Proof that each element of finite arbitrary group belongs to unique conjugacy class - question on step which takes inverse of a product $G$ is an arbitrary finite group. I need to show that each element lies in a unique conjugacy class.
My attempt:
suppose element $g$ lies in conjugacy class $A$ & $B$, but $A \neq B$,
then for some $a_x \in A, b_x \in B, a_x \not\sim b_x$.
$q, p \in G$
$g = p a_x p^{-1}$
$g=q b_x q^{-1}$
$a_x=p^{-1}gp=p^{-1}q b_x q^{-1}p$ $(*)$
then define: $m=p^{-1}q \implies m^{-1}=q^{-1}p$
Rewrite $(*)$: $a_x=mb_xm^{-1}$
$m \in G$ because $p, q \in G \implies p^{-1}, q^{-1} \in G \implies p^{-1}q, q^{-1}p \in G \iff m \in G \implies a_x\sim b_x$ 
Presupposition was wrong, $\implies A = B$.
So there are only disjunct or identical conjugacy classes, so each element belongs to a unique conjugacy class.
Is this proof above correct? If yes, how can I justify the step:
$m=p^{-1}q \implies m^{-1}=q^{-1}p$
I know it is true for matrices for example, but is this true for group elements in general?

Yes, it is correct. And, since\begin{align}q^{-1}p\overbrace{p^{-1}q}^{\phantom{m}=m}&=q^{-1}eq\\&=q^{-1}q\\&=e,\end{align}it is indeed true that the inverse of $m$ is $q^{-1}p$.
