Suppose $R$ is a commutative unital ring, $G$ a finite group, and $H$ a subgroup of $G$ whose order is invertible in $R$. Defining $e_H=|H|^{-1}\sum_{h\in H} h$, why is $RG/H\simeq e_HRG$?

This comes up in some reading showing that invariants and coinvariants of a representation can coincide.

I observed that $e_H$ is idempotent, and $e_Hh=e_H$ for all $h\in H$. I thought maybe the projection map $RG\to e_HRG:x\mapsto e_Hx$, would descend to an isomorphism somehow. It sends all of $H$ to $e_H$, but I'm not sure if this argument goes anywhere.


Define $\phi:R[G/H]\to e_HRG$ by $\phi(g+H)=e_Hg$ (extended by $R$-linearity). First, it is well defined because, as you noted, $e_Hh=e_H$ for every $h\in H$. Surjectivity of $\phi$ comes immediatly. I let you prove injectivity.

  • $\begingroup$ Thanks again Nitrogen. To show injectivity, is this the idea? If $e_Hg=e_Hg'$, then $\sum_{h\in H}gh=\sum_{h\in H} hg'$, or $\sum_{h\in H}hgg'^{-1}=\sum_{h\in H} h$. Since $G$ is a basis for $RG$, $hgg'^{-1}\in H$, so that we get $gg'^{-1}\in H$, finally giving $gH=g'H$. $\endgroup$ – Joie Hwang Oct 28 '15 at 17:31
  • 1
    $\begingroup$ Yep, that's correct. $\endgroup$ – Nitrogen Oct 28 '15 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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