the following result.

\[lem3\] With the notations as above, if $\operatorname{char}\kk\nmid n$ and $\gcd(j,n)=1$, then $y_j^n\in \mfr(R,G)$.

Let $A$ be the subalgebra of $R$ generated by $x_1^2,\dots,x^2_n$. Then $A$ is a subalgebra of the center of $R$. Note that $R$ is finitely generated as an $A$-module. For $j=1,\dots, n$, let $Y_j=\sum_{i=1}^n\xi^{ij}x_i^2$. Then $Y_1,\dots,Y_n$ are linearly independent generators of $R$ (cf. [@BHZ]). Let $A'=A/(A\bigcap \mfr(R,G))$. It is clear that $A'$ is a subalgebra of $\mathfrak{R}(R,G)$ and $\mathfrak{R}(R,G)$ is finitely generated as an $R'$-module since $R$ is finitely generated as an $A$-module. So, $\operatorname{GKdim}(\mathfrak{R}(R,G))=\operatorname{GKdim}(R')$. For $j=1,\dots,n$, we have $y_j^2=\sum_{i=1}^n\xi^{2ij}x_i^2$. For $n+1\leq t\leq 2n$, we set $Y_t=Y_{t-n}$. Then we see $y^2_j=Y_{2j}$ for $j=1,\dots,n$. If $\gcd(j,n)=1$, then $y_j^n\in \mfr(R,G)$ by Lemma \[lem3\], which implies $y_j^{2n}\in \mfr(R,G)$, and hence $Y_{2j}^n\in \mfr(R,G)$. So, we obtain that $Y_{2j}^n=0$ in $R'$ if $\gcd(j,n)=1$. Now let $T$ be the subalgebra of $A'$ generated by elements of the set $\{Y_1,\dots,Y_n\}\setminus\{Y_{2j}|\gcd(j,n)=1\}$. Then $A'$ is finitely generated as a $T$-module. Hence $\operatorname{GKdim}(A')=\operatorname{GKdim}(T)$. Let $t$ be the cardinality of the set $\{Y_{2j}|\gcd(j,n)=1\}$. The commutative subalgebra $T$ is generated by $n-t$ elements. Hence $\operatorname{GKdim}(T)\leq n-t$. Hence $$\Pty(R,G)=n-\operatorname{GKdim}(\mathfrak{R}(R,G))=n-\operatorname{GKdim}(T)\ge t.$$ Note that $$t=\begin{cases} \phi(n), & \hbox{if }4\nmid n;\\
\frac{\phi(n)}{2}, &\hbox{if } 4|n,\end{cases}$$ where $\phi(n):=n \prod_{{\text{all primes 