# Finitely generated torsion abelian group

I am trying to solve the following problem:

Given is a finitely generated abelian group $G$. Show that: $G$ is torsion group ($T(G)=G$) if and only if $G$ is a finite group.

The $\Rightarrow$ direction: Let $S=\left \{ g_{1}, \dots,g_{r} \right \}$ be a set of generators of $G$. According to the definition of finitely generated abelian group (namely that $\langle S \rangle=\left \{ g_{1}^{e_{1}}\cdot \dots \cdot g_{r}^{e_{m}}\mid e_{i}\in \mathbb{Z} \right \}$ ), i can write the elements of $G$ in the following form $\left \{ g_{1}^{e_{1}}\cdot \dots \cdot g_{r}^{e_{r}}\mid e_{i}\in \mathbb{Z} \right \}$. Since $G$ is a torsion group, then every $g_{i}$ must have a finite order, let's say $n_{i}$ for $1\leq i\leq r$. Then it follows that $G=\left \{ g_{1}^{e_{i}}\cdot ...\cdot g_{r}^{e_{r}} \mid 0\leq e_{i}\leq n_{i} \right \}$ for $1\leq i\leq r$. So $\mid G\mid < \infty$.

Is it correct? Can somebody help me with the other direction of the proof, where i have to show that every finitely generated abelian group of finite order is a torsion group i.e every element of $G$ must have finite order? Thank you in advance! I appreciate any hints and comments.

• The other direction is the easy one: every element of a finite group is torsion. – Qiaochu Yuan Oct 27 '15 at 22:50
• As Qiaochu says, you're already mostly done. You can quote lagrange's theorem if you really want to tie it in a bow – Adam Hughes Oct 28 '15 at 0:16

In what you've done, you even may suppose $$0\le e_i.
The other way, observe that, if a group $$G$$ contains a non-torsion element $$g$$, the subgroup $$\langle\,g\,\rangle$$ is isomorphic to $$\mathbf Z$$, hence $$G$$ contains an infinite subgroup.