User:J D Bowen/Math735 HW2

Katies stuff, needs texing:

1.6

is a homomorphism iff G is abelian. Suppose the map from G to itself defined by ϕ(g) = g-1 is a homomorphism. Then by the definition of homomorphism we know that ϕ(ab) = ϕ(a)ϕ(b) ∀a,b ∈ G. However, according to the map, ϕ(ab) = (ab)-1 = b-1a-1 = ϕ(b)ϕ(a). Therefore, we have ϕ(a)ϕ(b) = ϕ(b)ϕ(a). Thus, G is abelian. Now suppose G is abelian. Then ab = ba ∀a,b ∈ G. We know that our map is defined as ϕ(ab) = (ab)-1. Since G is abelian, (ab)-1= b-1a-1 = a-1b-1 = ϕ(a)ϕ(b). Therefore, ϕ(ab) = ϕ(a)ϕ(b), which means the map from G to itself is a homomorphism.
 * 1) 17. Let G be a group. Prove that the map from G to itself defined by

. Prove that for any fixed integer k < 1, the map from G to itself defined by
 * 1) 19. Let G =

is a surjective homomorphism but is not an isomorphism. Let z1 and z2 be elements of G. Then ϕ( z1 z2) = (z1 z2)k = z1k z2k = ϕ(z1)ϕ(z2). Therefore, the map is a homomorphism. Now consider the element zk in the image of G. Since G is being

mapped to itself, there exists some

that for some . Therefore, , so z is an element of G. Thus, the map is a surjective homomorphism. Next, to prove that the map is not an isomorphism, we need to show that it is not injective. Consider the elements . Clearly, . Now set , and observe that ϕ(1) = 14 = 1 and ϕ(i) = i4 = 1. Therefore, ϕ(1) = ϕ(i). Thus the map is not injective, and not an isomorphism.