# 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?

• Alternative route. The relation can be described by $a\sim b\iff\exists g\in G\;bg=ga$. So it is enough to prove that this is an equivalence relation. Sep 21, 2018 at 14:09

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 $$q^{-1}p$$.

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.