D({\O}_s)=\<a\o e(s\cdot g_i)|i=1,...,k, a\in G\>.$$ Then $D(s)=S(s)\oplus N(s)$.

Decompose $\S$ into a disjoint union of orbits ${\S}=\bigcup_{j\in J}{\O}_j$. Let $s_j$ be a representative element of ${\O}_j.$ Then ${\O}_j=\{s_j\cdot h|h\in G\}$ and $ {\A}_{\a}(G,{\S})=\bigoplus_{j\in J}D({\O}_{s_j}).$

\[l3.2\] Let $s\in \S$ and $G=\cup_{i=1}^kG_sg_i.$ Then

1\) $S(s)$ is a subalgebra of ${\A}_{\a}(G,{\S})$ isomorphic to $\C^{\a_{s}}[G_s]$ where $\C^{\a_{s}}[G_s]$ is the twisted group algebra.

2\) $N(s)$ is a 2-sided nilpotent ideal of $D(s)$ and $D(s)\cdot
N(s)=0.$

3\) $D({\O}_s)=\bigoplus_{i=1}^kD(sg_i)$ is a direct sum of left ideals.

4\) Each $D({\O}_s)$ is a 2-sided ideal of ${\A}_{\a}(G,{\S})$ and ${\A}_{\a}(G,{\S})=\oplus_{j\in J}D({\O}_{s_j}).$ Moreover, $D({\O}_s)$ has identity element $\sum_{t\in {\O}_s} 1\otimes e(t).$

1\) Clearly, $S(s)$ is a subalgebra of ${\A}_{\a}(G,{\S}).$ Let $\r:S(s)\rightarrow \C^{\a_{s}}[G_s]$ be a linear isomorphism determined by $\r(a\o e(s))=a$. Then $$\begin{aligned}
& & \ \ \ \ \r(a\o e(s)\cdot b\o e(s))\\
& &=\r(\a_{s}(a,b)ab\o e(sb)e(s))\\
& &=\r(\a_{s}(a,b)ab\o e(s))\\
& &=\a_{s}(a,b)ab\\
& &=\r(a\o e(s))\r(b\o e(s)).\end{aligned}$$ So, $\r$ is an algebra isomorphism.

2\) Since for $b\in G\setminus  G_{s},$ $sb\ne s$. We immediately have $ a\o e(s)\cdot b\o e(s)=\a_{s}(a,b)ab\o e((s)b)e(s)=0$ for $a\o e(s)\in D(s)$ and $b\o e(s)\in N(s)$.

It remains to show that $N(s)$ is right ideal. Take $a\o e(s)\in N(s)$ and $b\o e(s)\in S(s).$ Then $a\o e(s)\cdot b\o e(s)=\a_{s}(a,b)ab\o e((s)b)e(s)=\a_{s}(a,b)ab\o e(s).$ If $ab$ 