About Rim's theorem for projective modules over group rings

Question

Let $G$ be a finite group. A theorem of Rim (Proposition 4.9 here) states that a $\mathbb{Z}G$-module $M$ is projective if and only if $M$ is $\mathbb{Z}P$-projective for all Sylow subgroups $P$ of $G$.

Which (or what kind of) rings can we substitute for $\mathbb{Z}$ above such that the statement remains valid?

Answer 1

This question lies in the notion of cohomologically trivial modules, i.e. those $G$-modules $M$ which have $\hat{H}^*(H,M)=0$ for all subgroups $H\subseteq G$. In particular, any free $kG$-module is cohomologically trivial and hence so is any projective $kG$-module (arbitrary commutative ring $k$).

$(\ast)$ Now if $G$ is a $p$-group and $k$ is a field of characteristic $p$, then a $kG$-module $M$ is projective iff $M$ is cohomologically trivial. This is a consistent-but-vacuous generalization of Rim's result.

For a general finite group $G$, $M$ is cohomologically trivial iff its restriction to a Sylow p-subgroup $P$ is cohomologically trivial for all primes $p$ dividing $|G|$. Note that for a field $k$ of characteristic $p$, cohomology vanishes for groups $H$ with $gcd(|H|,p)=1$. So unless I am being very careless, we can apply $(\ast)$ to get your desired extension-result for any field of characteristic $p$.

Answer 2

It works for any ring in which at most one prime is not invertible (in particular it works for any field). In other words, if there is a ring homomorphism $\mathbb{Z}_{(p)} \rightarrow R$ for some prime $p$ (where $\mathbb{Z}_{(p)}$ is the ring of $p$-local integers), then $\mathbb{Z}$ can be replaced with $R$ in Rim's theorem.

The reason is the following: Let $H \leq G$ and consider the restriction functor $$\text{Res}_H^G: RG\text{-Mod} \rightarrow RH\text{-Mod}$$ and its left adjoint $$\text{Ind}_H^G: RH\text{-Mod} \rightarrow RG\text{-Mod} .$$ Then we have

Proposition: If $|G:H|$ is invertible in $R$, then the counit of the adjunction $\text{Ind}_H^G \dashv \text{Res}_H^G$ has a right inverse.

Proof. Let $E$ denote the set of left cosets of $H$ in $G$. For every $RG$-module M, define $$\alpha_M: M \rightarrow \text{Ind}_H^G\text{Res}_H^G(M) = RG \otimes_{RH} M$$ by $$\alpha_M(m) = \frac{1}{|G:H|}\sum_{gH \in E} g \otimes g^{-1}m$$ From here it is straightforward to check that $\alpha_M$'s are well-defined, $RG$-linear, natural in $M$ and the resulting natural transformation $\alpha$ is a right inverse to the counit $\varepsilon: \text{Ind}_H^G\text{Res}_H^G \rightarrow 1_{RG\text{-Mod}}$.

Corollary: If $|G:H|$ is invertible in $R$, then an $RG$-module $M$ is projective if and only if $\text{Res}_H^G(M)$ is projective.

Proof. Only if direction is clear. If $\text{Res}_H^G(M)$ is projective, so is the induced module $\text{Ind}_H^G\text{Res}_H^G(M)$, which has a direct summand isomorphic to $M$ by the above proposition. Hence $M$ is projective.

Now if every prime other than $p$ is invertible in $R$, let $H \in \text{Syl}_p(G)$ and apply the above corollary; we get the analogue of Rim's theorem for $R$.