Prove that the size of every conjugacy class of a finite group divides the order of the group We are effectively asked to show that a map is a bijection. 
$$\phi: (G:C_G(a))\rightarrow (a)$$
$$ C_G(a)x \rightarrow x^{-1}ax$$
Need to show that the above is well-defined, 1-1 and onto first and then answer the title question. 
I'm a bit stuck on where to even start with this...
Edit...
The notation is as follows:
$C_G(a)$ is the centraliser of $a$ in $G$, $(G:C_G(a))$ is the set of right cosets of $C_G(a)$ in $G$ and $(a)$ is the conjugacy class containing $a$.
 A: While your notation is a bit unclear, the solution to your problem is the orbit-stabilizer theorem.
A: We want to ensure that when we count the (right) cosets of $C_G(a)$ we get the same number of elements as distinct conjugates of $a$.
First we need to show "well-definedness", since the domain of our map is cosets, not elements of $G$ (so we need to be sure we get the same result no matter which element $g \in C_G(a)x$ we use).
So suppose $C_G(a)x = C_G(a)y$. This means that $xy^{-1}$ commutes with $a$ ($xy^{-1} \in C_G(a)$). So:
$xy^{-1}a = axy^{-1}\\xy^{-1}ay = ax\\y^{-1}ay = x^{-1}ax.$
Now it is clear that for any conjugate $g^{-1}ag$ we have the pre-image under our map, of $C_G(a)g$, which shows our map is onto (surjective).
On the other hand, suppose $g^{-1}ag = h^{-1}ah$. This means (doing the steps above "in reverse"):
$ag = gh^{-1}ah\\agh^{-1} = gh^{-1}a,$
that is, $gh^{-1} \in C_G(a)$, so that $C_G(a)g = C_G(a)h$, so our mapping is injective.
This means, then, that the index $[G:C_G(a)]$ (which is the number of right (or left) cosets of $C_G(a)$ in $G$) is precisely the number of distinct conjugates of $a$, that is the cardinality of the conjugacy class of $a$.
(In all fairness to Matt Samuel, this is the basic structure of the proof for the orbit-stabilizer theorem, just "instantiated" for the particular action of conjugation).
