Decomposing a product set into orbits

Question

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.

Answer 1

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.

Answer 2

replaced $u_1$, $u_2$, $u_3$ by 1, 2, 3

()U         1      2      3
(1 2)U      2      1      3
(2 3)U      1      3      2
(3 1)U      3      2      1
(1 2 3)U    2      3      1
(1 3 2)U    3      1      2
           ↓      ↓      ↓
       ｛1,2,3｝｛1,2,3｝｛1,2,3｝

on $U$ are transitive

so $S_3$ is divided into 1 piece, $S_3$ itself

for (a), $U×V$ is divided by whole $S_3$, and is still $U×V$

and for (b), orbits on $V$ are {1}, {2, 3}

()V         **1**      2      3
(1 2)V        2        1      3
(2 3)V      **1**      3      2
(3 1)V        3        2      1
(1 2 3)V      2        3      1
(1 3 2)V      3        1      2

so {(), (2 3)} is a part of separation, so do {(1 2), (1 3 2)}, {(3 1) (1 2 3)}

replaced $(u_i, v_j)$ by $(i, j)$

()(U×V)      (1, 1) (1, 2) (1, 3)
             (2, 1) (2, 2) (2, 3)
             (3, 1) (3, 2) (3, 3)

(2 3)(U×V)   (1, 1) (1, 3) (1, 2)
             (3, 1) (3, 3) (3, 2)
             (2, 1) (2, 3) (2, 2)

(1 2)(U×V)   (2, 2) (2, 1) (2, 3)
             (1, 2) (1, 1) (1, 3)
             (3, 2) (3, 1) (3, 3)

(1 3 2)(U×V) (3, 3) (3, 1) (3, 2)
             (1, 3) (1, 1) (1, 2)
             (2, 3) (2, 1) (2, 2)

(3 1)(U×V)   (3, 3) (3, 2) (3, 1)
             (2, 3) (2, 2) (2, 1)
             (1, 3) (1, 2) (1, 1)

(1 2 3)(U×V) (2, 2) (2, 3) (2, 1)
             (3, 2) (3, 3) (3, 1)
             (1, 2) (1, 3) (1, 1)

as we see, in each separation, (1, 1) is fixed, (1, 2), (1, 3) are exchanged each other, and so on

so here may be the answer of (b)

{($u_1$, $v_1$)},

{($u_1$, $v_2$), ($u_1$, $v_3$)},

{($u_2$, $v_1$), ($u_3$, $v_1$)},

{($u_2$, $v_2$), ($u_3$, $v_3$)},

{($u_2$, $v_3$), ($u_3$, $v_2$)}