Is the order of the product of an infinite order element and finite order element always infinite?

Question

Let G be an abelian group, and let g, h ∈ G. Show that, if g has infinite order and h has finite order, then gh has infinite order.

I was trying a proof by contradiction by letting (gh)^n = 1 so g^nh^n=1. But I don't know how to proceed/know if I am going down the right path. Am I going down the right path?

Answer 1

Suppose that $|h|=n$ and $|gh|=m$.
Then we have $$1=(gh)^m=g^mh^m\implies g^{-m}=h^m$$ Hence $|g^m|=|g^{-m}|=|h^m|=\frac{n}{(n,m)}$.
This is a contradiction because $|g|$ is infinite implies that $|g^m|$ is infinite.