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‎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?

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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?

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