1. m m ) ) m m For x H, x = rn where r Z. Then x = rn = rm G. Then 2 2 n n 2 m n m x = x = x, so is onto. Now, suppose x, y G. Then n m n 2 ⇣ ⌘ ⇣ ⌘ n n n (x + y) = (x + y) = x + y = (x) + (y), m m m so addition is preserved and G H. But, n ⇡ n n (xy) = (xy) = x y = (x)(y), m 6 m m so multiplication is not preserved by⇣. ⌘⇣ ⌘ ⇤ 6. ISOMORPHISMS 77 Theorem (3). is an equivalence relation on the set G of all groups. ⇡ Proof. (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 G, ⇡1 ! 1 2 (g) = h, then : H G where (h) = g is 1–1 and onto. ! Suppose a, b H. Since is onto, ↵, G (↵) = a and () = b. Then (↵) =2 (↵)() = ab, so 9 2 3 1 1 1 (ab) = (↵)() = (↵) = ↵ = 1 1 1 1 (↵) () = (a) (b). 1 Thus is an isomorphism and H G. ⇡ (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 G, 2 ( )(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 G. Deﬁne a function 1 2 : G G by (x) = axa . a ! a 1 1 Suppose (x) = (y). Then axa = aya = x = y by left and right a a ) cancellation. Thus a is 1–1. 1 Now suppose y G. We need to ﬁnd x G axa = y. But that will be 1 2 2 3 so if x = a ya, for then 1 1 1 1 1 a(x) = a(a ya) = a(a ya)a = (aa )y(aa ) = eye = y, so a is onto. Finally, for all x, y G, 2 1 1 1 1 a(xy) = axya = axeya = axa aya = a(x)a(y).

Thus a is an isomorphism from G onto G. Recall U(8) = 1, 3, 5, 7 . Consider . Since 7 is its own inverse, { } 7 (1) = 7 1 7 = 1, 7 · · (3) = 7 3 7 = 5 7 = 3, 7 · · · (5) = 7 5 7 = 3 7 = 5, 7 · · · (7) = 7 7 7 = 1 7 = 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. Proof. 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 G, the function T : G G by T (x) = gx x G. 2 g ! g 8 2 Now T (x) = T (y) = gx = gy = x = y by left cancellation, so T is 1–1. g g ) ) g Given y G, by Theorem 2 (solutions of equations) a unique solution 2 9 x G gx = y or T (x) = y. Thus T is onto and so is a permutation on G. 2 3 g g Now define G by T g G . For any g, h G, { g| 2 } 2 (T T )(x) = T T (x) = T (hx) = g(hx) = (gh)(x) = T (x) x G, g h g h g gh 8 2 1 so G is closed under composition. Then T is the identity and T = T 1. e g g Since composition is associative, G is a group. Now define : G G by (g) = T g G. ! g 8 2 (g) = (h) = T = T = T (e) = T (e) = ge = he = g = h, ) g h ) g h ) ) so is 1–1. is clearly onto by construction. Finally, for g, h G, 2 (gh) = Tgh = TgTh = (g)(h), so is an isomorphism. ⇤ 80 6. ISOMORPHISMS Example. U = 1, 3, 5, 7 . { } 1 3 5 7 1 3 5 7 1 3 5 7 1 3 5 7 T = , T = , T = , T = . 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1     Then U(8) = T , T , T , T . { 1 3 5 7} U(8) 1 3 5 7 U(8) T1 T3 T5 T7 1 1 3 5 7 T1 T1 T3 T5 T7 3 3 1 7 5 T3 T3 T1 T7 T5 5 5 7 1 3 T5 T5 T7 T1 T3 7 7 5 3 1 T7 T7 T5 T3 T1

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. Proof. (e) = (ee) = (e)(e). Also, since (e) G, (e) = e(e) = 2 ) e(e) = (e)(e) = e = (e) by right cancellation. ) ⇤ n (2) For all n Z and for all a G, (an) = (a) . Proof. 2 2 n ⇥ ⇤ For n Z, n 0, (an) = (a) from the definition of an isomorphism and induction.2 Forn < 0, n > 0, so ⇥ ⇤ n n n n n n n n e = (e) = (a a ) = (a )(a ) = (a ) (a) = (a) = (a ). ) ⇥ ⇤ ⇥ ⇤ ⇤ (3) For all a, b G, ab = ba (a)(b) = (b)(a). Proof. 2 () ab = ba (ab) = (ba) (a)(b) = (b)(a). () () ⇤ 6. ISOMORPHISMS 81 (4) G = a G = (a) . Proof.h i () h i Let G = a . By closure, (a) G. Since is onto, for any element b G, h i h i ✓ k 2 ak G (ak) = b. Thus b = (a) = b (a) . Thus G = (a) . 9 2 3 ) 2 h i h i Now suppose G = (a) . Clearly, ⇥a ⇤ G. For all b G, (b) (a) . h i h i ✓ 2 2 h i Thus k Z (b) = (a)k = (ak). Since is 1–1, b = ak. Thus a = 9G. 2 3 h i ⇤ (5) a = (a) a G (Isomorphisms preserve order). | | | | 8 2 Proof. n an = e (an) = (e) (a) = 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 G, xk = b has the same number of solutions in G as does xk = (b) in G2. Proof. Suppose a is a solution of xk = b in G, i.e., ak = b. Then k (ak) = (b) = (a) = (b), ) k so (a) is a solution of x = (b) in G. ⇥ ⇤ Now suppose y is a solution of xk = (b) in G. Since is onto, a G (a) = y. Then 9 2 3 k (a) = (b) (ak) = (b) = ak = b () ) k since is 1–1. Th⇥ us a⇤ is a solution of x = 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. Proof. Follows directly from property (5). ⇤ 82 6. ISOMORPHISMS Example. Is C⇤ R⇤ with multiplication the operation of both groups? Solution. ⇡ 2 x = 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. Proof. 1 The inverse of is 1–1 since is. Let g G. (g) = g = is onto. 2 ) Now let x, y G. Then 2 1 1 1 1 1 1 (xy) = (x) (y) (xy) = (x) (y) () () 1 1 xy = (x) (y) = xy. 1 Thus operations are preserved and is an isomorphism. ⇤ (2) G is Abelian G is Abelian. Proof. () Follows directly from Theorem 6.2(3) since isomorphisms preserve commutivity. ⇤ (3) G is cyclic G is cyclic. Proof. () Follows directly from Theorem 6.2(4) since isomorphisms preserve order. ⇤ 6. ISOMORPHISMS 83 (4) If K G, then (K) = (k) k K G. Proof. { | 2 }  Follows directly from Theorem 6.2(4) since isomorphisms preserve order. ⇤ 1 (5) If K G, then (K) = g G (g) K G. Proof. { 2 | 2 }  Follows directly from (1) and (4). ⇤ (6) Z(G) = Z(G). Proof. 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.) Solution. is clearly 1–1 and onto. Suppose a + bi, c + di C. 2 (a + bi) + (c + di) = (a + c) + (b + d)i = (a + c) (b + d)i (a bi) + (c di) = (a + bi) + (c + di), ⇥ ⇤ ⇥ ⇤ so is an automorphism of C under addition. Now suppose a + bi, c + di C⇤. 2 (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 R2. 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

(a,b)

Consider (a, b) not on y = 2x. The line through (a, b) y = 2x is ? 1 y b = x a . To ﬁnd P (using substitution): 2 ⇣1 ⌘ 5 1 1 2 2 4 2x b = x a = x = a + b = x = a + b = y = a + b. 2 ) 2 2 ) 5 5 ) 5 5 Thus the direction⇣ vector⌘ v = P (a, b) takes (a, b) to P . 4 2 2 1 v = a + b, a b . 5 5 5 5 Deﬁne⇣ ⌘ (a, b) = (b, a). is clearly 1–1 and onto. 6. ISOMORPHISMS 85 Finally, (a, b) + (c, d) = (a + c, b + d) = ⇥ ⇤ (b + d, a + c) = (b, a) + (d, c) = (a, b) + (c, d), so is an isomorphism. Definition (Inner Automorphism Induced by a).

1 Let G be a group, a G. The function a deﬁned by a(x) = axa x G is called the inner automorphism2 of G induced by a. 8 2

Example. The action of the inner automorphism of D4 induced by R90 is given in the following table. 86 6. ISOMORPHISMS Problem (Page 145 # 45). Suppose g and h induce the same inner auto- 1 morphism of a group G. Prove h g Z(G). 2 Proof. Let x G. Then 2 1 1 1 1 (x) = (x) = gxg = hxh = h gxg h = x = g h ) ) ) 1 1 1 1 1 (h g)x(h g) h = x = (h g)x = x(h g) = ) ) 1 h g Z(G) since x is arbitrary. 2 ⇤ 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. Proof. 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. 1 Suppose ↵ Aut(G). ↵ is clearly 1–1 and onto. Now 2 1 1 1 ↵ (xy) = ↵ (x)↵ (y) ( = by 1–1) () ( 1 1 1 1 1 ↵ ↵ (xy) = ↵ ↵ (x)↵ (y) xy = ↵ ↵ (x) ↵ ↵ (y) () () ⇥ ⇤ ⇥ ⇤ ⇥ ⇤ ⇥ xy⇤ = xy. 1 Thus ↵ is operation preserving, and Aut(G) is a group. 6. ISOMORPHISMS 87 Now let , Inn(G) Aut(G). For all x G, g h 2 ✓ 2 1 1 1 1 gh(x) = g(hxh ) = ghxh g = (gh)x(gh) = gh(x), so = Inn(G), and Inn(G) is closed under composition. gh g h 2 1 Also, = 1 since g g 1 1 1 1 1 1 1 1 (x) = 1(gxg ) = g gxg (g ) = g gxg g = exe = (x), g g g e and so also 1 = e. Thus Inn(G) Aut(G) by the two-step test. gg ✓ ⇤ Example. Find Inn(Z12). Solution.

Since Z12 = 0, 1, 2, . . . , 11 , Inn(Z12) = 0, 1, 2, . . . , 11 , but the second { } { } list may have duplicates. That is the case here. For all n Z12 and for all 2 x Z12, 2 (x) = n + x + ( n) = n + ( n) + x = 0 + x = 0 + x + 0 = (x), n 0 the identity automorphism. Thus Inn(Z12) = 0 . { } ⇤ 88 6. ISOMORPHISMS

Example. Find Inn(D3). Solution.

D = ", (1 2 3), (1 3 2), (2 3), (1 3), (1 2) as permutations. Then 3 { } Inn(D3) = ", (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(Zn) = U(n)). For all n N, Aut(Zn) = U(n). 2 Proof.

Let ↵ Aut(Zn). Then 2 ↵(k) = ↵(1 + 1 + + 1) = (↵(1) + ↵(1) + + ↵(1)) = k↵(1). · · · · · · k terms k terms Now 1 = n = ↵(1) = n, so ↵(1) is a generator of since 1 is also a | {z } | {z Zn} generator.| | Now)consider| |

T : Aut(Zn) U(n) where T (↵) = ↵(1). ! Suppose ↵, Aut(Zn) and T (↵) = T (). Then ↵(1) = (1), so k Zn, ↵(k) = k↵(1)2= k(1) = (k). Thus ↵ = and T is 1–1. 8 2

[To show T is onto.] Now suppose r U(n) and consider ↵ : Zn Zn deﬁned 2 ! by ↵(s) = sr mod n s Zn. 8 2 [To show ↵ Aut(Zn).] Suppose ↵(x) = ↵(y). Then xr = yr mod n. 2 1 1 1 But r exists modulo n = xrr = yrr mod n = x 1 = y 1 mod n = ) ) · · ) x = y mod n, so ↵ is 1–1.

Suppose x Zn. By Theorem2, s Zn ↵(s) = sr = x, so ↵ is onto. 2 9 2 3 Now suppose x, y Zn. 2 ↵(x + y) = (x + y)r mod n = (xr + yr) mod n = xr mod n + yr mod n = ↵(x) + ↵(y), so ↵ Aut(Zn). 2 Since T (↵) = ↵(1) = r, T is onto. 90 6. ISOMORPHISMS Finally, let ↵, Aut(Zn). Then 2 T (↵) = (↵)(1) = ↵ (1) = ↵(1 + 1 + + 1) · · · (1) terms ⇥ ⇤ ↵(1) + ↵(1) + + ↵(1) = ↵(1)(1) = T (↵)T (), | · ·{z· } (1) terms so Aut(Zn) U(n). ⇡ | {z } ⇤ Example. Aut(Z10) U(10). The multiplication tables for the two groups are given below: ⇡

The isomorphism is clear.