automorphisms $a_{1}$, $a_{2}$ and $a_{3}=a_{1}a_{2}$ acts with exactly four fixed points each one, and the corresponding regular branched cover with $H$ as deck group is $$\pi:S \to \widehat{\mathbb C}: (z,w_{1},w_{2},w_{3}) \mapsto \frac{\lambda_{3} \mu z +1}{z}.$$

The subgroups of $H$, in this case ($k=2$ is the only option) are $$H_{1,2}=\langle a_{1}a_{2}\rangle$$ $$H_{1,3}=\langle a_{1}a_{3}\rangle=\langle a_{2} \rangle$$ $$H_{2,3}=\langle a_{2}a_{3}\rangle=\langle a_{1}\rangle$$

The quotients $S/\langle a_{1} \rangle$, $S/\langle a_{2} \rangle$ and $S/\langle a_{1}a_{2} \rangle$ are of genus one and they correspond, respectively, to the elliptic curves $E_{\lambda_{2}}$, $E_{\lambda_{1}}$ and $E_{\lambda_{3}}$.

Proof of Theorem \[coro2\]
==========================

Let us first consider the case $r \geq 3$ odd. Write $r=2s-3$, where $s \geq 3$, and let us fix $(\lambda_{1},\ldots,\lambda_{r}) \in \Delta_{r}$. Set $\lambda=\lambda_{1}$ and, for $j=1,\ldots,s-2$, we set $\mu_{j,2}=\lambda_{j+1} \mu_{j,1}$ and $\mu_{j,1}$ a root of the polynomial $$\lambda_{j+1}(1-\lambda_{s-2+j})\mu_{j,1}^{2}+(\lambda_{s-2+j}-\lambda_{j+1}-\lambda_{1}+\lambda_{1}\lambda_{j+1}\lambda_{s-2+j})\mu_{j,1}+(1-\lambda_{1}\lambda_{s-2+j})=0.$$

Note that a value $\lambda_{j}$ may be changed to some other value $\lambda_{j}'$ (inside an infinite set of values) so that $E_{\lambda_{j}}$ and $E_{\lambda_{j}'}$ are isogenous. This observation permits to ensure that the constructed tuple $(\lambda,\mu_{1,1},\mu_{1,2},\mu_{2,1},\mu_{2,2},\ldots,\mu_{s-2,1},\mu_{s-2,2}) \in \Delta_{2s-3}$.

Let us consider the Riemann surface $S$ constructed in Theorem \[construccion\]. The jacobian variety of $S$ is isogenous to a product of certain explicit jacobian varieties. It can be seen, from the 