can take $\phi(g)=g$ and $\C^{\a_M}[G]=\C[G].$

Let $\Lambda_{G,\a_M}$ be the set of all irreducible characters $\lambda$ of $\C^{\a_M}[G]$. We denote the corresponding simple module by $W_{\l}.$ Again if $M=V$, $\Lambda_{V,\a_V}=\hat G$ is the set of all irreducible characters of $G.$ Note that $M$ is a semisimple $\C^{\a_M}[G]$-module. Let $M^{\lambda}$ be the sum of simple $\C^{\a_M}[G]$-submodules of $M$ isomorphic to $W_{\l}.$ Then $M=\oplus_{\lambda\in \Lambda_{G,\a_M}}M^{\lambda}$. Moreover, $M^{\lambda}=W_{\lambda}\otimes M_{\l}$ where $M_{\l}=\hom_{\C^{\a_M}[G]}(W_{\l},M)$ is the multiplicity of $W_{\l}$ in $M.$ As in \[DLM1\], we can, in fact, realize $M_{\l}$ as a subspace of $M$ in the following way. Let $w\in W_{\l}$ be a fixed nonzero vector. Then we can identify $\hom_{\C^{\a_M}[G]}(W_{\l},M)$ with the subspace $$\{f(w) |f\in \hom_{\C^{\a_M}[G]}(W_{\l},M)\}$$ of $M^{\l}.$

We also set $V^G=\{v\in V|gv=v, g\in G\}.$ Then $V^G$ is a simple vertex operator subalgebra of $V$ (see \[DM2\]) and $\phi(g)Y_M(v,z)\phi(g)^{-1}=Y_M(gv,z)
=Y_M(v,z)$ for all $v\in V^G.$ Thus the actions of $\C^{\a_M}[G]$ and $V^G$ on $M$ are commutative. This shows that both $M^{\l}$ and $M_{\l}$ are ordinary $V^G$-modules.

\[corect\] If $M_{\l}\ne 0$ then $M_{\l}$ is an irreducible $V^G$-module.

By Theorem \[l2.3\] 3) it is enough to prove that $M_{\l}(n)=M_{\l}\cap M(n)$ is a simple $A_n(V^G)$-module for all $n\geq 0.$ It is equivalent to show that for any $n\in \Z, 