Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 4.1 Exercise 4.1.4
Let $S_3$ act on the set $A = \{ (i,j) \ |\ 1 \leq i,j \leq 3 \}$ by $\sigma \cdot (i,j) = (\sigma(i), \sigma(j))$.
(1) Find the orbits of $S_3$ on $A$.
(2) Fix a labeling of the elements of $A$. For each $\sigma \in S_3$, find the cycle decomposition of $\sigma$ under the permutation representation $S_3 \rightarrow S_9$.
(3) For each orbit $\mathcal{O}$ of this action, choose some $a \in \mathcal{O}$ and compute $\mathsf{stab}(a)$.
Solution:
(1) We have $$1 \cdot (1,1) = (2\ 3) \cdot (1,1) = (1,1),$$ $$ (1\ 2) \cdot (1,1) = (1\ 2\ 3) \cdot (1,1) = (2,2),$$ and $$(1\ 3) \cdot (1,1) = (1\ 3\ 2) \cdot (1,1) = (3,3).$$ Thus the orbit containing $(1,1)$ is $\mathcal{O}_1 = \{ (1,1), (2,2), (3,3) \}$.
We also have $$1 \cdot (1,2) = (1,2), (1\ 2) \cdot (1,2) = (2,1),$$ $$(1\ 3) \cdot (1,2) = (3,2),$$ $$(2\ 3) \cdot (1,2) = (1,3),$$ $$(1\ 2\ 3) \cdot (1,2) = (2,3),$$ and $$(1\ 3\ 2) \cdot (1,2) = (3,2).$$ Thus the orbit containing $(1,2)$ is $\mathcal{O}_2 = \{ (1,2), (2,1), (3,2), (1,3), (2,3), (3,2) \}.$ These two orbits exhaust the elements of $A$.
(2) Fix the labeling $$\alpha_{1} = (1,1)\ \alpha_{4} = (2,1)\ \alpha_{7} = (3,1)$$ $$\alpha_{2} = (1,2)\ \alpha_{5} = (2,2)\ \alpha_{8} = (3,2)$$ $$\alpha_{3} = (1,3) \ \alpha_{6} = (2,3)\ \alpha_{9} = (3,3)$$Under the permutation representation $S_3 \rightarrow S_9$, we have$$1 \mapsto 1$$ $$(1\ 2) \mapsto (\alpha_1\ \alpha_5)(\alpha_2\ \alpha_4)(\alpha_3\ \alpha_6)(\alpha_7\ \alpha_8)$$ $$(1\ 3) \mapsto (\alpha_1\ \alpha_9)(\alpha_2\ \alpha_8)(\alpha_3\ \alpha_7)(\alpha_4\ \alpha_6)$$ $$(2\ 3) \mapsto (\alpha_2\ \alpha_3)(\alpha_4\ \alpha_7)(\alpha_5\ \alpha_9)(\alpha_6\ \alpha_8)$$ $$(1\ 2\ 3) \mapsto (\alpha_1\ \alpha_5\ \alpha_9)(\alpha_2\ \alpha_6\ \alpha_7)(\alpha_3\ \alpha_4\ \alpha_8)$$ $$(1\ 3\ 2) \mapsto (\alpha_1\ \alpha_9\ \alpha_5)(\alpha_2\ \alpha_7\ \alpha_6)(\alpha_3\ \alpha_8\ \alpha_4)$$(3) If $\sigma \in \mathcal{O}_1$, we have $[S_3 : \mathsf{stab}(\sigma)] = 3$, so that $|\mathsf{stab}(\sigma)| = 2$. Consider $(2,2) \in \mathcal{O}_1$; since $(1\ 3) \cdot (2,2) = (2,2)$, we have $\mathsf{stab}((2,2)) = \langle (1\ 3) \rangle$.
If $\sigma \in \mathcal{O}_2$, then $[S_3 : \mathsf{stab}(\sigma)] = 6$. Hence $|\mathsf{stab}(\sigma)| = 1$, and we have $\mathsf{stab}(\sigma) = 1$.
