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.