This is a problem I have recently encountered.

Let $c$ be a conjugacy class in $G$. Let $x$ and $y$ be any elements of $c$

  1. Show that $Z(x)= \{g \in G | gx=xg \}$ and $Z(y)$ have the same number of elements. Call this number $|z_c|$.

  2. Prove that $1=\sum_{classes} \frac{1}{|z_c|}. $

Here is my attempt:

  1. Let the conjugacy class $c$ be the set $ Gz= \{ gzg^{-1} | g \in G \}$. Hence $x=aza^{-1}$ for some $a \in G$. Notice that the set $Z(x)$ is equivalent to the set of $g$ in $G$ such that $ gxg^{-1}=x $.

We have $$ gxg^{-1}=gaza^{-1}g^{-1}=x=aza^{-1} $$

which gives

$$ (a^{-1}ga) z (a^{-1}ga)^{-1} = z $$ Consider the set $Z(z)= \{ h \in G | hz=zh \} $.

Then $|Z(z)|$ would be equivalent to $|Z(x)|$ because $Z(z)= \{ h \in G | hzh^{-1} =z \}$.

This can be shown by letting $g=aha^{-1}$ in the previous equation, giving a bijection.

Similarly, we can show that $|Z(z)|= |Z(y)|$. Hence, $|z_c|=|Z(z)|=|Z(x)|=|Z(y)|.$

  1. For the second part, we use the orbit-stabilizer theorem. Noting that


$$|\text{Stab}_G(z)|= |Z(z)|$$ we have

$$ \frac{|Gz|}{|G|} = \frac{1}{|z_c|},$$

since orbits partition $G$, the sum of all this equals $1$.

Can anyone show me how to solve the problem and explain if my solution is correct?

Thank you so much for your help!


General comment. Let me recall that for a finite group $G$ acting on a set $\Omega$, and $\omega\in\Omega$ the orbit-stabiliser theorem states

$|\mathrm{Orb}(\omega)| = \frac{|G|}{|\mathrm{Stab}_G(\omega)|}$.

Also, for the sake of a reader interested in learning about this topic let me notice that if $\alpha,\beta$ lie in the same orbit then $\mathrm{Orb}(\alpha)=\mathrm{Orb}(\beta)$ (can you work out why?).

In your case we let $G$ act on itself by conjugation. Hence $\Omega=G$.

On your question. Why not use the orbit-stabiliser theorem (ost) also for 1?

We have that $Z(x)=\{g\in G : g^{-1}xg=x\}=\mathrm{Stab}(x)$. If $x,y$ lie in the same orbit then $\mathrm{Orb}(x)=\mathrm{Orb}(y)$ hence, by the ost, their stabilisers have same order.

I think your solution for the second part is good.


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