Intersection of conjugacy class with a subgroup

I want to prove that, if $$x^G$$ denotes the conjugacy class of $$x$$ in a finite group $$G$$ and $$H$$ is a subgroup of $$G$$ then $$|\{g\in G : g^{-1}xg\in H \}|=|x^G\cap H|\cdot |C_G(x)|,$$ where $$C_G(x)$$ denotes centralizer of $$x$$ in $$G$$.

If $$x^G\cap H=\emptyset$$ then nothing to prove.

But, if $$x^G\cap H\neq \emptyset$$, I could not see a clever way to get the factor $$|C_G(x)|$$ in the right side when we try to count number of $$g$$'s such that $$g^{-1}xg\in H$$.

Can one help (or suggest any other clever way) to prove above equality?

3 Answers

Fix $$h\in H\cap x^G$$. Suppose that $$g_1^{-1}xg_1=h=g_2^{-1}xg_2$$. Then $$g_1g_2^{-1}\in C_G(x)$$ or equivalently $$g_1=cg_2$$ with $$c\in C_G(x)$$. Hope this helps.

More generally.

Suppose $$G$$ acts transitively on a set $$\Omega$$, and let $$x \in G$$. Let $$H$$ be a point stabilizer.

Then you can prove: $$\frac{|x^G \cap H|}{|x^G|} = \frac{|\Omega^x|}{|\Omega|},$$ where $$\Omega^x = \{ \alpha \in \Omega : x \cdot \alpha = \alpha\}$$, the fixed point set of $$X$$.

For your question, apply this with $$\Omega = G/H$$ (left coset action). Then $$\Omega^x = \{gH : g^{-1}xg \in H\}$$, also $$|x^G| = [G:C_G(x)]$$ and $$|\Omega| = [G:H]$$.

So $$|x^G \cap H| |C_G(x)| = |\Omega^x| |H|$$.

Note that $$|\Omega^x | |H| = | \{ g \in G : g^{-1}xg \in H\}|$$ to get the result you are asking for.

• The relation you mention is quite interesting, however you formulate it in a manner which presupposes finiteness on many of the objects involved (specifically that of $\Omega$ and of $\left(G \colon \mathrm{C}_G(x)\right)$. I have managed to obtain a proof of its validity in a slightly more general setting -- that of arbitrary group $\Gamma$ acting transitively on arbitrary set $A$ with fixed element $\alpha \in \Gamma$ such that at least one of the orders $\left \lvert \mathrm{C}_{\Gamma}(\alpha)\right \rvert$ or $\lvert H \rvert$ is strictly less than $\lvert \Gamma \rvert$.
– ΑΘΩ
Commented Nov 16, 2022 at 13:03
• (cont.) And interestingly enough, the proof uses the unconditionally general result presented in my answer above. Would you happen to be aware of a proof not relying on additional cardinal assumptions for the result you mention?
– ΑΘΩ
Commented Nov 16, 2022 at 13:07
• Yes, I think it is true in general that $x^G \times \Omega^x$ is in bijection with $\Omega \times (x^G \cap H)$, and it follows as in the finite case.
– spin
Commented Nov 16, 2022 at 14:28
• For conjugate $x,y$ you have $|\Omega^x| = |\Omega^y|$, so $x^G \times \Omega^x$ is in bijection with $$\bigcup_{g \in x^G} \{g\} \times \Omega^g.$$ On the other hand, this union is equal to $$\bigcup_{\beta \in \Omega} (x^G \cap G_{\beta}) \times \{ \beta\}.$$ By transitivity all stabilizers $G_{\beta}$, $G_{\beta'}$ are $G$-conjugate, so $|x^G \cap G_{\beta}| = |x^G \cap H|$ for all $\beta$. Hence $\cup_{\beta \in \Omega} (x^G \cap G_{\beta}) \times \{ \beta\}$ is in bijection with $\Omega \times (x^G \cap H)$.
– spin
Commented Nov 16, 2022 at 14:33
• Thank you for clarifying the situation. Indeed, a very nice double counting argument!
– ΑΘΩ
Commented Nov 17, 2022 at 13:07

There is a very elegant way to prove a very general statement, in relation to which your problem above occurs as a particular case. Consider an arbitrary group $$\Gamma$$ (no finiteness required) with a left action $$\cdot\ \colon \Gamma \times A \rightarrow A$$ on an arbitrary set $$A$$. Let us adopt the following syntactic (notational) conventions:

• given arbitrary subset $$\Lambda \subseteq \Gamma$$ of operators and arbitrary $$X \subseteq A$$, we write: $$\Lambda X \stackrel{\textrm{def}}{=} \cdot \left[\Lambda \times X\right]=\left\{\lambda x\right\}_{\substack{\lambda \in \Lambda\\x \in X}}$$ to denote the external product of the two subsets (external as it stems from an external operation, the action itself). When either of the two subsets in question is a singleton, we also adopt the abbreviations: $$\lambda X\stackrel{\textrm{def}}{=}\{\lambda\}X$$ for any $$\lambda \in \Gamma$$ and $$X \subseteq A$$ respectively $$\Lambda x \stackrel{\textrm{def}}{=}\Lambda \{x\}$$ for any $$\Lambda \subseteq \Gamma$$ and $$x \in A$$.
• given arbitrary subsets $$X, Y \subseteq A$$ we write: $$\left(Y \colon X\right)_\Gamma \stackrel{\textrm{def}}{=} \left\{\lambda \in \Gamma \mid \lambda X \subseteq Y\right\}$$ to refer to the quotient between the two subsets (the subscript $$\Gamma$$ serves to specify the action with respect to which the quotient is considered). As in the previous convention, we also adopt the abbreviations: $$\left(Y \colon x\right)_{\Gamma} \stackrel{\textrm{def}}{=} \left(Y \colon \{x\}\right)_{\Gamma}$$ for arbitrary $$Y \subseteq A$$ and $$x \in A$$ respectively $$\left(y\ \colon X\right)_{\Gamma} \stackrel{\textrm{def}}{=}\left(\{y\}\colon X\right)_{\Gamma}$$ for arbitrary $$X \subseteq A$$ and $$y \in A$$.
• given arbitrary binary relation $$R \in \mathscr{Rel}(A)$$ on $$A$$ together with arbitrary subset $$M \subseteq A$$, we write $$R_{|M} \stackrel{\textrm{def}}{=} R \cap \left(M \times M\right)$$ for the restriction to $$M$$ of relation $$R$$; if $$R \in \mathscr{Eq}(A)$$ is an equivalence on $$A$$ then the restriction $$R_{|M} \in \mathscr{Eq}(M)$$ is correspondingly an equivalence on $$M$$. Furthermore, in the particular case when $$M$$ is saturated with respect to equivalence $$R$$ - which means that for any given $$x \in M$$ we automatically have $$R\langle x \rangle \subseteq M$$, in other words together with any element $$M$$ contains (includes) the entire $$R$$-class of that element - the following relation holds: $$M/R_{|M}=\left\{X \in A/R \mid X \subseteq M\right\}=A/R \cap \mathscr{P}(M).$$ This relation means that the $$R_{|M}$$-classes of $$M$$ are none other than the $$R$$-classes (in the ambient set $$A$$) included in $$M$$.
• for arbitrary set $$M$$ we write $$\Delta_M$$ for its diagonal and given an arbitrary map $$f \colon A \to B$$ we denote its canonical equivalence by $$\mathrm{Eq}(f)\stackrel{\textrm{def}}{=}\left(f \times f\right)^{-1}\left[\Delta_B\right]$$; this equivalence relation can also be described by the (propositional) equivalence: $$\left(\forall x, y\right)\left(x\mathrm{Eq}(f)y \Leftrightarrow x, y \in A \wedge f(x)=f(y)\right).$$
• given arbitrary group $$F$$ and subgroup $$H \leqslant F$$, we shall write $$\mathrm{Cg}^F_{\mathrm{s}}(H)$$ for the left congruence on $$F$$ modulo $$H$$; this is the equivalence relation described by the (propositional) equivalence: $$x\mathrm{Cg}^F_{\mathrm{s}}(H)y \Leftrightarrow x, y \in G \wedge x^{-1}y \in H.$$ We also write $$\left(F/H\right)_{\mathrm{s}}\stackrel{\textrm{def}}{=}F/\mathrm{Cg}^F_{\mathrm{s}}(H)$$ to denote the quotient set consisting of all left congruence classes with respect to $$H$$ (in case anyone might be wondering about the subscript $$\mathrm{s}$$, it is the initial of the word left in both Latin as well as in my mother tongue).

Returning to the general setting of a given left action of $$\Gamma$$ on $$A$$, let us fix $$x \in A$$ and $$X \subseteq A$$. We shall establish the following:

Proposition. The cardinal relation $$\left\lvert \left(X \colon x\right)_{\Gamma}\right\rvert=\left\lvert \Gamma x \cap X\right\rvert \left\lvert \mathrm{Stab}_{\Gamma}(x)\right\rvert$$ holds, where the notation $$\mathrm{Stab}$$ is the traditional one for stabilisers of elements.

Proof : for simplicity let us introduce the notations $$\Lambda \stackrel{\textrm{def}}{=} \left(X \colon x\right)_{\Gamma}$$, $$H \stackrel{\textrm{def}}{=}\mathrm{Stab}_{\Gamma}(x)$$ and $$R \stackrel{\textrm{def}}{=} \mathrm{Cg}_{\mathrm{s}}^{\Gamma}(H)$$. It is clear that for any $$\lambda \in \Lambda$$ it is the case that $$\lambda x \in \Gamma x \cap X$$, which enables us to consider the following map: \begin{align} \varphi \colon \Lambda &\to \Gamma x \cap X \\ \varphi(\lambda)&=\lambda x. \end{align} The following are immediate observations:

1. $$\varphi$$ is surjective
2. the canonical equivalence of $$\varphi$$ is the restriction to $$\Lambda$$ of the left congruence modulo $$H$$, formally speaking $$\mathrm{Eq}\left(\varphi\right)=R_{|\Lambda}$$
3. $$\Lambda$$ is saturated with respect to $$R$$, which means explicitly that given any element $$\lambda \in \Lambda$$ we have $$\lambda H \subseteq \Lambda$$, i.e the entire left $$H$$-class of $$\lambda$$ (with respect to $$R$$) is included in $$\Lambda$$.

By virtue of observation 1) together with the fundamental theorem for quotient maps in set theory, we infer that: $$\left\lvert \Gamma x \cap X \right\rvert=\left\lvert \Lambda/\mathrm{Eq}\left(\varphi\right) \right\rvert. \tag{i}$$ By virtue of observation 2) and the above we gather that: $$\left\lvert \Gamma x \cap X \right\rvert=\left\lvert \Lambda/R_{|\Lambda} \right\rvert. \tag{ii}$$ By virtue of observation 3) together with the remark concerning saturated subsets made in one of the above paragraphs, it follows that $$\Lambda/R_{|\Lambda} \subseteq \left(\Gamma/H\right)_{\mathrm{s}}$$. This entails the fact that: $$\left(\forall \Theta\right)\left(\Theta \in \Lambda/R_{|\Lambda} \Rightarrow \left\lvert \Theta \right\rvert=\left\lvert H\right\rvert\right), \tag{iii}$$ in other words all the equivalence classes in the quotient set $$\Lambda/R_{|\Lambda}$$ have cardinality equal to that of $$H$$, since they are all left $$H$$-congruence classes of $$\Gamma$$. Taking into account relations $$\textrm{(i)-(iii)}$$ above we can produce the following relation: $$\left\lvert \Lambda \right\rvert=\left\lvert \bigcup \Lambda/R_{|\Lambda} \right\rvert=\sum_{\Theta \in \Lambda/R_{|\Lambda}}\left\lvert \Theta \right\rvert=\sum_{\Theta \in \Lambda/R_{|\Lambda}} \left\lvert H \right\rvert=\left\lvert \Lambda/R_{|\Lambda} \right\rvert \left\lvert H \right\rvert=\left\lvert \Gamma x \cap X \right\rvert \left\lvert H \right\rvert,$$ none other than the relation we had set out to prove. $$\Box$$

This general proposition has an obvious analogue referring to the dual case of right actions. Your particular formulation of the problem is the special case when $$\Gamma=A$$ and $$\Gamma$$ acts on itself by left (or right) conjugation. The subgroup $$H$$ is in the role of the generic subset $$X \subseteq \Gamma$$ in my formulation above (at this point it is worthy of remark that the particular subset $$H$$ - in your notation - being a subgroup plays absolutely no part in the proof) and the stabiliser of any $$x \in \Gamma$$ with respect to conjugation (whether left or right, regardless) is precisely the centraliser $$\mathrm{C}_{\Gamma}(x)$$.

To make one more technical remark, the subset featured in the left-hand term of your formulation of the cardinal relation above is - as you define it - the quotient between $$X$$ and $$x$$ with respect to right conjugation, so an application of the right-action version of the general proposition is most straight-forward.