$G$ 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. $Y=\{y_1,\dots,y_r\}=\{d_1x_1,\dots,d_rx_r\}$.

  • $\begingroup$ can you use the fundamental theorem of finitely generated abelian groups? $\endgroup$ – Ittay Weiss Mar 23 '13 at 8:09
  • $\begingroup$ This is the theorem 10.19 of Rotman's book, An introduction to thetheory of groups. $\endgroup$ – Mikasa Mar 23 '13 at 8:36
  • $\begingroup$ In fact this is one of the steps in the proof for the classification of f.g. abelian groups. Torsionfree f.g. abelian groups are free. $\endgroup$ – Martin Brandenburg Mar 23 '13 at 11:20
  • $\begingroup$ Thank Ittay (I can use FTFGAG), @Babak, and Martin for your reference and hint. $\endgroup$ – Sam Mar 23 '13 at 16:22
  • $\begingroup$ @Sam: If you like I can bring you here the proof from that book. $\endgroup$ – Mikasa Mar 23 '13 at 18:23

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.

  • $\begingroup$ Thank you very much, @YACP. I like your answer. Could you explain a little about how to get this? $G\simeq \mathbb{Z}^n/\ker(f)\simeq \mathbb Zy_1\oplus\cdots\oplus\mathbb Zy_n/\mathbb Zd_1y_1\oplus\cdots\oplus\mathbb Zd_ry_r$. $\endgroup$ – Sam Mar 24 '13 at 5:19
  • 1
    $\begingroup$ Actually we have $\mathbb Z^n=\mathbb Zy_1\oplus\cdots\oplus\mathbb Zy_n$ and $\ker(f)=\mathbb Zd_1y_1\oplus\cdots\oplus\mathbb Zd_ry_r$ since for a free abelian group $F$ with a basis $e_1,\dots,e_n$ we have $F=\mathbb Ze_1+\cdots+\mathbb Ze_n$ (because $e_1,\dots,e_n$ is a system of generators for $F$) and the sum is direct because $e_1,\dots,e_n$ are linealy independent. $\endgroup$ – user26857 Mar 24 '13 at 5:42
  • $\begingroup$ Thank you so much, @YACP. Now I am clear. $\endgroup$ – Sam Mar 24 '13 at 5:51

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