Exercise 4.1.5 (Action of S_3 on some specified sets)

Proof. We obtain similarly the following results.

(a)

For $\mathrm{\Omega }={[[1,3]]}^3$, we choose the numbering $$f(k)=(a+1,b+1,c+1)$$

where $a,b,c$ are the digits of $k$ in base $3$.

We use the same method as in Exercise 4,i.e., $S_3$ acts on $\mathrm{\Omega }=[[1,27]]$ by

$$\sigma \cdot k=f^{-1}(\sigma \cdot f(k)),\sigma \in S_3,k\in [[1,27]].$$

To avoid repetitive computations, we give a notebook Sagemath program to compute to corresponding permutation $\tilde{\sigma }$ for every $\sigma \in S_3$:

     def f( k):
         j = k - 1
         l = []
         for _ in range(3):
             l.append(j % 3 + 1);
             j = j // 3
         l.reverse()
         return tuple(l)
     
     def f_inv(t):
         a, b, c = t[0] - 1, t[1] - 1, t[2] - 1
         return 1 + c + 3 * b + 3^2 * a
     
     G = SymmetricGroup(3); G
          
         Symmetric group of order 3! as a permutation group
     
     def tilde(sigma):
         t_sigma = lambda k : f_inv((sigma(f(k)[0]), sigma(f(k)[1]), sigma(f(k)[2])))
         tau = Permutation([t_sigma(i) for i in range(1,3^3 + 1)])
         return tau.to_permutation_group_element()
     
     for sigma in G:
         print sigma, ’\t=>\t’, tilde(sigma)
     
     ()      =>  ()
     (1,2)   =>  (1,14)(2,13)(3,15)(4,11)(5,10)(6,12)(7,17)(8,16)
                       (9,18)(19,23)(20,22)(21,24)(25,26)
     (1,2,3) =>  (1,14,27)(2,15,25)(3,13,26)(4,17,21)(5,18,19)
                       (6,16,20)(7,11,24)(8,12,22)(9,10,23)
     (1,3,2) =>  (1,27,14)(2,25,15)(3,26,13)(4,21,17)(5,19,18)
                       (6,20,16)(7,24,11)(8,22,12)(9,23,10)
     (2,3)   =>  (2,3)(4,7)(5,9)(6,8)(10,19)(11,21)(12,20)(13,25)
                       (14,27)(15,26)(16,22)(17,24)(18,23)
     (1,3)   =>  (1,27)(2,26)(3,25)(4,24)(5,23)(6,22)(7,21)(8,20)
                       (9,19)(10,18)(11,17)(12,16)(13,15)
     

This gives the following correspondence, for the chosen numbering:

$\sigma$	$\tilde{\sigma }$
()	()
(1 2)	(1 14)(2 13)(3 15)(4 11)(5 10)(6 12)(7 17)(8 16)(9 18)(19 23)(20 22)(21 24)(25 26)
(1 2 3)	(1 14 27)(2 15 25)(3 13 26)(4 17 21)(5 18 19)(6 16 20)(7 11 24)(8 12 22)(9 10 23)
(1 3 2)	(1 27 14)(2 25 15)(3 26 13)(4 21 17)(5 19 18)(6 20 16)(7 24 11)(8 22 12)(9 23 10)
(2 3)	(2 3)(4 7)(5 9)(6 8)(10 19)(11 21)(12 20)(13 25)(14 27)(15 26)(16 22)(17 24)(18 23)
(1 3)	(1 27)(2 26)(3 25)(4 24)(5 23)(6 22)(7 21)(8 20)(9 19)(10 18)(11 17)(12 16)(13 15)

There are three orbits for the action of $S_3$ on ${[[1,27]]}^3$. The stabilizers are conjugates of

$$G_1=\{(),(23)\},G_2=G_6=\{()\}.$$

(b)

In this part, $S_3$ acts on $A=𝒫(\{1,2,3\})-\{\varnothing \}$ by $$\sigma (X)=\{\sigma (x)\mid x\in X\},(X\in A,\sigma \in S_3).$$

We choose the following numbering for $A$:

$k$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$f(k)$	$\{1\}$	$\{2\}$	$\{3\}$	$\{1,2\}$	$\{1,3\}$	$\{2,3\}$	$\{1,2,3\}$

Then $\sigma$ acts on $[[1,7]]$ by

So every permutation $\sigma \in S_3$ induces a permutation $\tilde{\sigma }\in S_7$, defined by

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

So

Similarly, we obtain

There are $3$ orbits for the action of $S_3$ on $[[1,7]]$. The stabilizers are conjugates of

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