Abstract Algebra Theory and Applications - Computer Science ...

8.1 DEFINITION AND EXAMPLES 139the group operation is preserved.Example 2. We can define an isomorphism φ from the additive group ofreal numbers (R, +) to the multiplicative group of positive real numbers(R + , ·) with the exponential map; that is,φ(x + y) = e x+y = e x e y = φ(x)φ(y).Of course, we must still show that φ is one-to-one and onto, but this can bedetermined using calculus.Example 3. The integers are isomorphic to the subgroup of Q ∗ consistingof elements of the form 2 n . Define a map φ : Z → Q ∗ by φ(n) = 2 n . Thenφ(m + n) = 2 m+n = 2 m 2 n = φ(m)φ(n).By definition the map φ is onto the subset {2 n : n ∈ Z} of Q ∗ . To show thatthe map is injective, assume that m ≠ n. If we can show that φ(m) ≠ φ(n),then we are done. Suppose that m > n and assume that φ(m) = φ(n). Then2 m = 2 n or 2 m−n = 1, which is impossible since m − n > 0. Example 4. The groups Z 8 and Z 12 cannot be isomorphic since they havedifferent orders; however, it is true that U(8) ∼ = U(12). We know thatU(8) = {1, 3, 5, 7}U(12) = {1, 5, 7, 11}.An isomorphism φ : U(8) → U(12) is then given by1 ↦→ 13 ↦→ 55 ↦→ 77 ↦→ 11.The map φ is not the only possible isomorphism between these two groups.We could define another isomorphism ψ by ψ(1) = 1, ψ(3) = 11, ψ(5) = 5,ψ(7) = 7. In fact, both of these groups are isomorphic to Z 2 × Z 2 (seeExample 14 in Chapter 2).Example 5. Even though S 3 and Z 6 possess the same number of elements,we would suspect that they are not isomorphic, because Z 6 is abelian andS 3 is nonabelian. To demonstrate that this is indeed the case, suppose thatφ : Z 6 → S 3 is an isomorphism. Let a, b ∈ S 3 be two elements such that