Let $X,Y$ be two varieties, and $G$ be a group together with actions on $X$ and $Y$. Moreover, we assume the action on $X$ is free. I would like to know if the following statement is true:
There is always an isomorphism $(X \times Y)/ G \simeq X/G \times Y$, where $G$ acts on $X\times Y$ by diagonal.
I guess this is not true in general, as it seems not easy to define a morphism between them. However, I don't have a counter-example in mind. It would be nice if there are some general facts about this.
Thanks in advance!
In general there is no such an isomorphism. Here an example: Let $X=Y=C$ be a Riemann surface of genus $g(C)=3$, and $G\cong \mathbb Z/2\mathbb Z$ (there do exist Riemann surfaces with such an action, by the Riemann existence theorem).
By Künneth formula (since G acts freely on $C$ and diagonal on $C\times C$) you have $$ \begin{array}{l}H^0((C\times C)/G, \Omega^1)&=& H^0((C\times C), \Omega^1)^G= H^0(C, \Omega^1)^G \oplus H^0(C, \Omega^1)^G \\&=& H^0(C/G, \Omega^1)\oplus H^0(C/G, \Omega^1)\end{array}$$ In particular $q((C\times C)/G)= \dim H^0((C\times C)/G, \Omega^1)=2\cdot g(C/G)=2\cdot 2$ (by the Hurwitz formula).
Arguing in the same way you see that $q(C/G \times C)= \dim H^0(C/G\times C, \Omega^1)= 2+ 3\neq 4$.