Same conjugacy implies same isotropy?

Question

I'm struggling at the class formula in Serge Lang's algebra. The formula is a consequence of $$|S| = \sum_{i} |Gs_i| = \sum_{i} (G : G_{s_i})$$ Here, $G$ is some group that acts on $S$. Since the set of all orbits of $S$ under $G$ form partition, the first equality holds. The second holds simply because order of a orbit of $s_i$ is equal to the index of its isotropy group.

What I don't get is with the specific case where $G$ is a group and $S$ is $G$ itself and we consider conjugation on $G$. Then, $$|G| = \sum_{s \in C} (G : G_{s})$$ where- $s$ is some distinct elements in conjugate class. He has not mentioned this concept so I searched it up:

Given $G$ as a group, $a$ and $b$ are conjugate if and only if exists $g \in G$ such that $gag^{-1} = b$.

I guess he basically assume that "same conjugacy, same isotropy". My attempt: Let us prove that $x \sim y$ implies that $G_x = G_y$. We start with $$\exists g \in G: gxg^{-1} = y$$ Then let $z \in G_y$(proving $G_y \subseteq G_x$). We have $$zyz^{-1} = y \Rightarrow z(gxg^{-1})z^{-1} = gxg^{-1}$$ Then what? How is $zxz^{-1} = x$ true?

Answer 1

This is wrong and irrelevant. And Lang does not assume that "same conjugacy, same isotropy."

(They are called conjugacy classes, not "conjugate classes")

The correct statement is that if $x$ and $y$ are conjugate, then the isotropy groups are also conjugate. Explicitly:

Let $G$ act on $X$, and let $x,y\in X$. If $g\cdot x=y$, then $G_y=gG_xg^{-1}$.

(In fact, Lang proves this; bottom of page 27 in the Addison-Wesley 3rd Edition, same page and location in the Springer-Verlag revised 3rd Edition).

Indeed, note that if $g\cdot x=y$, then $x=g^{-1}\cdot y$. Now let $h\in G_x$. Then $ghg^{-1}\in G_y$, since $$\begin{align*} (ghg^{-1})\cdot y &= g\cdot(h\cdot(g^{-1}\cdot y))\\ &= g\cdot(h\cdot x)\\ &= g\cdot x\\ &= y. \end{align*}$$ Thus, $gG_xg^{-1}\subseteq G_y$. Applying this again swapping $x$ and $y$ and using $g^{-1}\cdot y=x$, we get $g^{-1}G_yg\subseteq G_x$. Multiplying on the left by $g$ and on the right by $g^{-1}$ we get $G_y\subseteq gG_xg^{-1}$, so we get the desired equality.

Applying this to the action by conjugation we get that if $x$ and $y$ are conjugate, then their isotropy groups are conjugate, not equal. So their indices are equal.

None of this is actually relevant, though. You have $G$ partitioned into orbits, say the orbits of the representatives $s\in C$. That is, the set $C$ contains one and only one element from each orbit of the action. By the Orbit-Stabilizer Theorem, the size of the orbit of $s$ equals the index of its stabilizer/isotropy group. The orbits are the equivalence classes under the relation of conjugation, i.e., the conjugacy classes. So the size of $G$ equals the sum of the sizes of the orbits, which equals the sum of the indices of the isotropy groups of the selected orbit representatives.

I repeat what I and others have already told you. Lang is not a good book to try to learn the material from, and you should be looking at a better book that is meant to be a textbook, not a reference. Lang is more a reference than a textbook,

Answer 2

In general, any $G$-action on any set $S$ induces an equivalence relation$^1$ on $S$ via: $$t\sim s\iff \exists g\in G\mid t=g\cdot s$$ The equivalence class of $s\in S$ is then: \begin{alignat}{1} [s] &= \{t\in S\mid t\sim s\} \\ &= \{t\in S\mid t=g\cdot s,\text{ for some }g\in G\} \\ &= \{g\cdot s\mid g\in G\} \\ &=: Gs \end{alignat} referred to also as "the orbit of $s$". The isotropy subgroup of $s$, referred to also as "the stabilizer of $s$", is defined by: $$G_s=\{g\in G\mid g\cdot s=s\}$$ Lemma. For every $s\in S$, there is a bijection, $\varphi$, between the left quotient set $G/G_s$ and $Gs$, defined by: $$\varphi(gG_s)=g\cdot s$$ Proof:

• good definition: let $h\in gG_s$; then, $g^{-1}h\cdot s=s$, hence: $\varphi(hG_s)=$ $h\cdot s=$ $g\cdot s=$ $\varphi(gG_s)$;
• surjectivity: by definition of $Gs$, for every $t\in Gs$, there is some $g\in G$ such that $t=g\cdot s=$ $\varphi(gG_s)$;
• injectivity: we get $\varphi(gG_s)=\varphi(g'G_s)\Longrightarrow$ $g\cdot s=g'\cdot s\Longrightarrow$ $s=g^{-1}\cdot(g'\cdot s)\Longrightarrow$ $s=(g^{-1}g')\cdot s\Longrightarrow$ $g^{-1}g'\in G_s\Longrightarrow$ $g'\in gG_s\Longrightarrow$ $g'G_s=gG_s$.

$\Box$

Note that, if $t\in Gs$, then $t=g\cdot s$ for some $g\in G$; then: \begin{alignat}{1} G_t &=\{g'\in G\mid g'\cdot t=t\} \\ &= \{g'\in G\mid g'\cdot (g\cdot s)=g\cdot s\} \\ &= \{g'\in G\mid (g^{-1}g'g)\cdot s=s\} \\ &\stackrel{(g'':=g^{-1}g'g)}{=}\{gg''g^{-1}\in G\mid g''\cdot s=s\} \\ &= g\{g''\in G\mid g''\cdot s=s\}g^{-1} \\ &= gG_sg^{-1} \\ \end{alignat} namely: the stabilizers of two points in one same orbit are conjugate each other via the element of $G$ "bringing one point to the other". So, in general, they are nor equal, and hence your from "We start with" onwards is wrong.

Now back to your "struggling with the class formula". By the Lemma, in particular, if $S$ is finite and $R=\{s_i, i=1,\dots,k\}$ is a complete set of orbits' representatives, we get: $$|S|=\sum_{i=1}^k|Gs_i|=\sum_{i=1}^k[G:G_{s_i}]$$ If $G$ is finite either, then: $$|S|=\sum_{i=1}^k|Gs_i|=\sum_{i=1}^k\frac{|G|}{|G_{s_i}|}\tag1$$ When $S=G$, there is the special case of the action by conjugation, where:

• $Gs$ is the conjugacy class of $s\in G$;
• $G_s$ is $C_G(s)$, the centralizer of $s\in G$ in $G$, and
• if further $G(=S)$ is finite, $(1)$ yields the Class Equation: $$|G|=|Z(G)|+\sum_{\text{noncentral reps }g_i}\frac{|G|}{|C_G(g_i)|}$$ being each $z\in Z(G)$ lonely in its singleton conjugacy class.

---

$^1$The proof uses the existence of the identity and the inverses in a group, and group action definition.

Answer 3

As @art and @kan't point out, the even more general fact is that elements in the same orbit (which in the case of the conjugation action are conjugate elements) have conjugate isotropy groups.

Now just for emphasis, conjugate subgroups are certainly not in general the same. Consider for example different Sylow subgroups, which are always conjugate.

Conjugate groups are always isomorphic (conjugation is an isomorphism), but can be distinct, and often are.