Is a projective $R[G]$-module a projective $R[H]$-module if $H$ is a subgroup of $G$?

Question

I have a ring $R$ of characteristic $0$ and a finite group $G$. Let $H$ be a subgroup of $G$.

Question: If $M$ is a projective $R[G]$-module where $R[G]$ is the usual group ring then is $M$ also projective as an $R[H]$-module?

This seems easy but I'm having trouble showing that if $M$ is a summand of a free $R[G]$-module then it's also the summand of a free $R[H]$-module.

I would really appreciate some help.

Answer 1

The mentioned assumptions are not necessary. If $R$ is any ring and $H$ is a subgroup of any group $G$, then $R[G]$ is a free $R[H]$-module (both from left and from right). This is basically because $G$ is a free $H$-set. We have $G = \coprod_i H g_i$ for some $g_i \in G$, and hence $R[G] = \bigoplus_i R[H] g_i$.

Since $R[G]$ is a free $R[H]$-module, any free $R[G]$-module restricts to a free $R[H]$-module. Hence, the same holds for projective modules.