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Is there a possibility to get the simple $R[G]$-modules, if $R$ is the ring $\mathbb{Z}/n\mathbb{Z}$, $G$ a finite group and $\operatorname{ord}(G)$ and $n$ are relatively prime? For which groups would their number be finite?

I could solve this just for special $n$, but not in general. Maybe there is some literature where one can find examples for special groups?

Thanks and best regards.

Edit: Can one solve this maybe for $n$ a power of a prime?

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Every ring with identity has simple modules: you just take $R/M$ where $M$ is a maximal right ideal (or left ideal, if you want a left module.)

If $G$ is finite then this group ring is Artinian, but this paper leads me to believe that not all Artinian rings are of finite representation type.

If $G$ is finite and the order $|G|$ is a unit in $R$, the ring is semisimple by Maschke's theorem, and hence has finite representation type.

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    Maybe this could be easier to solve for special cases. I tried to start which $n$ as a power of a prime $l$ to use Maschke's theorem. But this does not give me a classification or the number of the simple modules, just that there are finitely many. Whould $R/M$ be all simple modules?2012-11-01
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    Since Kolja asked for simple modules, there is no restriction on $n$ (other than it be a nonzero integer) for $\mathbb{Z}/n\mathbb{Z}[G]$ to have finitely many simple modules. The problem with representation type are the non-simple indecomposables.2012-11-01
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    Its simple modules are the simple modules for $\mathbb{Z}/p\mathbb{Z}[G]$ where $p$ divides $n$.2012-11-01
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    Thank you Jack Schmidt. Is this true for all finite groups? Can I find the proof somewhere? (Just the ideas, not the details if it is very complicated. I'm just an undergraduate student)2012-11-01
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    @kolja: yes, every finite group. If M is Z/nZ[G] module and p is prime then M_p = { m in M : pm = 0 } is a submodule. If M is simple, then M_p is M or 0. Since M_n = M it is easy to see M_p cannot always be 0. But M_p is always a Z/pZ[G] module, simply because pm=0.2012-11-02