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!
 A: Suppose $G$ is finite. Let $g \in G$ be arbitrary. Then there exists some $h \in G$ such that $h^{\lvert G \rvert} = g$. But $h^{\lvert G \rvert} = e$ by Lagrange's theorem! Since $g$ was arbitary, we conclude that $G = \{e\}$ is trivial.
A: 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 $g=e$.
A: 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.
A: If $G$ is not the trivial group, take $g\in G\setminus\{e_G\}$. There is some $h_1\in G$ such that $h_1^{\,2}=g$ and there is some $h_2\in G$ such that $h_2^{\,3}=g$. But then $h_1\ne h_2$, since this would imply that $h_1^{\,2}=h_1^{\,3}$, which in turn would imply that $h_1=e_G$, which is impossible, since $g\ne e_G$. Now, take $h_3\in G$ such that $h_3^{\,7}=g$. Then $7\equiv1\pmod2$ and $7\equiv1\pmod3$. So, by the same argument as before, $h_3\ne h_1$ and $h_3\ne h_2$. And so on…
