How many orbits does this induced action of $S$ on $X\times X\times X$ have?

Question

Let $X$ denote the set $\{1,2,\cdots,n\}$ of the first $n$ positive integers (we assume $n\geq 3$). Let $\mathcal{G}$ be the group of bijections of $X$, with multiplication being the composition. The group $\mathcal{G}$ acts on $X$ in the standard manner: $\mathcal{G}\times X\rightarrow X$ being given by $(\varphi,i)\mapsto \varphi\cdot i:=\varphi(i)$, where $\varphi(i)$ denotes the image of $i$ under $\varphi$. Consider the induced action of $\mathcal{G}$ on $X\times X\times X$ by $\varphi\cdot(i,j,k):=(\varphi(i),\varphi(j),\varphi(k))$. How many orbits does this induced action of $\mathcal{G}$ on $X\times X\times X$ have?

My Attempt: Possible stabilizer of points in $X\times X\times X$ is $S_{n-3},S_{n-2},S_{n-1}$ so possible size of orbits are $n, n(n-1), n(n-1)(n-2)$. How can I compute number of orbits from here? Give some hints.

Answer 1

$(i_1, j_1, k_1) $ and $(i_2, j_2, k_2) $ belongs to same orbit iff $\exists \phi\in \mathcal{G}$ such that

$(\phi(i_1), \phi(j_1), \phi(k_1))=(i_2, j_2, k_2) $

i.e $\phi(i_1) =i_2,\phi(j_1) =j_2, \phi(k_1) =k_2$

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Distinct orbits are :$\mathcal{O}(1, 1,1) ,\underbrace{\mathcal{O}(1, 1,2) ,\mathcal{O}(1, 2,1) , \mathcal{O}(2,1,1)},\mathcal{O}(1,2,3) $

Answer 2

With reference to the action on the Cartesian product: \begin{alignat}{1} \operatorname{Stab}((i,j,k))&=\{\varphi\in\mathcal{G}\mid\varphi\cdot(i,j,k)=(i,j,k)\} \\ &=\{\varphi\in\mathcal{G}\mid(\varphi(i),\varphi(j),\varphi(k))=(i,j,k)\} \\ &=\{\varphi\in\mathcal{G}\mid(\varphi(i)=i)\wedge(\varphi(j)=j)\wedge(\varphi(k)=k)\} \\ \end{alignat} whence:

• $\left|\operatorname{Stab}((i,j,k))\right|=(n-3)!$, if $i,j,k$ are pairwise distinct;
• $\left|\operatorname{Stab}((i,j,k))\right|=(n-2)!$, if any two amongst $i,j,k$ are equal;
• $\left|\operatorname{Stab}((i,j,k))\right|=(n-1)!$, if $i=j=k$.

By the orbit-stabilizer theorem, the orbits have accordingly size:

• $n(n-1)(n-2)$;
• $n(n-1)$;
• $n$.

Now, there are $n(n-1)(n-2)$ elements $(i,j,k)\in X^3$ such that $i,j,k$ are pairwise distinct, so they all build up $\frac{n(n-1)(n-2)}{n(n-1)(n-2)}=1$ orbit. Likewise, there are $3n(n-1)$ elements $(i,j,k)\in X^3$ such that any two amongst $i,j,k$ are equal$^\dagger$, so they all build up $\frac{3n(n-1)}{n(n-1)}=3$ orbits. Finally, there are $n$ elements $(i,i,i)\in X^3$, so they all build up $\frac{n}{n}=1$ orbit. Therefore, there are $1+3+1=5$ orbits overall.

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$^\dagger$There are $n(n-1)$ of the form $(i,i,j)$, other $n(n-1)$ of the form $(i,j,i)$, and finally $n(n-1)$ of the form $(j,i,i)$.