In a previous example we have seen that the stabilizer in S₅ of {4,5} is equal to H = S₃×<(4,5)>. The index |G/H| of this subgroup in S₅ is equal to 5!/(3!·2!) = 10. This is equal to the number of sets of {1, ..., 5} of size 2. Under the bijection of the theorem, the coset (1,2,3,4,5)H is mapped onto {1,5}. This image can be computed by use of any element from (1,2,3,4,5)H; for example

• (1,2,3,4,5) maps {4,5} to (1,2,3,4,5)({4,5}) = {1,5} , but
• (1,2,3,4,5)(1,2)(3,2) maps {4,5} to (1,2,3,4,5)(1,2)(3,2)({4,5}) = {1,5}.

Let G be a group of order 10. It acts by left multiplication on the set X consisting of the 45 subsets of G of size 2. By the theorem, the number of elements in an orbit is a divisor of |G| and hence equal to 1, 2, 5, or 10.

An orbit cannot be a singleton (do you see why?). As |X| is odd, there must be an orbit of size 5. The stabilizer of an element from this orbit as order 2. This establishes that G has subgroup and thus also an element of order 2.

Later, we shall repeat this argument in greater generality to show that if p is a prime dividing |G|, there is an element of order p in G.

Let x G. Then the centralizer C_G(x) of x in G is the stabilizer of x in the conjugation action. See a previous example. Hence the number of conjugates of x equals the index of C_G(x) in G and is a divisor of the order of G.

Let G = D_n acting on the n vertices of a regular n-gon. The group G is transitive on the n vertices. So, since the order of G is 2n, the stabilizer G_x consists of two elements. See a previous example.