of type $C_4\times C_2$.

Let $ A_{s,t} $ be defined as in . Then we have $$L[A_{s,t}]^G\cong F^4\times F(\alpha,\iota)^d \mbox{ as $ F $-algebras,}$$ where $\iota\in F^{\text{alg}}$ is such that $\iota^2=-1$ and $d=2/[F(\alpha, \iota):F(\alpha)]$.

Let $$\begin{aligned}
b_{0}&=\frac{1}{8}\big(1+\lambda(s)+\lambda(s^2)+\lambda(s^{-1})+\rho(t)^{-1}+\lambda(s)^{-1}\rho(t)+\rho(t)+\lambda(s)\rho(t)\big),\\
b_{1}&=\frac{1}{8}\big(1-\lambda(s)+\lambda(s^2)-\lambda(s^{-1})-\rho(t)^{-1}+\lambda(s)^{-1}\rho(t)-\rho(t)+\lambda(s)\rho(t)\big),\\
b_{2}&=\frac{1}{8}\big(1+\lambda(s)+\lambda(s^2)+\lambda(s^{-1})-\rho(t)^{-1}-\lambda(s)^{-1}\rho(t)-\rho(t)-\lambda(s)\rho(t)\big),\\
b_{3}&=\frac{1}{8}\big(1-\lambda(s)+\lambda(s^2)-\lambda(s^{-1})+\rho(t)^{-1}-\lambda(s)^{-1}\rho(t)+\rho(t)-\lambda(s)\rho(t)\big), \\
b_{4}&=\frac{1}{4}\big(1-\lambda(s^2)+\lambda(s)^{-1}\rho(t)-\lambda(s)\rho(t)\big),\\
b_{5}&=\frac{1}{4}\big(1-\lambda(s^2)-\lambda(s)^{-1}\rho(t)+\lambda(s)\rho(t)\big),\\
b_{6}&=\frac{1}{4}\big(\lambda(s)-\lambda(s^{-1})-\rho(t)^{-1}+\rho(t)\big),\\
b_{7}&=\frac{1}{4}\big(-\lambda(s)+\lambda(s^{-1})-\rho(t)^{-1}+\rho(t)\big).\end{aligned}$$ It is easily verified that these $ 8 $ elements of $L[A_{s,t}]$ are linearly independent over $L$ and so form an $ L $-basis of $L[A_{s,t}]$. Recall from that the non-trivial $G$-orbits of $A_{s,t}$, are $\{\lambda(s),\lambda(s^{-1})\}$, $\{\lambda(s)\rho(t),\lambda(s^{-1})\rho(t)\}$, both with stabiliser $\langle s \rangle$. From this we see that $ b_{0}, b_{1}, b_{2} $ and $ b_{3} $ are fixed by $ G $, that $ \,^{t}b_{4}=b_{5} $, and that $ \,^{t}b_{6}=b_{7} $. Therefore the following linear combinations of the above elements are all fixed by $G$, and in fact form a basis of $L[A_{s,t}]^G$ over $F$. $$\begin{aligned}
a_{0}&=b_{0}, \\
a_{1}&=b_{1}\\
a_{2}&=b_{2},\\
a_{3}&=b_{3}, \\
a_{4,0}&=b_{4}+b_{5}=\frac{1}{2}\big(1-\lambda(s^2)\big)=\mathfrak{e},\\
a_{4,1}&=\alpha(b_{4}-b_{5})=-\alpha\mathfrak{e}\lambda(s)\rho(t),\\
a_{4,2}&=b_{6}+b_{7}=\mathfrak{e}\rho(t),\\
a_{4,3}&=\alpha(b_{6}-b_{7})=\alpha\mathfrak{e}\lambda(s).\end{aligned}$$

We have $ {a}_i{a}_j = \delta_{i,j} {a}_i $ for $ i,j=0,1,2,3 $ and $a_{4,k}a_i={a}_i{a}_{4,k}=0$ for all $i=0,1,2,3$ and $k=0,1,2,3$. Finally, we consider the multiplication table of the ${a}_{4,k}$.

               $a_{4,0}$       $a_{4,1}$        $a_{4,2}$         $a_{4,3}$
  ----------- ----------- -------------------- ------------ ---------------------
   $a_{4,0}$   $a_{4,0}$       $a_{4,1}$        $a_{4,2}$         $a_{4,3}$
   $a_{4,1}$   $a_{4,1}$   $\alpha^2 a_{4,0}$   $a_{4,3}$    $\alpha^2 