# Prove that the quotient group $A/\varphi (A)$ is finite when $A$ is torsion-free abelian

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.

• I wonder whether $A$ needs to be finitely generated for this to hold.
– lhf
Commented Dec 28, 2019 at 19:03
• The statement holds in general, whether or not A is fin. gen. (See, for instance, Fuchs "Infinite Abelian Groups", volume 2, exercise 92.5) Commented Dec 28, 2019 at 19:18
• The exercise you mentioned holds for a "torsion-free group of finite rank"
– user291350
Commented Dec 28, 2019 at 19:51
• Ops, you are right! Thank you. I edited the question. Commented Dec 28, 2019 at 20:24

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 $$A$$.