# Orbits of the group action $g. (x,y) = (gx,gy)$ on cartesian product

Let $G$ be a group acting on a set $X$. Then we have a natural action on $X \times X$ in the following way: $g. (x,y) = (gx,gy)$. Then, suppose we have two points of interest $x_1,x_2 \in X$, and we know their orbits under $G$. Is there some procedure to help up calculate the orbit of $(x_1,x_2)$?

edit: for what it's worth, both $G$ and $X$ are finite.

• What do you mean by calculating the orbit? The orbit is the orbit. Do you mean the size of the orbit? The isomorphism classes of orbits (as $G$-sets) are in bijection with the coset spaces $G/H$ accompanying conjugacy classes of subgroups $H$, so you could ask for the class of $H$ corresponding to $(x,y)\in X\times Y$. – blue Jun 23 '14 at 2:40
• @blue Probably OP means, if you know the orbits of $x_1$ and $x_2$ (under $G$ in $X$), how do you construct the orbit of $(x_1, x_2)$ (under $G$ in $X \times X$), from those two? – M. Vinay Jun 23 '14 at 2:43
• @M.Vinay Typo fixed, thanks. Yes, constructing the orbit in that way is exactly what I mean. – nigel Jun 23 '14 at 2:47
• @nigelvr If you can represent $G$ with some (arbitrary) ordering, and $(g(x_1))_{g \in G} = (y_{1(g)})_{g \in G}$ (say) is the orbit of $x_1$ in $G$, written in this order, and $((g(x_2))_{g \in G} = (y_{2(g)})_{g \in G}$ is defined similarly, then the orbit of $(x_1, x_2)$ is $(y_{1(g)}, y_{2(g)})_{g \in G}$ [obtained by zipping the "lists" $(x_{1(g)})$ and $(x_{2(g)})$]. – M. Vinay Jun 23 '14 at 2:50
• @nigelvr Are $G$ and $X$ infinite? The notation would be much simpler if one or both were finite, or at least countable. – M. Vinay Jun 23 '14 at 2:52