I'm trying to solve the following exercise (Fuchs, "Infinite Abelian Groups", Vol. $2$, pp. $153$, Ex.$5$):
Let $A$ be a torsion-free abelian group of finite rank. If $\phi$ is an isomorphism of $A$ into itself, then $\phi A$ is of finite index in $A$. [Hint: if $F$ is a free subgroup of $A$ of the same rank as $A$, then $F\cap \phi F$ is of finite index in $F$; apply Ex. $4$ to $\phi A / (F\cap \phi F)\le A/(F\cap \phi F)$, noting that $A/F\cong \phi A/\phi F$.]
In order to get some intuition, I've considered the finitely generated free group $F=\oplus_{i=1}^n\langle x_i\rangle\cong \oplus_{i=1}^n \mathbb{Z}$. If $\phi$ is a monomorphism of A into intself, then $F\cap \phi F\cong\oplus_{i=1}^m \mathbb{Z}$ for some $m\le n$ since every subgroup of a free group is again free. But now $F/(F\cap\phi F)\cong\mathbb{Z}^n/\mathbb{Z}^m\cong \mathbb{Z}^{n-m}$, which doesn't have finite index if $m < n$. What am I missing?
Thanks in advance for your help.