The theory of representations of finite groups is the same over $\mathbb{C}$ or over $\mathbb{Q}_p$. If you don't know the theory over $\mathbb{C}$, then try to look at : Serre, 'Representations of finite group'. – user10676 Jul 3 '12 at 10:57
I know the theory over $\mathbb{C}$ an the article of I.Reiner about the irreducible representations over $\mathbb{Z}$ of the cyclic group of prime order. I never worked with $p$-adic integers before. Why sould it be the same as over $\mathbb{C}$? – Alex G. Jul 3 '12 at 10:59
@user10676: Dear user, Since $\mathbb Q_p$ is not algebraically closed, there are subtleties over $\mathbb Q_p$ that do not occur over $\mathbb C$, because of the distinction between irreducible and absolutely irreducible representations, and the possible occurence of Schur indices. Furthermore, I think that the OP is asking about representations over $\mathbb Z_p$, not $\mathbb Q_p$. Regards, – Matt E Jul 3 '12 at 11:04
Dear Alex, Yes, it certainly helps first to work over $\mathbb Q_p$. If you have a finite dim'l rep'n of the finite group $G$ over $\mathbb Q_p$, then you can always choose a basis so that it is defined over $\mathbb Z_p$. This $\mathbb Z_p$-model will typically not be unique, though --- this will depend on whether or not the reduction mod $p$ of your representation remains irreducible. Indeed, the key new feature of working over a ring like $\mathbb Z_p$, rather than a field, is that one really has two fields in play: the fraction field $\mathbb Q_p$, and the residue field ... – Matt E Jul 3 '12 at 11:46
... $\mathbb F_p$, and a big part of the story involves studying the reduction of your representation from char. $0$ to char. $p$. Regards, – Matt E Jul 3 '12 at 11:48