Hi, is there a possibility to classify irreducible representations of (finite) groups over $\mathbb{Z_p}$ with $p\neq 2$ a prime? Can one get the number of the irred. representations here? I tried to solve this for "easy" groups like the cyclic or symmetric groups, but did not really get a result. Maybe there is a connection to representation over $\mathbb{Z}/p\mathbb{Z}$?

Best regards

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Usually, "irreducible" means having no subrepresentations. In the integral context, there is no such thing, since for any $\mathbb{Z}_p[G]$-module $M$, $pM$ is a proper submodule. The right question is to try and classify the indecomposable representations, i.e. those that cannot be written as direct sums of subrepresentations.

To find all indecomposable $\mathbb{Z}_p$-representations of a given finite group is a hard problem, in general, and a lot of work has been done on special cases. If $G$ is a cyclic group of order $n$, then it is known that there is a finite number of indecomposable $\mathbb{Z}_p[G]$-modules if and only if $p$ divides $n$ at most to the power 2. If $p^3|n$, then there are infinitely many indecomposable modules over $\mathbb{Z}_p$. See Heller, Representations of Cyclic Groups in Rings of Integers, II in The Annals of Mathematics, Vol. 77, No. 2. Many other special cases have been treated in the literature, e.g. dihedral groups of order $2p$ (M. P. Lee, Integral representations of dihedral groups of order 2$p$, Trans. American Math. Soc. 110 no. 2 (1964), 213-231).

There is indeed a strong connection to modular representations. For example if $G$ is a finite group and $M$ is a $\mathbb{Z}_p[G]$-module, then $M$ is decomposable if and only if $M/p^kM$ is a decomposable $\mathbb{Z}/p^k\mathbb{Z}[G]$-module for some $k$. Also, two modules $M$ and $N$ are isomorphic if and only if $M/p^kM$ is isomorphic to $N/p^kN$ for some (explicitly computable) $k$. For this and more results, see e.g. Curtis and Reiner, Representation theory of finite groups and associative algebra, §76.

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Thank you. In which articles could I find this for the dihedral group? –  Matt E. Sep 6 '12 at 21:24
I have added a reference. –  Alex B. Sep 6 '12 at 21:28

What Alex says is correct, but there is another viewpoint, which is a bit easier in some ways. You can consider finitely generated $\mathbb{Z}_{p}$-free $\mathbb{Z}_{p}G$-modules which have no proper non-zero pure submodules. These are the modules which are irreducible when the ground ring is extended to $\mathbb{Q}_{p}.$ There are finitely many isomorphism types of such modules. It may be possible to deduce the structure of faithful such modules for the dihedral group using Clifford theory.

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