Abstract Algebra Theory and Applications - Computer Science ...

140 CHAPTER 8 ISOMORPHISMSab ≠ ba. Since φ is an isomorphism, there exist elements m and n in Z 6such thatHowever,φ(m) = aφ(n) = b.ab = φ(m)φ(n) = φ(m + n) = φ(n + m) = φ(n)φ(m) = ba,which contradicts the fact that a and b do not commute.Theorem 8.1 Let φ : G → H be an isomorphism of two groups. Then thefollowing statements are true.1. φ −1 : H → G is an isomorphism.2. |G| = |H|.3. If G is abelian, then H is abelian.4. If G is cyclic, then H is cyclic.5. If G has a subgroup of order n, then H has a subgroup of order n.Proof. Assertions (1) and (2) follow from the fact that φ is a bijection.We will prove (3) here and leave the remainder of the theorem to be provedin the exercises.(3) Suppose that h 1 and h 2 are elements of H. Since φ is onto, thereexist elements g 1 , g 2 ∈ G such that φ(g 1 ) = h 1 and φ(g 2 ) = h 2 . Therefore,h 1 h 2 = φ(g 1 )φ(g 2 ) = φ(g 1 g 2 ) = φ(g 2 g 1 ) = φ(g 2 )φ(g 1 ) = h 2 h 1 .We are now in a position to characterize all cyclic groups.Theorem 8.2 All cyclic groups of infinite order are isomorphic to Z.Proof. Let G be a cyclic group with infinite order and suppose that a is agenerator of G. Define a map φ : Z → G by φ : n ↦→ a n . Thenφ(m + n) = a m+n = a m a n = φ(m)φ(n).□