Question

(1) Let φ:G→H be an isomorphism between two groups G,H. Assume that G is abelian. Prove that H is abelian. (2) Consider the additive group Z and the multiplicative group G={1,i,−1,−i}, and define φ:Z→G by φ(n)=in (a) Show that φ is a homomorphism. (b) Is φ injective? surjective? (c) Find φ−1({i}) and φ−1({1}).

Answer 1

The first part demonstrates that if G is abelian and there exists an isomorphism φ from G to H, then H is also abelian. In the second part, we show that the specified φ from Z to G is a homomorphism, it is neither injective nor surjective, and we determine the pre-images of the elements i and 1 in G.

Step-by-step explanation:

Part (1): Proving H is Abelian

Given that φ:G→H is an isomorphism and G is abelian, we want to show that H is also abelian. Since G is abelian, for any a, b ∈ G, we have ab = ba. Considering the isomorphism φ, this implies that φ(ab) = φ(ba).

Applying the properties of isomorphisms, we get φ(a)φ(b) = φ(b)φ(a). Because φ(a) and φ(b) are elements of H, this proves that H is abelian as the product is commutative in H.

Part (2): Properties of φ:Z→G

(a) To show that φ is a homomorphism, we must show that φ(n+m) = φ(n)φ(m) for any n, m ∈ Z. We have φ(n+m) = i^{n+m} = i^n × i^m = φ(n)φ(m), which confirms that φ is a homomorphism.

(b) φ is not injective because there are distinct integers whose images are the same, for example, φ(0) = φ(4) = 1. It is not surjective because there are elements in G not reached by φ, such as ½ (not in Z).

(c) φ^{-1}({i}) = φ(n) = i = 4k+1 , and φ^{-1}({1}) = φ(n) = 1 = 4k .