The stabilizer of 3 consists of all permutations g with g(3) = 3. These are all permutations of {1,2,4,5}. Hence, the stabilizer is Sym({1,2,4,5}), which is isomorphic with S₄.

The stabilizer of the set {4,5} consists of all elements g of S₅ with g(4) {4,5} and g(5) {4,5}. In the disjoint cycle decomposition of such an element g, we find either the cycle (4,5), or no cycle at all in which 4 or 5 occurs. Thus, such an element g is either of the form h or h(4,5), for some h S₃. Hence, the stabilizer of {4,5} is the subgroup S₃<(4,5)> of S₅.

More precisely, the stabilizer is the image of the natural morphism

S₃ × <(4,5)> -> S₅ ,   (h,g) -> hg. Thus, the stabilizer has order 6 · 2 = 12.

According to Lagrange's Theorem, there are 120/12 = 10 cosets. This is no coincidence, as will become clear from the next theorem.

[It means that G is transitive on X. For, restrict the representation to the orbit containing {4,5}; then, by the theorem, the orbit has size 10, which equals |X|. So X is a single orbit.]

Let x be the first standard basis vector. Then G_x is the subgroup of G of all invertible matrices of the form

1 * * 0 * * 0 * *

.

Let x G. Then G_x is the subgroup of G of all g G with gx = xg. This subgroup is called the centralizer of x in G. It coincides with the centralizer of the set {x}.

Observe that G_e = G.

Let G be the group D_n acting on the n vertices of a regular n-gon. See a previous example. Let x be a vertex. Among the n rotations in G only the identity fixes x. The only reflection fixing x is the reflection in the axis through x and the center of the n-gon. So in this case G_x consists of two elements.