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 φ from a group G to a group G is a 1–1 mapping from G onto G that preserves the group operation. That is, φ(ab) = φ(a)φ(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 φ : G G. ! (2) Show φ is 1–1: assuming φ(a) = φ(b), show a = b. (3) Show φ is onto: g G. show g G φ(g) = g. 8 2 9 2 −3− (4) Show φ(ab) = φ(a)φ(b) a, b G. 8 2 Example. With a = 42, show a Z42. | | h i ⇡ Proof. n Define φ : a Z42 by φ(a ) = n mod 42. h i ! [Show 1–1.] Suppose φ(an) = φ(am). Then n = m mod 42 = 42 n m = an = am ) | − ) by Theorem 4.1. Thus φ is 1–1. n n [Show onto.] For all n Z42, a a and φ(a ) = n, so φ is onto. 2 2 h i [Show operation preservation.] For all an, am a , 2 h i φ(anam) = φ(an+m) = n+m mod 42 = n mod 42+m mod 42 = φ(an)+φ(am). Thus, by definition, a Z42. h i ⇡ ⇤ Example. Any finite cyclic group a with a = n is isomorphic to Zn under φ(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 φ : G G by φ(x) = ln(x). ! [Show 1–1.] Suppose φ(x) = φ(y). Then ln(x) = ln(y) = eln(x) = eln(y) = x = y, ) ) so φ is 1–1. [Show onto] Now suppose x G. ex > 0 and φ(ex) = ln ex = x, so φ is onto. 2 [Show operation preservation.] Finally, for all x, y G, 2 φ(xy) = ln(xy) = ln x + ln y = φ(x) + φ(y), so G G. ⇡ [Question: is φ the only isomorphism?] ⇤ Example. Any infinite cyclic group is isomorphic to Z with addition. Given a with a = , define φ(ak) = k. The map is clearly onto. If φ(an) = φ(am), nh =i m =| | an 1= am, so φ is 1–1. Also, ) φ(anam) = φ(an+m) = n + m = φ(n) + φ(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 φ : 2 Z defined by φ(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 φ(x) = 2 − . φ is 1–1 and onto but φ(1 + 2) = φ(3) = 4 = 2 = 1 2 = φ(1)φ(2), 6 · so φ is not an isomorphism. Can this mapping be adjusted to make it an isomorphism? Use φ(x) = 2x. If φ(x) = φ(y), 2x = 2y = log 2x = log 2y = x = y, so φ is 1–1. ) 2 2 ) log2 y Given any positive y, φ(log2 y) = 2 = y, so φ is onto. Also, for all x, y R,φ(x + y) = 2x+y = 2x2y = φ(x)φ(y). 2 Thus φ 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 φ : U(8) U(12) by 1 (1), 3 5, 5 7, 7 11. ! ! ! ! ! φ is clearly 1–1 and onto. Regarding operation preservation: φ(3 5) = φ(7) = 11 and φ(3)φ(5) = 5 7 = 11. · · φ(3 7) = φ(5) = 7 and φ(3)φ(7) = 5 11 = 7. · · φ(5 7) = φ(3) = 5 and φ(5)φ(7) = 7 11 = 5. · · φ(1 3) = φ(3) = 5 and φ(1)φ(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 φ : G H by φ(a + bp2) = . ! b a a 2b c 2d Suppose φ(a + bp2) = φ(c + dp2). Then = = b a d c ) a = c and b = d = a + bp2 = c + dp2 = φ is 1–1. ) ) a 2b a 2b For H, a + bp2 G and φ(a + bp2) = , so φ is onto. b a 2 2 b a Now suppose a + bp2, c + dp2 G. Then 2 a + c 2(b + d) φ (a + bp2) + (c + dp2) = φ (a + c) + (b + d)p2 = = b + d a + c a + c 2b + 2d a 2b c 2d = + = φ(a + bp2) + φ(c + dp2). b + d a + c = b a d c Thus φ is an isomorphism under addition. Also, ac + 2bd 2(ad + bc) φ (a+bp2)(c+dp2) = φ (ac+2bd)+(ad+bc)p2 = = ad + bc ac + 2bd ac + 2bd 2ad + 2bc a 2b c 2d = = φ(a + bp2) φ(c + dp2) . ad + bc ac + 2bd b a d c So φ preserves multiplication also. (We will later see that thismakes φ 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 φ : G H by φ(x) = x. Suppose φ(x) = φ(y). Then ! m n n x = y = x = y = φ 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 φ x = x = x, so φ is onto. Now, suppose x, y G. Then n m n 2 ⇣ ⌘ ⇣ ⌘ n n n φ(x + y) = (x + y) = x + y = φ(x) + φ(y), m m m so addition is preserved and G H. But, n ⇡ n n φ(xy) = (xy) = x y = φ(x)φ(y), m 6 m m so multiplication is not preserved by⇣φ. ⌘⇣ ⌘ ⇤ 6. ISOMORPHISMS 77 Theorem (3). is an equivalence relation on the set G of all groups. ⇡ Proof. (1) G G. Define φ : G G by φ(x) = x. This is clear. ⇡ ! (2) Suppose G H. If φ : G H is the isomrphism and, for g G, ⇡1 ! 1 2 φ(g) = h, then φ− : H G where φ− (h) = g is 1–1 and onto. ! Suppose a, b H. Since φ is onto, ↵, β G φ(↵) = a and φ(β) = b. Then φ(↵β) =2 φ(↵)φ(β) = ab, so 9 2 −3− 1 1 1 φ− (ab) = φ− φ(↵)φ(β) = φ− φ(↵β) = ↵β = 1 1 1 1 φ− φ(↵) φ− φ(β) = φ− (a)φ− (b). 1 Thus φ− is an isomorphism and H G. ⇡ (3) Suppose G H and H K with φ : G H and : H K the isomorphisms. Then⇡ φ : G ⇡K is 1–1 and onto.! ! ! For all a, b G, 2 ( φ)(ab) = φ(ab) = φ(a)φ(b) = φ(a) φ(b) = ( φ)(a) ( φ)(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 com- putation in one group, the isomorphism allows us to perform the analagous computation in the other group.