We say that a complex surface $S$ is isogenous to an (unmixed) product if there exists a finite group $G$, acting faithfully on two smooth projective curves $C_1$ and $C_2$ and freely on their product (with the diagonal action), so that $S$ is isomorphic to $(C_1 \times C_2)/G$.

If $S =(C_1 \times C_2)/G$ is any surface isogenous to a product, the two natural projections of $C_1 \times C_2$ induces two fibrations $f_1 \colon S \to C_1/G$ and $f_2 \colon S \to C_2/G$, whose smooth fibres $F_1$ and $F_2$ are isomorphic to $C_2$ and $C_1$, respectively. Moreover, since the action of $G$ on the product is free, all the singular fibres of $f_1$ and $f_2$ are multiple of smooth curves.

Let us consider now the group $\textrm{Tors(Pic}^0 S)$. Inside this group naturally lives the subgroup $H$ generated by $f_1^* \textrm{Tors(Pic}^0 C_1/G)$ and $f_2^* \textrm{Tors(Pic}^0 C_2/G)$.

Question. Are there conditions on $C_1$, $C_2$, $G$ and the actions ensuring that $H=\textrm{Tors(Pic}^0 S)?$ In other words, when the torsion part of the Picard group of $S$ is determined by the torsion parts of the bases $C_1/G$ and $C_2/G$ of the two natural fibrations?