every $1\neq g\in G$, since $g$ acts on $a_1,\dots,a_{n-1}$ diagonally, we have $a_{\widehat{[i_1i_2\cdots i_s]}}g((-1)^sq_{[i_1i_2\cdots i_s]}a_{[i_1i_2\cdots i_s]})=(-1)^s a_1\cdots g(a_{i_1})\cdots g(a_{i_s})\cdots a_{n-1}$. Then the same arguments as in the proof of Lemma \[lem2-1\] hold.

As a consequence, we have the following result, a part of which is indicated in [@BL Lemma 2.2] with different constructions.

\[prop2-1\] With the same hypothesis as in Lemma \[lem2-1\], we have $$\prod_{i=1}^{n-1}(g_i(a_i)-a_i)\in \mfr(R,G).$$

Note that the sum of the products of the corresponding elements in the pertinent sequences $\bigcup_{s=0}^{n-1}g_{\overline{[i_1i_2\cdots i_s]}}(a_{\overline{[i_1i_2\cdots i_s]}})$ and $\bigcup_{s=0}^{n-1}(a_{[i_1i_2\cdots i_s]})$ is exactly equal to $\prod_{i=1}^{n-1}(g_i(a_i)-a_i)$.

There is another way to construct elements in the radical of a group action by means of the determinant.

Let $G=\{1=g_0,g_1,\dots,g_{n-1}\}$ be a finite group acting on an algebra $R$. Taking any elements $a_1,\dots,a_n\in Z(R)$, we define an element $$\label{eq-det}
  \delta_G(a_1,\dots,a_n)=\begin{vmatrix}
a_1&a_2&\cdots& a_n\\
g_1(a_1)&g_1(a_2)&\cdots&g_1(a_n)\\
\vdots&\vdots&&\vdots\\
g_{n-1}(a_1)&g_{n-1}(a_2)&\cdots&g_{n-1}(a_n)
\end{vmatrix}.$$

\[prop2-6\] With the notions as above, we have $\delta_G(a_1,\dots,a_n)\in \mfr(R,G)$.

Consider the following two sequences $(a_1,\dots,a_n)$ and $(A_1,\dots,A_n)$, where $A_i$ $(i=1,\dots n)$ is the cofactor of the element $a_i$ in the determinant (\[eq-det\]). For every $1\neq g\in G$, we have $gg_j=1$ for some $1\leq j\leq n-1$. Hence $$g(A_i)=\begin{vmatrix}
gg_1(a_1)&\cdots&gg_1(a_{i-1})&gg_1(a_{i+1})&\cdots&gg_1(a_n)\\
\vdots&&\vdots&\vdots&&\vdots\\
a_1&\cdots&a_{i-1}&a_{i+1}&\cdots& a_n\\
\vdots&&\vdots&\vdots&&\vdots\\
gg_{n-1}(a_1)&\cdots&gg_{n-1}(a_{i-1})&gg_{n-1}(a_{i+1})&\cdots&gg_{n-1}(a_n)
\end{vmatrix}.$$ Then $\sum_{i=1}^n(a_ig(A_i))$ is a determinant in 