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Let $G$ be a finite group and let $R$ be a commutative ring.

I'd like to ask, if there is a theorem of the following kind:

The augmentation ideal $I_G$ is projective as RG-module, if and only if ... ?

This should happen only in rare cases, but I was wondering, if there exists an if-and-only-if criterion.

Thank you very much for the help.

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    $\begingroup$ I'm not sure about general rings, but this happens never for $R=\mathbb{Z}$, or for $R=\mathbb{Z}_p$ and $p$ a prime dividing the order of $G$. This is due to the fact that characters of projective $\mathbb{Z}_p[G]$-modules vanish on elements of order divisible by $p$. $\endgroup$
    – Achim Krause
    Commented Jun 16, 2021 at 20:14
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    $\begingroup$ The general result for arbitrary groups is mentioned in mathoverflow.net/questions/297043/… $\endgroup$
    – Benjamin Steinberg
    Commented Jun 16, 2021 at 23:17
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    $\begingroup$ Namely G has projective augmentation ideal iff it is the fundamental group of a graph of finite groups with order invertible in R. Equivalenty G acts on a tree and all the vertex and edge stabilizers are finite groups with order invertible in R $\endgroup$
    – Benjamin Steinberg
    Commented Jun 17, 2021 at 0:50
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    $\begingroup$ I mentioned the characterization via cohomological dimension in this post but it is nice to see the more concrete consequence in the solutions below. $\endgroup$
    – rschwieb
    Commented Jun 22, 2021 at 14:53

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Okay, this happens precisely in the obvious case, namely if all primes dividing the order $|G|$ are invertible in $R$.

To see this, note that $\operatorname{Ext}^*_{R[G]}(R,R)$ is group cohomology of $G$ with coefficients in $R$. If $I_G$ were projective, $I_G\to R[G]$ would be a projective resolution of $R$ and thus the cohomology would be $1$-dimensional. However, the cohomology of $G$ with coefficients in $\mathbb{Z}$ has $p$-torsion in arbitrarily high degrees for all primes dividing $|G|$ (this is a well-known fact about cohomology of finite groups). So by the universal coefficient theorem, the cohomology with $R$-coefficients is unbounded, except if $p$ is invertible in $R$ for each $p$ dividing $|G|$.

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