all $j=1,\dots,k-1$. Hence, for each $1\leq j\leq k-1$, there are pertinent sequences $(x_{j1},\dots,x_{jm_j})$ and $(y_{j1},\dots,y_{jm_j})$ in $R$ such that $\sum_{i=1}^na_i(g_j\cdot b_i)=\sum_{i=1}^{m_j}x_{ji}y_{ji}$. We claim that the sequences $(a_1,\dots,a_n)\vee(-x_{11},\dots, -x_{1m_1})\vee\cdots\vee(-x_{k-1,1},\dots, -x_{k-1,m_{k-1}})$ and $(b_1,\dots,b_n)\vee(g_1^{-1}\cdot y_{11},\dots,g_1^{-1}\cdot y_{1m_1})\vee\cdots\vee(g_{k-1}^{-1}\cdot y_{k-1,1},\dots,g_{k-1}^{-1}\cdot y_{k-1,m_{k-1}})$ are pertinent. Indeed, for $1\leq j\leq k-1$, we have $$\begin{array}{cl}
  &\displaystyle\sum_{i=1}^na_i(g_j\cdot b_i)-\sum_{t=1}^{k-1}\sum_{s=1}^{m_t}x_{ts}(g_jg_t^{-1}\cdot y_{ts})\\
  =&\displaystyle\sum_{i=1}^na_i(g_j\cdot b_i)-\sum_{s=1}^{m_j}x_{js}y_{js}-\sum_{t\neq j}\sum_{s=1}^{m_t}x_{ts}(g_jg_t^{-1}\cdot y_{ts})\\
  =&0.
\end{array}$$ Hence we have $\displaystyle\sum_{i=1}^na_ib_i-\sum_{t=1}^{k-1}\sum_{s=1}^{m_t}x_{ts}(g_t^{-1}\cdot y_{ts})\in \mfr(R,G)$. Since $(x_{t1},\dots,x_{tm_t})$ and $(y_{t1},\dots,y_{tm_t})$ are pertinent for all $t=1,\dots,k-1$, we have $\displaystyle\sum_{t=1}^{k-1}\sum_{s=1}^{m_t}x_{ts}(g_t^{-1}\cdot y_{ts})=0$. Hence $\sum_{i=1}^na_ib_i\in \mfr(R,G)$, that is, $\sum_{i=1}^n\overline{a}_i\overline{b}_i=0$ in $\mathfrak{R}(R,G)$. It follows that the $G$-module algebra $\mathfrak{R}(R,G)$ is $G$-semisimple.

By the third statement of the proposition above, we see that the pertinency algebra $\mathfrak{R}(R,G)$ is always $G$-semisimple for any finite group action. Indeed, the $G$-radical $\mfr(R,G)$ is the smallest ideal of $R$ closed under $G$-action such that the the quotient algebra is $G$-semisimple.

\[lem-min\] Let $G$ be a finite group acting on $A$ by automorphisms. If $I$ is a $G$-ideal of $R$ such that $R/I$ is $G$-semisimple, then $\mfr(R,G)\subseteq I$.

For an element $a\in R$, we write $\overline{a}$ for the image of $a$ in $R/I$. Let $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ be pertinent sequences in $R$. Then $\sum_{i=1}^na_i(g\cdot b_i)=0$ for all $1\neq g\in G$. Hence $\sum_{i=1}^n\overline{a_i}(\overline{g\cdot b_i})=\sum_{i=1}^n\overline{a_i}(g\cdot \overline{b_i})=0$ in $R/I$. Since 