$G$ acts on $X$ and $Y$, prove that $g((x, y)) = (g(x), g(y))$ defines an action of $G$ on $X\times Y$.
Then, since $G$ acts on $X$ and $Y$. Then we have both $X$ and $Y$ forms a bijection to itself, from which we could prove that $X\times Y$ also forms a bijection to itself. And we thus get a function $G\to S_{X\times Y}$.
Then I am going to prove that $G\to S_{X\times Y}$ is an homomorphism. However, I didn't really know how to go from $gh(x)=g(h(x))$ and $gh(y)=g(h(y))$ to $gh((x,y))= g(h(x,y))$.
So I am thinking about another definition of ACTION that
$G\times X \to X$, $(g,x)\mapsto gx\in X$
such that $gh(x)=g(hx)$
$ex=x $
which will be easy to get.
My confusions are : 1) How to prove homomorphism by the first definition (I think it's due to my vague understanding of action, I have read the books again and again, and the concept is almost there, but not 100% clear enough).
2) If I can't, can I simply use the equivalent definition to prove homomorphism? I am thinking about using the second definition to prove $G$ is an action on $X\times Y$, which I personally preferred. But if I use the second one, it seems that I can't really prove that the action is exactly $g((x, y)) = (g(x), g(y))$.