I want to prove the following.
Suppose $A$ is a torsion-free abelian group of finite rank (or, if you prefer, an additive subgroup of $\Bbb{Q}^n$, where $n\ge1$ is the rank of $A$)and $\varphi$ is an injective homomorphism of $A$ into itself (not necessarily surjective). Show that the quotient group $A/\varphi (A)$ is finite.
Using the Chinese Remainder theorem, I was able to prove it for the special case $\varphi (a)=ma$, where $m$ is a non zero integer.
What about the general case?
Thank you in advance for your help.
We can suppose that $A\subset\mathbf{Q}^n$ contains $\mathbf{Z}^n$; we can view $\varphi$ as an element of $\mathrm{GL}_n(\mathbf{Q})$. In addition, since you already did the case of $a\mapsto ma$, you can compose with such a map, to ensure that $\varphi(\mathbf{Z}^n)\subset\mathbf{Z}^n$; that is, $\varphi$ has integer entries and nonzero determinant $d$. Then $\varphi$ induces an endomorphism of $\mathbf{Q}^n/\mathbf{Z}^n$, preserving the image $\bar{A}=A/\mathbf{Z}^n$. On the one hand for each prime $p$ not dividing $d$, $\bar{\varphi}$ is invertible on the $p$-primary component of $\mathbf{Q}^n/\mathbf{Z}^n$, and more precisely has the property of being "locally of finite order", that is, has only finite orbits; this implies that it preserves every subgroup, and in particular preserves the $p$-primary part of $\bar{A}$. On the other hand for every prime divisor of $d$ (which are finitely many), the $p$-primary part of $\bar{A}$ is an artinian abelian $p$-group and hence the image of $\bar{\varphi}$ therein has finite index. So, $\bar{\varphi}(\bar{A})$ has finite index in $\bar{A}$. Since $\varphi(\mathbf{Z}^n)$ has finite index in $\mathbf{Z}^n$, we conclude that $\varphi(A)$ has finite index in $A$.