Proof that each element of finite arbitrary group belongs to unique conjugacy class - question on step which takes inverse of a product

Question

$G$ 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?

Answer 1

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 $q^{-1}p$.

Answer 2

I'm not sure what the problem is.

For $x,y\in G$, define $x\sim y$ if and only if there exists $z\in G$ such that $$ y=zxz^{-1} $$ (that is, $y$ is conjugate to $x$).

1. The relation $\sim$ is reflexive
2. The relation $\sim$ is symmetric
3. The relation $\sim$ is transitive

The above facts (that you should prove, if you haven't already) say that the relation $\sim$ is an equivalence relation.

If we consider $[x]_\sim=\{y\in G:x\sim y\}$, the conjugacy class of $x$, then the general theory of equivalence relations tells you that if $x,y\in G$, then $$ [x]_\sim=[y]_\sim \qquad\text{or}\qquad [x]_\sim\cap[y]_\sim=\emptyset $$ that is, two conjugacy classes are either equal or disjoint.

In particular, no element can belong to two distinct conjugacy classes.