This is part of Exercise 1.6.8 of Robinson's "A Course in the Theory of Groups (Second Edition)".
The Details:
The centraliser of an element $x$ in a group $G$ is defined as
$$C_G(x)=\{g\in G\mid xg=gx\}.$$
The conjugacy class of $x$ is defined as
$$Cl(x)=\{gxg^{-1}\mid g\in G\}.$$
The previous part of the exercise in question is as follows.
Lemma: Let $G$ be a finite group. The elements of the same conjugacy class have conjugate centralisers.
Proof: Let $x,y\in Cl(a)$ for some $a\in G$. Then there exist $g,h\in G$ such that $x=gag^{-1}$ and $y=hah^{-1}$. Note that $gh=f$ for $f\in G$, solved for $g$, has the unique solution $g=fh^{-1}$. We have
$$\begin{align}
C_G(x)&=\{k\in G\mid kx=xk\}\\
&=\{k\in G\mid kgag^{-1}=gag^{-1}k\}\\
&=\{k\in G\mid (kg)a(kg)^{-1}=gag^{-1}\}\\
&=\{k\in G\mid (kfh^{-1})a(kfh^{-1})^{-1}=(fh^{-1})a(fh^{-1})^{-1}\}\\
&=\{k\in G\mid (f^{-1}kf)h^{-1}ah(f^{-1}kf)^{-1}=h^{-1}ah\}\\
&=\{b=f^{-1}kf\in G\mid byb^{-1}=y\}\\
&=f\{k\in G\mid (fkf^{-1})y=y(fkf^{-1})\}f^{-1}\\
&=f\{K\in G\mid Ky=yK\}f^{-1}\\
&=fC_G(y)f^{-1},
\end{align}$$
because $fkf^{-1}$ runs through $G$ just as an arbitrary $K\in G$ does.$\square$
(I'm sorry about the garbled proof. I'm trying to piece together my understanding without recourse to external sources.)
The Question
If $c_1,\dots, c_h$ are the orders of the centralisers of elements of distinct conjugacy classes of a finite group, then $1/c_1+\dots+1/c_h=1$.
Thoughts:
This might require a combinatorial approach (in the numerical sense), since it involves counting, but of reciprocals; that's where I'm struggling: I can't use anything beyond the number of conjugacy classes being
$$\frac{1}{|G|}\sum_{\pi\in G}|{\rm Fix}(\pi)|,$$
by Exercise 1.6.2, where ${\rm Fix}(\pi)$ is the number of fixed points of the action of conjugacy on $\pi$. (I'm a little unsure of this. Have I got it right?)
I know that conjugation partitions the underlying set of $G$. This is mentioned earlier in the book.
Please help 🙂
Best Answer
Hint : The class equation is $$|G|=\sum_{g \in G}|Cl(g)|=\sum_{g \in G}|G:C_G(g)|=\sum_{g \in G}{|G|\over |C_G(g)|}$$ where we're taking only one $g$ from each conjugacy class.