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Put $S=R^G$, the ring of invariants. It is an integral domain because it is a subring of the domain $R$. Embed $\mathrm{Frac}S$ in $\mathrm{Frac}R$ in the evident way.

Suppose $q\in \mathrm{Frac}S$ is an integral element over $S$. Its monic equation has coefficients in $S\subseteq R$, so $q$ is integral over $R$. Since $R$ is a normal domain, $q\in R$.

Write $q=a/b$ with $a,b\in S$ and $b\ne 0$. Each ring automorphism in $G$ extends to the fraction field by acting on numerator and denominator, and fixes this fraction. Hence $q\in R^G=S$. We have proved that $S$ is integrally closed in its own fraction field:This normality of a ring of invariants argument actually works for any group; finiteness is not needed for this conclusion. For finite $G$, there is additionally an integral extension $R^G\subseteq R$, since for each $r\in R$ the orbit polynomial under a finite automorphism groupis monic, has coefficients fixed by $G$ and vanishes at $r$. Neither argument divides by $|G|$, so it is valid when the characteristic divides the group order. Here normality means integral closedness of a domain.

A subset closed under addition, additive inverses and multiplication, with the induced operations. Under the unital convention used here it also contains the identity of $R$. A subring of an integral domain is again an integral domain. The ring of invariants is one example: fixed elements remain fixed under all ring operations.