$B$-modules which are annihilated by $e$. The next proposition shows that the $G$-radical of $R$ is indeed the $\mathcal{M}_R$-radical in the classical radical theory (cf. [@Sz]).

\[prop-mrad\] With the notions as above, the following holds: $$\mfr(R,G)=\bigcap_{M\in\mathcal{M}_R}Ann_R(M)$$ where $Ann_R(M)$ is the annihilator of $M$ in $R$.

We always view $R$ as a subalgebra of $B$ through the map $\iota$. We see that two sequences $(a_1,\dots,a_n)$ and $(b_1,\dots,b_n)$ are pertinent under the $G$-action if and only if $\sum_{i=1}^na_ieb_i=\sum_{i=1}^na_ib_i\otimes 1$. Note that $BeB=ReR$ (cf. [@CFM Proposition 2.13]). We have $R\cap BeB=\mfr(R,G)$. For a module $M\in \mathcal{M}_R$, we have $Me=0$, and hence $BeB\subseteq Ann_R(M)$. Therefore $\mfr(R,G)\subseteq Ann_R(M)$. Since $\overline{B}\in \mathcal{M}_R$, we have $Ann_R(\overline{B})=R\cap BeB=\mfr(R,G)$. Therefore the proposition follows.

The following lemma is easy to check.

\[lem-pertinred\] Let $G$ be a finite group, and $R$ a left $G$-module algebra.

-   If $(a_1,\dots,a_n)\overset{G}\sim(c_1b_1,\dots,c_nb_n)$ where $c_1,\dots,c_n\in R^G$. Then $$(a_1c_1,\dots,a_nc_n)\overset{G}\sim(b_1,\dots,b_n).$$

-   Assume $(a_1,\dots,a_n)\overset{G}\sim(b_1,\dots,b_n)$. If $b_i=b_j$ for some $i,j$, then $$(a_1,\dots,a_i+a_j,\dots,\hat{a}_j,\dots,a_n)\overset{G}\sim(b_1,\dots,b_i,\dots,\hat{b}_j,\dots,b_n),$$ where $\hat{a}_j$ (and $\hat{b}_j$) means deleting the element $a_j$ ($b_j$,resp.).

-   Assume $(a_1,\dots,a_n)\overset{G}\sim(b_1,\dots,b_n)$. If $a_i=a_j$ for some $i,j$, then $$(a_1,\dots,a_i,\dots,\hat{a}_j,\dots,a_n)\overset{G}\sim(b_1,\dots,b_i+b_j,\dots,\hat{b}_j,\dots,b_n),$$ where $\hat{a}_j$ (and $\hat{b}_j$) means deleting the element $a_j$ ($b_j$,resp.).

We end this section with some sufficient conditions for a $G$-module algebra to be $G$-semisimple.

\[prop-ss\] Let $G$ 