# Finitely generated module with free submodule $S$ and $M/S$ torsion-free implies $M$ is free

Let $$R$$ be a PID and $$M$$ a module. Show the following:

(i) If $$M$$ is finitely generated and $$S$$ is a free submodule with $$M/S$$ torsion-free, then $$M$$ is free.

(ii) If $$M$$ is torsion-free and $$M/S$$ is finitely generated of torsion for all submodule $$S \neq 0$$ of $$M$$, then $$M \cong R$$. Deduce that if an infinite abelian group satisfies every non zero subgroup has finite index then $$G \cong \mathbb Z$$.

I think I could solve (i) but I am not so sure if it is correct so I'll write my solution:

I know that if $$R$$ is a PID and $$M$$ is a f.g. torsion-free module, then $$M$$ is free. So, using this, I've tried to show that $$M$$ is torsion-free. So suppose there is $$r \neq 0, m \in M$$ such that $$rm=0$$. Consider the projection $$\pi: M \to M/S$$ $$m \to \overline{m}$$

Then $$0=\pi(rm)=r\pi(m)$$. Since $$r \neq 0$$ and $$M/S$$ is torsion-free, it must be $$\pi(m)=0 \implies m \in S$$. But $$S$$ is a free submodule which implies $$S$$ is torsion-free, as $$rm=0$$ and $$r \neq 0$$, it follows $$m=0$$. It follows directly that $$M$$ is torsion-free.

I have no idea what to do in (ii). As for the last part of (ii), any abelian group is a $$\mathbb Z$$-module. I don't see why $$G/S$$ is of torsion (if we have this, then we are under the hypothesis of first part of (ii) and we can apply the proposition).

Any suggestions would be appreciated.

• (After fixing a typo) I find your proof to (i) pretty clear. – user26857 Dec 8 '14 at 8:50

If consider $S$ a non-zero cyclic submodule, from $M/S$ finitely generated you get $M$ finitely generated, so $M$ is free. Now you can use this answer.
For an abelian group $G$ and $S$ a non-zero subgroup, having finite index means that $G/S$ is finite, or a finite abelian group is finitely generated and torsion.