Explain why for any finite group G and any g in G...

the number of elements |K(g)| of the conjugacy class K(g) of the element g divides the order |G| of the group G.

If you are familiar with the terminology of group actions this is a consequence of a standard fact sometimes called the "orbit-stabilizer theorem". If a finite group G acts on a set S, then for any s in S, if you define orbit(s) = {g.s: g in G}, which is a subset of S, and stabilizer(s) = {g in G: g.s = s}, which is a subgroup of G, then it turns out that stabilizer(s) is always a subgroup of G and |orbit(s)| is equal to the index of this subgroup in G. Since Lagrange's theorem implies that the index of a subgroup divides the order of the group, you conclude |orbit(s)| divides |G| for any s in S. You get the statement that you want by letting your finite group G act on itself, by conjugation; orbit(s) is then your "K(s)."

If you aren't familiar with the terminology of group actions, or don't have this theorem, then one way to do this is to to prove as much of the content of this theorem as we need in this particular case. Let's do that.

Fix an element g in G. [I will write e for the identity element of G.]

Let H = {x in G: x^{-1} g x = g}. I claim that H is a subgroup of G.

Since e^{-1} g e = g, e is in H.

If x and y are in H then
(xy)^(-1) g (xy) = y^{-1} x^{-1} g x y
= y^{-1} g y [because x is in H]
= g [because y in H]
showing that xy is also in H.

If x is in H, then since x^(-1) g x = g, we have that x(x^(-1) g x)x^(-1) = xgx^(-1), or that g = xgx^(-1), or that g = (x^(-1))^(-1) g x^(-1), showing that x^(-1) is in H also. This completes the proof that H is a subgroup of G.

By Lagrange's theorem, |H| is a divisor of |G|. In fact, |G| = |H| [G:H] where [G:H] denotes the index of H in G, ie, the number of left cosets of H in G. We will show that |K(g)| divides G by showing that |K(g)| = [G:H]. We will show this by exhibiting a function f from the set of left cosets of G in H to K(g) and proving that it is a bijection.