Prove that the number of elements of every conjugacy class of a finite group $G$ divides the order of $G$.
I'm studying for my Group Theory exam and this was a question on a previous exam.
I know I have to use the counting orbit theory, but I have no idea how this would work. Furthermore I have some difficulties with understanding what conjugacy classes are. We use the book "Groups and Symmetry" by M.A. Armstrong which gives the following definition:
"Given elements x,y of a group G, we say that x is conjugate to y if $gxg^{-1}=y$ for some $g\in G$. The equivalence classes are called conjugacy classes."
But what does it mean that some elements are in the same conjugacy class? For me the deffiniton isn't clear enough. Could some one please try and explain some more? For example what the usage of conjugacy classes is.
Regarding to Andreas's post including the very important equation in permutation groups, one of the usage can be:
The proof is routine and is based on defining an action of $G$ on $\Omega=\{Hx\mid x\in G\}$. The action is transitive and since $|\Omega|>1$ so $$\exists h\in G, |Fix(h)|=0$$. Now if we assume that every elements of $G$ is a conjugate element inside $H$ so $$\exists x\in G, \exists t\in H, h=x^{-1}tx$$ and so $(Hx)^h=Hx$ which is a contradiction.