Let n 1 n as usual Prove that sigman acts on the power set

Let (n) = {1, ..., n}, as usual. Prove that sigma_n acts on the power set P(n) by sigma middot S:= sigma(S), where sigma: (n) rightarrow (n) and S (n). How many orbits are there? Prove your answer correct.

Solution

4. (a) <n> = {1,2,3,…,n}

P<n> denotes the power set of <n>, i.e. the collection of all sunsets of <n>

_n be the symmetric group on {1,2,3,…,n}

Then, the group _n acts on the power set P<n> defined by .S = (S) where S is a subset of <n>, i.e. S P<n> and : <n> <n> is a function, i.e. _n

(Proved)

(b) The symmetric group _n acting on the nonempty set P<n>

The equivalence class {.S | _n } is called orbit of _n containing S P<n>

Since there are n! elements in the symmetric group _n, the number of orbits is n!