Showing that a group has infinite order Let $G$ be an abelian group such that for all $g \in G$ and for all positive integers $k$, there exists some $h \in G$ such that $h^k = g$.
Now, I would like to show that the group is either the trivial group, or of infinite size. The trivial case is straightforward, but I am having trouble proving the infinite cardinality.
I start by supposing that $G$ has finite cardinality with $|G| = m>1$ where $m$ is some integer. I now pick an abitrary $g$ and $k=m+1$, and the corresponding $h$ so that we have $h^k=g$.
If I am on the right track, I'm lost as to where to go to from here. If I am not on the right track, well then I am quite lost!
Thank you for any help!

If $m:=|G|$ is finite, then $h^m=e$ for all $h\in G$. So with $k=m$ (not $m+1$), we conclude from the existence of $h$ with $h^k=g$ that $g=e$ (and so $m=1$).

Alternative approach:
Suppose $G$ is not trivial and let $1\ne a\in G$. If $a$ has inifnite order, $G$ is infintei and we are done. So assume $a^m=1$ for some $m>1$. Then $x\mapsto x^m$ is not injective, but by assumption is surjective. This is possible only for infinite sets.
