# Does $M$ satisfy the descending chain conditions on $\mathbb{Z}G$-retracts?

‎Let $$H$$ be a subgroup of $$G$$‎. ‎Then a homomorphism $$r:G\to H$$ is said to be a retraction if the inclusion homomorphism $$i:H\hookrightarrow G$$ is a right inverse of $$r$$‎, ‎i.e‎. ‎$$r(x)=x$$ for all elements $$x\in H$$‎. ‎Then $$H$$ is called a retract of $$G$$‎ and denoted by $$H<_r G$$ (we can define retracts of $$R$$-modules similarly which I call them $$R$$-retracts).

A group $$G$$ is said to satisfy the descending chain conditions on retracts if for every chain of $$G_1 >_r G_2 >_r \cdots$$ of retracts of $$G$$ there is an ineger $$n$$ such that $$G_i =G_n$$ for all $$i\geq n$$.

My question : Assume that $$G$$ satisfies the descending chain conditions on retracts and $$M$$ is a finitely generated $$\mathbb{Z}G$$-module. Does $$M$$ satisfy the descending chain conditions on $$\mathbb{Z}G$$-retracts?

Let $$G=\mathbb{Z}/3^\infty=\displaystyle\lim_\to(\mathbb{Z}/3\hookrightarrow\mathbb{Z}/3^2\hookrightarrow\cdots)$$, and let $$M$$ be $$\mathbb{F}_2G$$ as a $$\mathbb{Z}G$$-module. Then $$G$$ has no non-trivial retracts, so presumably it vacuously satisfies the descending chain condition on them. But $$M$$ has a descending chain of retracts as follows. The first given by taking $$e_1$$ equal to the sum of the group elements as an idempotent in $$\mathbb{F}_2\mathbb{Z}/3$$, and $$M_1=e_1M$$. Then we refine it by taking $$e_2$$ equal to the sum of the group elements as a primitive idempotent in $$\mathbb{F}_2\mathbb{Z}/3^2$$, so that $$e_1e_2=e_2e_1=e_2$$. Then $$M_2=e_2M$$ is a summand of $$M_1$$, and so on. Or have I misunderstood something?