n\geq 0$ $M_{\lambda}(n)$ is generated by any nonzero vector in $M_{\lambda}(n)$ as an $A_{n}(V^G)$-module.

Note from the definition of $A_n(V)$ that $A_n(V)$ is a $G$-module via $g(v+O_n(V))=gv+O_n(V)$ for $g\in G$ and $v\in V.$ Also $\End M(n)$ is an $G$-module via $gf=\phi(g)f\phi(g)^{-1}$ for $g\in G$ and $f\in
\End M(n).$ Then the algebra homomorphism $v+O_n(V)\mapsto o_M(v)$ from $A_n(V)$ to $\End M(n)$ is a $G$-homomorphism as $\phi(g)o_M(v)\phi(g)^{-1}=o_M(gv).$ From Proposition \[4.19\] and Theorem \[l2.3\] 3) we see that $\End M(n)=\{o_M(v)|v\in V\}.$ Since $G$ is a finite group we immediately have $\psi(A_n(V)^G)=(\End M(n))^G=\{o_M(v)|v\in V^G\}.$

Let $x,y\in M_{\l}(n)$ be linearly independent. Since $M(n)$ is a $\C^{\a_M}[G]$-module, we can write $M(n)$ as a direct sum $W_{\lambda}\otimes x\oplus W_{\lambda}
\otimes y\oplus W $ where $W$ is a $\C^{\a_M}[G]$-submodule of $M(n).$ Define a map $\beta\in\End M(n)$ such that $\beta(u\otimes x+v\otimes y+w)=
v\otimes x+u\otimes y+w$ for $u,v\in W_{\lambda}$ and $w\in W.$ Then $\beta\in(\End M(n))^{G}.$ Thus there exists vectors $v\in V^G$ such that $o_M(v)x=y.$ This implies that $M_{\l}(n)$ is an simple $A_n(V)^G$-module. Since $M_{\l}(n)$ is also an $A_n(V^G)$-module with the same action, we see that $M_{\l}(n)$ is a simple $A_n(V^G)$-module. This can be also explained by noting that the identity on $V^G$ induces an onto algebra homomorphism from $A_n(V^G)$ to $A_n(V)^G.$

