Let $G$ be a finitely presented group and $M$ be a finitely generated $ZG$-module. By a retract of $M$, I mean a submodule $N$ of $M$ for which there is a homomorphisms $f:M\to N$ such that $fi=id_N$, where $i$ is the inclusion.
My question is that for which groups $G$ there is an integer $n$ such that for every chain $\cdots \subset M_3 \subset M_2 \subset M_1 \subseteq M$ of retracts of $M$, we have $M_i =M_n$ for all $i\geq n$? Can we find such an $n$ for free groups of finite ranks?
For example, If $G=1$, then $M$ be a finitely generated abelian groups and $n$ can be the number of direct summands in the canonical form of $M$.
For finite groups I believe this is still true.
In the case $G=1$, all that you are using is that every finitely generated $\mathbb ZG$-module is also a finitely generated abelian group.
But that fact remains true if $G$ is a finite group, for the following reasons. Let $\{m_1,...,m_K\} \subset M$ be a finite generating set for $M$ as a $\mathbb ZG$ module. Enumerating $G=\{g_1,...,g_L\}$, it follows that $\{g_l \cdot m_k \mid l=1,\ldots,L, \, k=1,\ldots,K\}$ is a finite generating set for $M$ as an abelian group. So once again, one may use the number of direct summands of $M$.