Let $A = (1 2 3 \dots n)$ be an element of $P_n$. So that the group $P_n$ acts on itself by means of the action of conjugation, for $B \in P_n$,
$$B · A = B A B^{-1}.$$
Stabilizer of $A$ is subgroup; $$\{A^{c} \mid c = 0, \dots , n-1\}.$$
I want to find the cardinality of $Orb_{P_n}(A)$.
So by applying the Orbit stabilizer theorem:
Then for A to act in itself by conjugation then I have that $Orb_{P_n}(A)=ABA^{-1}$ : where $A$ is an element of $P_n$.
[Provided by my friend Andreas]
Proof of the stabilzer of A:
By applying the orbit-stabilizer theorem, you can count how many n-cycles there are.
The result is $n! / n = (n-1)!$, as each $n$-cycle can be written in $n$ different ways,
$$
(12\dots n) = (23\dots n 1) = \dots.
$$
Since the conjugacy class of $(12\dots n)$ consists of all $n$-cycles, orbit-stabilizer tells you that the stabilizer you are looking for (which is called the centralizer of $A$) has order $n$.
Since there are n distinct powers of A, and they clearly stabilize A, and so forth.
Is this correct :O
-nomad609
Best Answer
It would be this I think:
Suppose that the finite group G acts on the set X, and that x ∈ X. Then the cardinality of G is the product of the cardinality of the orbit of x and the cardinality of the isotropy group at x. Writing |S| for the cardinality of a finite set S, this is |G| = |G · x| |Gx|.
If anyone can verify this please :)