# $\mathbb{Z}/p^k \mathbb{Z}[G]$-modules in repr. theory

My question is referring to the answer in this link

especially to this sentence:

"if $G$ is a finite group and $M$ is a $Z_p[G]$-module, then $M$ is (in)decomposable if and only if $M/p^kM$ is a (in)decomposable $Z/p^kZ[G]$-module for some $k$."

Why is it easier to determine the number of the indecomposable $Z/p^kZ[G]$-modules for some $k$ which have the form $M/p^kM$. Do I really have to search first for the indecomposable $M/p^kM$-modules for each $k=1,\cdots,d$, with $d$ minimal in $1=p^d$. And then look if one can write this modules in the form $M/p^kM$? This seams quite difficult for me.

Thank you for hints how to solve this.

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well... one can start thinking to the case when $G$ is trivial. You can probably feel how $\mathbb Z/p^k\mathbb Z$-modules are easier then $\mathbb Z_p$-modules. In particular, the torsion $\mathbb Z_p$-modules are the abelian $p$-groups, while the $\mathbb Z/p^k\mathbb Z$-modules are the $p^k$-bounded abelian $p$-groups. It seems more than reasonable to start classifying the indecomposable bounded groups in order to classify the torsion ones... when $G$ is not trivial, the situation is similar.

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Could you give me please an example how to do this, when $G$ is not trivial, but e.j. cyclic of order $q$, where $q \neq p$ prime. What whould the number of the indecomp. modules be here? – Mike Beyman Oct 28 '12 at 17:06
well... if you are just looking for a number of indecomposables, then it wuold always be not finite. In fact, even for $G=0$, all the $\mathbb Z_p$-moules of the form $\Z/p^k\Z$ are indecomposables. If you want to take $G$ non-trivial, say $G=\mathbb Z/q\mathbb Z$ as in your comment, you always have the modules of the form $\Z/p^k\Z \otimes_{\mathbb Z_p} \mathbb Z_p[G]$, which are indecomposable. – Simone Virili Oct 30 '12 at 10:31
(sorry for the missprints... by $\Z$ I mean $\mathbb Z$) – Simone Virili Oct 30 '12 at 10:31