Answer to Question #206174 in Abstract Algebra for Sujata

Question

Prove that Homz(Z,M) is isomorphic to M where M is an abelian group.

Answer 1

Define a function $\phi:\text{Hom}(\mathbb{Z},G) \to G$

such that $\phi(f)=f(1) \forall f \in \text{Hom}(\mathbb{Z},G).$

This function is well-defined since $f(n) = [f(1)]^n \forall n \in \mathbb{Z}.$

Let $f,g \in \text{Hom}(\mathbb{Z},G).$ Consider

$\phi(fg)=(fg)(1)=f(1)g(1)=\phi(f)\phi(g).$

Hence $\phi$ is a homomorphism.

Next, suppose $\phi(f)=\phi(g).$ Then $f(1)=g(1).$ So $\forall n \in \mathbb{Z}, f(n) = [f(1)]^n\\ = [g(1)]^n=g(n).$

Thus $f = g.$ Hence $\phi$ is one-to-one.

Finally, let $x \in G.$ Define a function $\psi:\mathbb{Z}\to G$ such that $\psi(n)=x^n\forall n \in \mathbb{Z}.$ Let $m,n \in \mathbb{Z}.$ Consider

$\psi(m+n) = x^{m+n}=x^mx^n = \psi(m)\psi(n).$

So $\psi$ is a homomorphism.

Hence $\phi(\psi) = \psi(1)=x.$ Thus $\psi$ is onto.

Hence $\phi$ is an isomorphism.