F_{1}, F_{2}\rangle \cong {\mathbb Z}_{2}^{2}$ and that each of the invoilutions $F_{1}$, $F_{2}$ and $F_{1}\circ F_{2}$ acts with exactly $4$ fixed points on $R$.

The quotients $R/\langle F_{1} \rangle$, $R/\langle F_{2} \rangle$ and $R/\langle F_{1} \circ F_{2}\rangle$ have genus one. We may apply Proposition \[coroKR\] to $R$ using the three cyclic groups of order two in order to see that $JR$ is isogenous to the product of three elliptic curves.

Proof of Theorem \[coro1\]
==========================

Given the triple $(\lambda_{1},\lambda_{2},\lambda_{3}) \in \Delta_{3}$ and $\mu$ a root of $$\lambda_{2}\lambda_{3} \mu^{2}-(\lambda_{1}\lambda_{2}+\lambda_{2}\lambda_{3}+\lambda_{1}\lambda_{3}-\lambda_{1}-\lambda_{3}+1) \mu +\lambda_{1}\lambda_{2}=0,$$ then we set $\lambda=\lambda_{1}$, $\mu_{1,1}=\mu$ and $\mu_{1,2}=\lambda_{3}\mu$. It can be seen that $(\lambda,\mu_{1,1},\mu_{1,2}) \in \Delta_{3}$ and that $$F_{1}: y^{2}=x(x-1)(x-\lambda_{1}) := E_{\lambda_{1}},$$ $$F_{2}: y^{2}=(x-1)(x-\lambda_{1})(x-\mu)(x-\lambda_{3}\mu):\cong E_{\lambda_{2}},$$ $$F_{3}: y^{2}=x(x-\mu)(x-\lambda_{3}\mu) :\cong E_{\lambda_{3}}.$$

Theorem \[construccion\], applied to the triple $(\lambda,\mu_{1,1},\mu_{1,2})$, asserts that $JS$ is isogenous to the product $F_{1} \times F_{2} \times F_{3}$, so isogenous to the product $E_{1} \times E_{1} \times E_{3}$.

In this case the Riemann surface $S$ is described by the curve $$\left\{ \begin{array}{lcl}
w_{1}^{2}&=&\mu(\lambda_{3} \mu -1)(\lambda_{3} \mu -\lambda_{1})(\lambda_{3} - 1)
z \left( z-\dfrac{1}{\lambda_{1}-\lambda_{3}\mu} \right) \left(z-\dfrac{1}{\mu (1-\lambda_{3})} \right)\\
w_{2}^{2}&=&-\lambda_{3}\mu^{2}(\lambda_{3}-1)z\left(z+\dfrac{1}{\lambda_{3} \mu}\right)  \left(z-\dfrac{1}{1-\lambda_{3}\mu} \right)\\
w_{3}^{2}&=&-\lambda_{3}\mu^{2}(\lambda_{3}\mu-1)(\lambda_{3}-1)z^{2}\left(z+\dfrac{1}{\lambda_{3} \mu}\right) \left(z-\dfrac{1}{\mu(1-\lambda_{3})}\right) \\
\end{array}
\right\},$$ the group $H=\langle a_{1},a_{2}\rangle \cong {\mathbb Z}_{2}^{2}$ is generated by $$a_{1}(z,w_{1},w_{2},w_{3})=(z,w_{1},-w_{2},-w_{3}),$$ $$a_{2}(z,w_{1},w_{2},w_{3})=(z,-w_{1},w_{2},-w_{3}),$$ the three 