can be written in the desired algebraic form.

If $\Phi_{1}:D \to {\mathbb C}^{2^{s-1}}$ is the map whose coordinates are $z$ and $w_{\alpha}$, where $\alpha \in V_{s}$, then $\Phi_{1}(D)=C$.

If $\alpha=(\alpha_{1},\ldots,\alpha_{s})$ and $j=1,\ldots,s-1$, then the induced automorphisms $a_{j}$ acts by multiplication by $-1$ at coordinates $w_{\alpha}$ if $\alpha_{j}=1$ and acts by the identity on the rest of coordinates.

The map $$\pi:C \to \widehat{\mathbb C}: (z,\{w_{\alpha \in V_{s}}\}) \mapsto \frac{1+\mu_{s-2,2}z}{z}$$ is a regular branched cover with $H$ as its deck group and its satisfies that $P=\pi \circ \Phi_{1}$. The branch locus of $\pi$ is the set $$\{\infty,0,1,\lambda, \mu_{1,1}, \mu_{1,2},\ldots,\mu_{s-2,1}, \mu_{s-2,2}\}.$$

\[lema01\] The only non-trivial elements of $H$ acting with fixed points are $a_{1},\ldots,a_{s-1}$ and $a_{s}=a_{1}a_{2} \cdots a_{s-1}$. Moreover, $$\pi({\rm Fix}(a_{1}))=\{\infty,0\}, \quad \pi({\rm Fix}(a_{2}))=\{1,\lambda\},$$ $$\pi({\rm Fix}(a_{k}))=\{\mu_{k-2,1}, \mu_{k-2,2}\},  \quad k=3,\ldots, s-1,$$ $$\pi({\rm Fix}(a_{s}))=\{\mu_{s-2,1}, \mu_{s-2,2}\}.$$ It can be seen that, for $j=1,\ldots,s$, ${\rm Fix}(a_{j})$ has cardinality $2^{s-1}$.

This follows directly from all the above.

If $2 \leq k \leq s$ is even and $1 \leq i_{1}<i_{2}< \cdots < i_{k} \leq s$, then we consider the subgroup $$H_{i_{1},i_{2},\ldots,i_{k}}=\langle a_{i_{1}}a_{i_{2}}, a_{i_{1}}a_{i_{3}},\ldots,a_{i_{1}}a_{i_{k}},a_{j}; j\in\{1,\ldots,s\}-\{i_{1},\ldots,i_{k}\}\rangle \cong{\mathbb Z}_{2}^{s-2}.$$

In the case $k=2$ we have $s(s-1)/2$ such subgroups. Between them are the following ones $$H_{1,2}=\langle a_{1}a_{2},a_{3},\ldots,a_{s}\rangle$$ $$H_{2,3}=\langle a_{2}a_{3},a_{4},\ldots,a_{1}\rangle$$ $$H_{3,4}=\langle a_{3}a_{4},a_{5},\ldots,a_{2}\rangle$$ $$\vdots$$ $$H_{s-1,s}=\langle 