Decomposing a product set into orbits The symmetric group S3 operates on two sets U and V of order 3. 
Decompose the product set $U × V$ into orbits for the “diagonal action” $g(u,v) = (gu,gv)$ when 
$(a)$ The operations on U and V are transitive
$ (b) $ The operation on U is transitive and the orbits for the operation on V are {v1} and {v2,v3}. 
I don't really have any idea how to do this. Any help that could be given would be appreciated.
 A: Let's look at (a). The elements of the symmetric group $S_3$ are


*

*The identity $e=()$,

*The transpositions $(12)$, $(23)$ and $(31)$,

*The $3$-cycles $(123)$, $(132)$.


You should already know how $S_3$ acts on $\{1,2,3\}$ transitively: it's the standard action. (Every permutation is just a function you apply to this set.)
So then, the elements of $U\times V$ are the ordered pairs
$$ \begin{array}{ccc} (1,1) & (1,2) & (1,3) \\ (2,1) & (2,2) & (2,3) \\ (3,1) & (3,2) & (3,3) \end{array} $$
The "diagonal" action is when we apply a permutation to both components simultaneously. So for example if $\pi=(123)$ then $\pi\cdot(1,2)=(\pi(1),\pi(2))=(2,3)$.
Start with $(1,1)$. Apply different permutations to it. Experiment! Test things out! Now, after some exploration, can you form a hypothesis as to what the orbit is? (The range of ordered pairs that can be gotten to from $(1,1)$ by applying a permutation to it.) Now prove it's an orbit.
Next, pick an ordered pair not in the orbit you already found. Can you take it from there?

For part (b) we'll want to understand how $S_3$ acts on $V=\{\circ,\oplus,\ominus\}$:


*

*The element $\circ$ is fixed. That is, $\pi\cdot\circ=\circ$ for all $\pi$.

*$e,(123),(132)$ fix both of $\oplus$ and $\ominus$, whereas $(12),(23),(31)$ all swap $\oplus\leftrightarrow\ominus$.


Now write down the elements of $U\times V$ again and go spelunking.
