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Let $G$ be a finite group. Let $\mathcal{O}$ be a suitably large finite extension of the $p$-adic integers, with residue field $\mathbf{F}_q$.

The Grothendieck group of the category of finitely-generated $\mathbf{F}_q[G]$-modules is naturally identified with a group of $\mathbf{C}$-valued functions on the set of conjugacy classes of elements of $G$ of order prime to $p$, via Brauer characters.

What is the Grothendieck group of finitely generated $\mathcal{O}[G]$-modules?

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  • $\begingroup$ I've added a tag, since you assume $G$ is finite. Note that the irreducible $p$-adic representations of an arbitrary $G$ are poorly understood, so it's not clear how far the full Grothendieck group can be characterized. Also, the standard texts of Serre and Curtis-Reiner emphasize (following Brauer) the "intermediate" role of the Grothendieck group of f.g. projective modules instead: projectives for $\mathbb{F}_q [G]$ lift nicely to projective $\mathcal{O}[G]$-modules. What is your motivation? $\endgroup$
    – Jim Humphreys
    Commented May 31, 2014 at 14:58
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    $\begingroup$ Note that surprinsingly many modules are zero in the Grothendieck group of $\mathcal{O}[G]$ because of exakt sequences like $0 \to M \xrightarrow{p} M \to M/pM \to 0$ for any $M$ which is free as an $\mathcal{O}$-module. In particular: The projective $\mathbb{F}_q[G]$-modules (considered via the canonical map as $\mathcal{O}[G]$-modules) are zero in the Grothendieck group because they lift to projective and hence torsionfree $\mathcal{O}[G]$-modules. $\endgroup$
    – Johannes Hahn
    Commented May 31, 2014 at 15:34

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If I remember correctly, it is isomorphic to the group of ordinary (virtual) characters over $K=\mbox{Frac}(\mathcal{O})$, the field of fractions of $\mathcal{O}$. That is, the Grothendieck group of $KG$. An isomorphism is given by simply tensoring with $K$, so that $\mathcal{O}G$-modules $M,N$ are identified in $K_0(\mathcal{O}G)$ if and only if $K\otimes_{\mathcal{O}}M$ and $K\otimes_{\mathcal{O}}N$ afford the same character. In particular, all torsion modules are zero in the Grothendieck group.

This was proved by Swan in 'The Grothendieck ring of a finite group', Topology 2, 85-110, 1963. He proves a more general result over an arbitrary integral domain there, and Theorem 3 is the one you're after.

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