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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 $k<n$, 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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You do not have that $H$ is generated by the $x_i$. The generators will generally be linear combinations of these. –  Aaron Oct 10 '12 at 19:58
    
@Aaron: So, all above will collapse? –  Babak S. Oct 10 '12 at 20:01
1  
Yes, it would. To give an example, let $F=\langle x_1,x_2 \rangle$ and $H=\langle 2x_1 \rangle$. There is no way to extend a basis of $H$ to a basis for $F$. More generally, given a basis for $H$, there is a linear map sending the initial elements of a bass of $F$ to the basis of $H$: just take a square matrix whose first few columns are the basis of $H$ (expressed in terms of the basis of $F$). Over a field, this can always be made to be an invertible, over $\mathbb Z$ (i.e., with an abelian group), many fewer matrices are invertible. –  Aaron Oct 10 '12 at 23:39
    
I've noticed lately that someone has been "serial upvoting" your questions almost daily, in excess, to the point where you are losing points. *I would raise this issue with a moderator, so they can look into it. (I will do so, if you'd like.) Yesterday, for example: It looks like 19 of your questions were upvoted one right after the other...If I recall, a week or two ago, the same thing happened, to you, but it was excessive, successive upvoting of answers. (Trust me, I take care not to overdo upvotes) but someone seems to be doing the serial upvoting regularly with your questions/answers. –  amWhy Apr 18 '13 at 3:45
    
@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. :^) –  Babak S. Apr 18 '13 at 13:32

1 Answer 1

up vote 1 down vote accepted

Let $F$ be a free abelian group of rank $n$ and $H$ a subgroup of rank $r<n$. 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. –  Babak S. Oct 10 '12 at 20:26

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