# If $H\leq G$ and $|H|$ is invertible in a ring $R$, why does $RG/H\simeq e_HRG$?

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.

## 1 Answer

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.

• 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 $gH=g'H$. – Joie Hwang Oct 28 '15 at 17:31
• Yep, that's correct. – Nitrogen Oct 28 '15 at 17:43