$p\mid n$}}}
(1-\frac{1}{p})$ is the Euler’s totient function (cf. [@BHZ]).

Summarizing the above narratives, we have provided a simpler proof for [@BHZ Theorem 5.7(iii)].

[@BHZ] Assume $\operatorname{char}\kk\nmid n$ $(n\ge2)$. Let $R=\kk_{-1}[x_1,\dots,x_n]$, and $G$ the subgroup of $\operatorname{Aut}(A)$ generated by the automorphism $\sigma$ defined by $\sigma(x_i)=x_{i+1}$ for $1\leq i\leq n-1$ and $\sigma(x_n)=x_1$. Then we have $$\Pty(R,G)\ge \begin{cases} \phi(n), &\hbox{if }4\nmid n;\\
\frac{\phi(n)}{2}, & \hbox{if }4\mid n,\end{cases}$$ where $\phi(n):=n \prod_{{\text{all primes $p\mid n$}}}
(1-\frac{1}{p})$ is the Euler’s totient function.

Now we consider the down-up algebra: $R=\kk\langle x,y\rangle/(r_1,r_2)$, where $r_1=x^2y-\alpha xyx-\beta yx^2$ and $r_2=x y^2-\alpha yxy-\beta y^2 x$. The pertinencies of finite group actions on $R$ were computed in [@BHZ2] when $\beta\neq-1$ or $\beta=-1$ but $\alpha=2$. We have the following result.

Let $R$ be a down-up algebra such that $\beta=-1$ and $\alpha\neq 2$. Let $G=\{1,\sigma\}$ where $\sigma$ is the automorphism of $R$ defined by $\sigma(x)=ay$ and $\sigma(y)=a^{-1}x$ with $a\neq 0$. If $\operatorname{char}\kk=0$, then $\Pty(R,G)=3$.

We see that the sequences $(1,-x)$ and $(ay,1)$ are pertinent under the $G$-action. Hence $ay-x\in \mfr(R,G)$. Therefore $x=ay$ in the pertinency algebra $\mathfrak{R}(R,G)$. By definition, $x^2y=\alpha xyx-yx^2$ in $R$, and hence it holds in $\mathfrak{R}(R,G)$. Combining the relations $x=ay$ and $x^2y=\alpha xyx-yx^2$ in $\mathfrak{R}(R,G)$, we see $y^3=0$ in $\mathfrak{R}(R,G)$ 