Let the symmetric group $S_4$ act on $X = \{1,2,3,4 \}.$ This gives an action on the power set $\mathcal P(X).$ Count the number of orbits for the action of $S_4$ on $\mathcal P(X).$
My attempt $:$ By orbit counting theorem (i.e. by Burnside's lemma) we have $$\textbf {Number of orbits}\ = \dfrac {1} {\left \lvert S_4 \right \rvert} \sum\limits_{\sigma \in S_4} \left \lvert \mathcal P (X)^{\sigma} \right \rvert\ \ \ \ \ \ \ \ \ \ (*)$$ where $\mathcal P (X)^{\sigma} = \{A \subseteq X\ |\ \sigma \cdot A = A \},\ \sigma \in S_4$ and where $\sigma \cdot A = \{\sigma (a)\ |\ a \in A\},\ \sigma \in S_4.$
Now it is clear that identity will fix every element of $\mathcal P (X).$ Hence $\left \lvert \mathcal P (X)^{\text {id}} \right \rvert = \left \lvert \mathcal P (X) \right \rvert = 2^4 = 16.$ Now take a transposition $(1\ 2).$ Suppose $A \subseteq X$ be a fixed point of $(1\ 2).$ Then $1 \in A \iff 2 \in A.$ So the only possible subsets of $X$ which are fixed by $(1\ 2)$ are $\varnothing, \{1,2\}, \{1,2,3\},\{1,2,4 \},\{1,2,3,4\}, \{3\},\{4\},\{3,4\},$ which are eight in numbers. So $\left \lvert \mathcal P (X)^{(1\ 2)} \right \rvert = 8.$ Similarly for any transposition $\sigma \in S_4$ we have $\left \lvert \mathcal P (X)^{\sigma} \right \rvert = 8.$ Now let us take a $3$-cycle in $S_4$ say $(1\ 2\ 3).$ If $A \subseteq X$ is fixed by $(1\ 2\ 3)$ then either $\{1,2,3\} \subseteq A$ or $A \cap \{1,2,3\} = \varnothing.$ This gives us the four possible fixed points of $(1\ 2\ 3)$ which are $\varnothing, \{4\}, \{1,2,3\}, \{1,2,3,4 \}.$ Similarly for any $3$-cycle $\sigma \in S_4$ we have $\left \lvert \mathcal P (X)^{\sigma} \right \rvert = 4.$ Now let us take a $4$-cycle in $S_4$ say $(1\ 2\ 3\ 4).$ If $A \subseteq X$ is fixed by $(1\ 2\ 3\ 4)$ then either $A = \{1,2,3,4 \}$ or $A = \varnothing.$ So for any $4$-cycle $\sigma \in S_4$ we have $\left \lvert \mathcal P (X)^{\sigma} \right \rvert = 2.$ By similar argument one can show that for any $2$–$2$ cycle $\sigma \in S_4$ we have $\left \lvert \mathcal P (X)^{\sigma} \right \rvert = 4.$ For instance if $\sigma = (1\ 2) (3\ 4)$ then $\mathcal P (X)^{\sigma} = \varnothing, \{1,2\},\{3,4\},\{1,2,3,4\}.$ This exhausts all the elements of $S_4.$ Now there are $1$ $\text {id},$ $6$ transpositions, $8$ $3$-cycles, $6$ $4$-cycles and $3$ $2$–$2$ cycles in $S_4.$ So the RHS of $(*)$ becomes $$\dfrac {(1 \times 16) + (6 \times 8) + (8 \times 4) + (6 \times 2) + (3 \times 4)} {24} = \dfrac {120} {24} = 5.$$ Please check my reasoning above. Thanks in advance.
Best Answer
In general if the symmetric group $S_n$ is acting on $\mathcal P (X),$ where $X = \{1,2, \cdots ,n\},$ then the orbits under this action are precisely of the form $\mathcal P_{d} (X),$ for $0 \leq d \leq n,$ where $$\mathcal P_{d} (X) : = \left \{A \subseteq X\ |\ \text {Card}\ (A) = d \right \}.$$