If $c_1,\dots, c_h$ are the orders of the centralisers of elements of distinct conjugacy classes of a finite group, then $1/c_1+\dots+1/c_h=1$

Question

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

$$Cl(x)=\{gxg^{-1}\mid g\in G\}.$$

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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 :)

Answer 1

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.

Answer 2

If we consider the action of conjugacy, the stabilizer of an element is exactly its centralizer. Let $\gamma_i$ be the cardinality of the conjugacy class of the i-th conjugacy class. Since the product of the orbit of an element and its stabilizer is $|G|$, we have that $\gamma_i c_i = |G| $. Hence

$$ \sum_{i=1}^h 1/c_i = \sum_{i=1}^h \gamma_i / |G| = \frac{1}{|G|} \sum_{i=1}^h \gamma_i = 1$$

Since the sum of the cardinalities of all the conjugacy classes sum up to $|G|$ (the conjugacy classes are a partition of $G$).