& &=\d_{sh,t}\a_t(g,h)\a_u(gh,k)ghk\o e(tk)e(u)\\
& &=\d_{sh,t}\d_{tk,u}\a_t(g,h)\a_u(gh,k)ghk\o e(u)\\
& &=\d_{sh,t}\d_{tk,u}\a_{uk^{-1}}(g,h)\a_u(gh,k)ghk\o e(u)\\
& &=\d_{sh,t}\d_{tk,u}\a_{u}(g,hk)\a_u(h,k)ghk\o e(u)\end{aligned}$$ and $$\begin{aligned}
& &\ \ \ \ g\o e(s)\cdot(h\o e(t)\cdot k\o e(u))\\
& &=g\o e(s)\cdot(\a_u(h,k)hk\o e(tk)e(u))\\
& &=g\o e(s)\cdot(\d_{tk,u}\a_u(h,k)hk\o e(u))\\
& &=\d_{tk,u}\a_u(g,hk)\a_u(h,k)ghk\o e(s(hk))e(u)\\
& &=\d_{tk,u}\d_{s(hk),u}\a_u(g,hk)\a_u(h,k)ghk\o e(u).\end{aligned}$$ Thus $(g\o e(s)\cdot h\o e(t))\cdot k\o e(u)=g\o e(s)\cdot(h\o e(t)\cdot k\o e(u))$ and associativity holds.

A straightforward verification $$\begin{aligned}
& &\ \ \ \ g\o e(s)\cdot\sum_{t\in {\S}}1\o e(t)\\
& &=\sum_{t\in {\S}}\a_t(g,1)g\o e(s)e(t)\\
& &=g\o e(s)\\ \end{aligned}$$ and $$\begin{aligned}
& &\ \ \ \ (\sum_{t\in {\S}}1\o e(t))\cdot g\o e(s)\\
& &=\sum_{t\in {\S}}\a_s(1,g)g\o e(tg)e(s)\\
& &=g\o e(s)\end{aligned}$$ shows that $\sum_{t\in {\S}}1\o e(t)$ is an identity on ${\A}_{\a}(G,{\S})$.

For ii) since $\b$ is cohomologous to $\a$, so there exists a map $\l:G\rightarrow{\U}(\C{\S})$ such that $\b(x,y)=\a(x,y)\l(x)^y\l(xy)^{-1}\l(y)$ for all $x,y\in G$. For each $x\in G$ , we can rewrite $\l(x)$ as $\sum_{s\in {\S}}\l_s(x)e(s)$. Therefore, we have $\b_s(x,y)=\a_s(x,y)\l_{sy^{-1}}(x)\l_s(xy)^{-1}\l_{s}(y)$, for any $s\in {\S}, x,y\in G.$ Let $f:{\A}_{\a}(G,{\S})\rightarrow{\A}_{\b}(G,{\S})$ defined by $f(g\o e(s))=\l_{s}(g)^{-1}(g\o e(s)).$ Since $$\begin{aligned}
& &\ \ f(g\o e(s)\cdot h\o e(t))\\
& &=f(\a_t(g,h)gh\o e(sh)e(t))\\
& &=\d_{sh,t}\l_{t}(gh)^{-1}\a_t(g,h)gh\o e(t)\end{aligned}$$ and $$\begin{aligned}
& &\ \ f(g\o e(s))\cdot f(h\o e(t))\\
& &=\l_{s}(g)^{-1}g\o e(s)\cdot \l_{t}(h)^{-1}h\o e(t)\\
& &=\l_{s}(g)^{-1}\l_{t}(h)^{-1}\b_t(g,h)gh\o e(sh)e(t),\end{aligned}$$ ${\A}_{\a}(G,{\S})$ and ${\A}_{\b}(G,{\S})$ are isomorphic as algebras.

\[ra\] [The algebra $A_{\alpha}(G,\S)$ is essentially the crossed product 