Timeline for $G$-module structure of the relation module for a presentation of a finite group $G$

6 events

when toggle format	what		by	license	comment
Feb 23 at 19:21	comment	added	Benjamin Steinberg		Well I'm not so certain because I guess I need to be able to cancel a copy of ZG
Feb 23 at 18:59	comment	added	Benjamin Steinberg		In fact you just need $G$ is invertible in the ring $k$. Then a $kG$-module is projective iff it is projective over $k$, which is the case for the chain complex and the homology.
Feb 23 at 12:42	comment	added	Andy Putman		If the characteristic of $k$ does not divide the order of the group, then the proof in my answer works just fine. All that is needed for it is that representations of $G$ over $k$ are semisimple.
Feb 23 at 10:17	comment	added	user522465		@MaxHorn: As a starting point, $R/R’$ is naturally a module over the integers $\mathbb{Z}$, with a $G$-action derived from conjugation. This makes it a $\mathbb{Z}G$-module, where $\mathbb{Z}G$ is the group ring of $G$ over the integers. To consider $R/R’$ in the context of a field $F$ (especially when discussing characteristics not dividing the order of $G$), we often look at $F \otimes_{\mathbb{Z}} (R/R’)$. This operation turns $R/R’$ into an $F[G]$-module
Feb 23 at 7:42	comment	added	Max Horn		A naive but genuine question from someone not doing much representation theory, but: isn't $R/R'$ only a $\mathbb{Z}G$-module? Or should one read $R/R'$ as shorthand for $F\otimes (R/R')$? Or perhaps there is a meaning to "isomorphic as $G$-module" that I've just not seen before?
Feb 23 at 6:28	history	answered	user522465	CC BY-SA 4.0