$G$ is a finitely generated abelian group. Any element of $G$ (except $0$) is of infinite order. Then $G$ is a free abelian group.
I cannot finish the proof. Any hint is appreciated.
Let $X=\{x_1,\dots,x_n\}$ be a system of generators for $G$. Define a map $f:\mathbb{Z}^n\to G$ by $(a_1,\dots,a_n)\mapsto a_1x_1+\cdots+a_nx_n\Rightarrow f$ is a group epimorphism $\Rightarrow \mathbb{Z}^n/\ker(f)\cong G$ by FIT. Since $\mathbb{Z}^n$ is free abelian and $\ker(f)\le\mathbb{Z}^n\Rightarrow \ker(f)$ is free abelian $\Rightarrow\exists$ a basis $Y=\{y_1,\dots,y_r\}$ for $\ker(f)$ and positive integers $d_1\mid d_2\mid\cdots\mid d_r$ s.t. $Y=\{y_1,\dots,y_r\}=\{d_1x_1,\dots,d_rx_r\}$.
Best Answer
Actually you wrote something which is not quite correct. The correct assertion looks like this:
Since $\mathbb{Z}^n$ is free abelian and $\ker(f)\le\mathbb{Z}^n\Rightarrow \ker(f)$ is free abelian $\Rightarrow\exists$ a basis $Y=\{y_1,\dots,y_n\}$ for $\mathbb Z^n$ and positive integers $d_1\mid d_2\mid\cdots\mid d_r$ s.t. $\{d_1y_1,\dots,d_ry_r\}$ is a basis of $\ker(f)$. Then $G\simeq\mathbb Zy_1\oplus\cdots\oplus\mathbb Zy_n/\mathbb Zd_1y_1\oplus\cdots\oplus\mathbb Zd_ry_r\simeq\mathbb Zy_1/\mathbb Zd_1y_1\oplus\cdots\oplus\mathbb Zy_r/\mathbb Zd_ry_r\oplus\mathbb Z^{n-r}$. But $\mathbb Zy_i/\mathbb Zd_iy_i\simeq \mathbb Z/\mathbb Zd_i$ and thus $G$ has elements of finite order unless $r=0$ and in this case $G$ is free.