most recent 30 from http://mathoverflow.net 2013-06-19T04:03:49Z http://mathoverflow.net/feeds/question/110910 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110910/mathbbz-pk-mathbbzg-modules-in-repr-theory Mike Beyman 2012-10-28T15:39:00Z 2013-06-10T04:22:00Z

<p>My question is referring to the answer in <a href="http://mathoverflow.net/questions/106543/p-adic-representations-of-groups" rel="nofollow">this link</a></p> <p>especially to this sentence:</p> <p>"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 $k$."</p> <p>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.</p> <p>Thank you for hints how to solve this.</p>

http://mathoverflow.net/questions/110910/mathbbz-pk-mathbbzg-modules-in-repr-theory/110919#110919 Simone Virili 2012-10-28T16:55:09Z 2012-10-28T16:55:09Z

<p>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.</p>