Let $G$ is an abelian infinte group such that for all nontrivial subgroups $H$ $\forall H\leq G, \left|\frac{G}{H}\right|<\infty$ Prove that $G\cong\mathbb Z$.
What I have done:
Clearly, it is enough to show that $G$ is a cyclic group. Moreover, I see as $G/H$ is a finite group for all subgroups $H$, then $G$ cannot have any elements with the finite order. This means to me that if $H \leq G$ and $H$ is cyclic then $H$ cannot be finite. Help your friend for the rest. Thanks. :)