as $D({\O}_s)$-modules. For this purpose we define a linear map $$\begin{aligned}
\r: D(s)\bigotimes_{S(s)} d_{1,s}N &\rightarrow & N\\
  d_{b,s}\o_{S(s)}d_{1,s}n&\mapsto& d_{b,s}n.\end{aligned}$$ Noting that $\r(d_{b,s}\o_{S(s)}d_{1,s}n)=d_{b,s}d_{1,s}n$ we immediately see that $\r$ is well defined. The fact that $\r$ is a $D({\O}_s)$-module homomorphism follows from $$\begin{aligned}
& &\ \ \ \ \r(d_{a,sg_i}d_{b,s}\o_{S(s)}d_{1,s}n)\\
& &=\r(\d_{sg_ib,s}\a_s(a,b)d_{ab,s}\o_{S(s)} d_{1,s}n)\\
& &=\d_{sg_ib,s}\a_s(a,b)d_{ab,s}n\\\end{aligned}$$ and $$\begin{aligned}
& &\ \ \ \ d_{a,sg_i}\r (d_{b,s}\o_{S(s)}d_{1,s}n)\\ 
& &=d_{a,sg_i}d_{b,s}n\\
& &=\d_{sg_ib,s}\a_s(a,b)d_{ab,s}n\end{aligned}$$ for $a,b\in G,$ $1\leq i\leq k$ and $n\in N.$

It remains to show that $\r$ is a bijection. As $\sum_{i=1}^kd_{1,sg_i}$ is the identity of $D(\O_s),$ we have $n=\sum_{i=1}^k d_{1,sg_i}n$. For fixed $i$ one can easily see that $$\r(d_{g_i^{-1},s}\o_{S(s)}\a_{sg_i}(g_i^{-1},g_i)^{-1}d_{g_i,sg_i}n)=d_{1,sg_i}n.$$ Thus $\r$ is onto.

Since $d_{1,sg_i}d_{1,sg_j}=\delta_{i,j}d_{1,sg_i}$ we have $$N=\bigoplus_{i=1}^kd_{1,sg_i}N.$$ Let $b=g_i^{-1}a\in g_i^{-1}G_s.$ Then $\r(d_{b,s}\o_{S(s)}d_{1,s}n)= d_{1,sg_i}d_{b,s}n\in d_{1,sg_i}N$ for $n\in N.$ In order to prove that $\r$ is one to one, it is enough to show that if $\r(\sum_{p}d_{b_p,s}\o_{S(s)}d_{1,s}n_p)=0$ for some $b_p\in g_i^{-1}G_s$ and $n_p\in N,$ then $\sum_{p}d_{b_p,s}\o_{S(s)}d_{1,s}n_p=0.$ Using the relation $d_{g_i^{-1}a,s}=\a_s(g_i^{-1},a)^{-1}d_{g_i^{-1},s}d_{a,s}$ for $a\in G_s$ we can rewrite $\sum_{p}d_{b_p,s}\o_{S(s)}d_{1,s}n_p$ as $d_{g_i^{-1},s}\o_{S(s)}d_{1,s}n$ for some $n\in N.$ Thus $\r(d_{g_i^{-1},s}\o_{S(s)}d_{1,s}n)=d_{g_i^{-1},s}n=0.$ Applying $\a_s(g_i,g_i^{-1})^{-1}d_{g_i,sg_i}$ to $d_{g_i^{-1},s}n$ yields $d_{1,s}n=0.$ This shows that $\r$ is one to one. This completes the proof that $\r$ is an isomorphism.

\[t50\] We have

i\) Algebra $D({\O}_s)$ is semisimple for 