Let X be the set consisting of the 15 subsets of S₃ having exactly two elements. The map L : S₃ -> Sym(X) with L_g({a,b}) = {g(a),g(b)} is a permutation representation (it was treated once before).

The orbit of {e, (1,2)} consists of 3 elements:

{e,(1,2)}, {(1,3), (1,2,3)}, and {(2,3),(1,3,2)}.

Can you describe the other orbits of the given action L?

The group GL_n(R), with n > 1, is not transitive on the set of all vectors of Rⁿ. For, the (set consisting of the) zero vector 0 can only be transformed into itself.

The group is transitive on the set of nonzero vectors: if v₁ and w₁ are two such vectors, then v₁ (respectively w₁) can be extended to a basis v₁, ... , v_n (respectively w₁, ... , w_n) of Rⁿ and determine an invertible linear map a : Rⁿ -> Rⁿ by

a(r₁v₁ + ··· + r_nv_n) = r₁w₁ + ··· + r_nw_n.

The map a belongs to GL_n(R) and satisfies Av₁ = w₁, and so indeed v and w are in the same orbit.

Conclusion: there are precisely two orbits, viz., Rⁿ\{0} and {0}.

The orbits in a group G of the group G acting by conjugation on itself are the so-called conjugacy classes. Since {e} is a single orbit, the action is transitive only if G is the trivial group {e}.

We work the conjugacy classes out for G = S₃. They are

e

(1,2), (2,3), (1,3)

(1,2,3), (1,3,2)

More generally, for S_n the conjugacy classes consist of all elements of a given cycle type. See Section 5.2 Above, the cycle structure 1,1,1 belongs to e, the type 2,1 to the class of (1,2) and the type 3 to the class of (1,2,3).

The cycle structures are nothing but the partitions of n. For n = 4, the partitions are

4,   31,   22,   211,   1111.

Representative elements from the corresponding conjugacy classes are:

(1,2,3,4),  (1,2,3),   (1,2)(3,4),   (1,2),   e.