Theorem
Let G be a finite group and let a be an element of G . Then,$ |cl(a)|=|G:C(a)|.$
Attempt
Consider the function$T$ that sends the coset $xC(a)$ to the conjugate $xax^{-1}$.
$xC(a)=yC(a)$ which implies $y^{ -1}x \in C(a)$ which inturn implies $y^{-1}xa=ay^{-1}x$.Rearranging we get $xax^{-1}=yay^{-1}$. So T is well defined. By reversing the argument ,we can see T is one-one. Clearly T is onto.
Doubt
Is my proof correct?
How to prove T preserves operation? Or does it preserves operation. (It is not required for the proof)