Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

2 Answers

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 $p$.

share|improve this answer
    
Thank you. I didn't understand how to apply (*): Let $k$ be a field of positive characteristic and G a finite group. If a $kG$-module $M$ is $kP$-projective for every Sylow subgroup $P$ of $G$, I understand it follows that $M$ is cohomologically trivial as a $G$-module. How can we deduce $kG$-projectivity now? –  Cihan Jun 9 '12 at 8:53
    
Ah sorry, no it doesn't quite work, I only looked at one direction of the proof... Hmmm, I'm not sure we're gonna get any better here, but at least the relation of Rim's result is seen to be related to cohomological-triviality. –  Chris Gerig Jun 9 '12 at 19:47
add comment

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 $R$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.