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Here is a problem:

If $F$ is a free abelian group of rank $n$ and $H$ is a subgroup of rank $k, then $F/H$ has an element of infinite order.

What I did:

I assume $F=\langle x_1,x_2,…,x_n\rangle$; we can have $H=\langle x_1,x_2,…,x_k\rangle$. And I suppose, in contrast, that all elements of $F/H$ be of finite order, so for $f+H\in\frac{F}{H}$, there is an integer $n$ that $nf\in H$. I see that: $nf\in H\Longrightarrow\exists m_1,m_2,…m_k\in\mathbb Z,nf=\sum_{i=1}^km_ix_i$ So by taking $f=x_{k+1}, x_{k+2},...x_{n}$, I have: $t_1x_{k+1}=\sum_{i=1}^km_{i1}x_i\\t_2x_{k+2}=\sum_{i=1}^km_{i2}x_i\\ \vdots \\t_nx_{n}=\sum_{i=1}^km_{in}x_i $ wherein $t_1,t_2,...,t_n\in\mathbb Z$. I think by adding the sides of above equalities, I can find a contradiction to independency of $x_1,x_2,…,x_n$. Is above idea for proofing right? Thanks for your time.

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    @amWhy: Thanks. I saw that. It is OK to me and as you, I am here not to up and down. :-) I dont want to be up when I know my Maths knowledge. Thanks Amy. My Tel line is good now and I hope see you and other Masters again. Believe me, bretheang here between you all has been my old wish. :^)2013-04-18

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Let $F$ be a free abelian group of rank $n$ and $H$ a subgroup of rank $r. Let $x_1,\dots,x_n$ be a basis of $F$. Suppose $F/H$ is torsion, then for some positive $m_1,\dots,m_n$ we have $m_ix_i \in H$. Let $m=m_1\cdots m_n$ then $m x_1,\dots , mx_n \in H$. Note that the $mx_i$ are linearly independent since if $\sum_{i=1}^n mc_ix_i=0 \Rightarrow m\sum_{i=1}^n c_ix_i=0,$ but this implies the rank of $H$ is at least $n$, so $F/H$ is not torsion.

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    The magic point is to take $m$ as above. Now I see what I had a mistake. Thanks Jacob.2012-10-10