if $\Pty(R,G)=1$.

By [@BHZ2 Lemma 7.2], we see that $\operatorname{gldim}(R^G)<\infty$ if and only if $\Pty(R,G)\leq1$. Proposition \[prop2-2\] applies to obtain the desired result.

If $R$ is not commutative, we have the following result.

\[prop2-5\] Let $R$ be a neotherian AS-regular algebra which is a domain and Cohen-Macaulay, and let $G\subseteq\operatorname{Aut}(R)$ be a finite subgroup. Assume that $R$ and $G$ are in one of the cases in Proposition \[prop2-2\]. If $\operatorname{gldim}(R^G)<\infty$, then $\Pty(R,G)=1$.

By [@KKZ Lemmas 1.10 and 1.11], if $R^G$ has finite global dimension, then $R$ is free as a graded right $R^G$-module. Since both $R$ and $R^G$ are connected graded, $R$ can be written as $R\cong R^G\oplus M$ where $M\cong\oplus_{s=1}^n R^G(i_s)$ for some $n$ and $i_s\leq -1$, where $R^G(i_s)$ is a graded $R^G$-module by a shift of degree $i_s$. Hence as a graded algebra $\operatorname{End}_{R^G}(R)$ contains nonzero component of negative degree. So, $\operatorname{End}_{R^G}(R)$ can not be isomorphic to $R*G$ as graded algebras. By [@BHZ Theorem 0.3], $\Pty(R,G)\leq 1$. It follows from Proposition \[prop2-2\] that $\Pty(R,G)=1$.

Group actions with $\Pty(R,G)>1$ {#set4}
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Let $G$ be a finite group, and let $R$ be a noetherian left $G$-module algebra. It is possible that the pertinency algebra $\mathfrak{R}(R,G)$ is trivial, that is, $\mfr(R,G)=R$. 