Exercise 4.1.4 (Action of $S_3$ on $[\![1,3]\!]^2$)

Proof. $G=S_3$ acts on $\mathrm{\Omega }={[[1,3]]}^2$ by

Indeed, for all $(i,j)\in \mathrm{\Omega }$, and for all $\sigma ,\tau \in S_3$, if $1=1_{S_3}=()$,

• $1\cdot (i,j)=(1_{S_3}(i),1_{S_3}(j))=(i,j)$,
• $\sigma \cdot (\tau \cdot (i,j))=\sigma \cdot (\tau (i),\tau (j))=(\sigma (\tau (i)),\sigma (\tau (j)))=(\mathit{𝜎𝜏})\cdot (i,j).$

Since $S_3=\{(),(12),(13),(23),(123),(132)\}$, the orbit of $(i,j)$ is

so

and since $\mathrm{\Omega }={𝒪}_{(1,1)}\cup {𝒪}_{(1,2)}$, there is no other orbit.

We choose an arbitrary numbering $f$ of $\mathrm{\Omega }$:

where $f:[[1,9]]\to \mathrm{\Omega }$ is a bijection:

Then $S_3$ acts on $[[1,9]]$ by

If $\sigma ,\tau \in S_3$, and $1\le k\le 9$,

• $1\cdot k=f^{-1}(1\cdot f(k))=f^{-1}(f(k))=k$,
  

• $$\begin{array}{llll}\hfill \sigma \cdot (\tau \cdot k)& =\sigma \cdot f^{-1}(\tau \cdot f(k))& \hfill & \\ \hfill & =f^{-1}(\sigma \cdot f(f^{-1}(\tau \cdot f(k)))& \hfill & \\ \hfill & =f^{-1}(\sigma \cdot (\tau \cdot f(k))& \hfill & \\ \hfill & =f^{-1}((\mathit{𝜎𝜏})\cdot f(k))& \hfill & \\ \hfill & =(\mathit{𝜎𝜏})\cdot k.& \hfill & \end{array}$$

  So every permutation $\sigma \in S_3$ induces a permutation $\tilde{\sigma }\in S_9$, defined by

  $$\tilde{\sigma }(k)=\sigma \cdot k=f^{-1}(\sigma \cdot f(k)),(k\in [[1,9]]).$$

(Any other numbering of $\mathrm{\Omega }$ leads to conjugate permutations to $\tilde{\sigma }$.)

We give the cycle decomposition of $\tilde{\sigma }$ for every $\sigma \in S_3$

For instance, if $\sigma =(12)$, then

So

Similarly, we obtain

By the first part, the two orbits of $S_3$ acting on $[[1,9]]$ are

For any $\sigma \in S_3$,

and

So the stabilizers of $1$ and $2$ are