TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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$G/C_G(x)\to\operatorname{Cl}_G(x)$ is a bijection, so $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ whenever these cardinalities are finite

Statement

For a group $G$ and $x\in G$ , the map

$G/C_G(x)\longrightarrow\operatorname{Cl}_G(x),\qquad gC_G(x)\longmapsto gxg^{-1},$

is a well-defined bijection. Consequently

$|\operatorname{Cl}_G(x)|=[G:C_G(x)]$

whenever these cardinalities are finite, in particular when $G$ is finite.

Facts & Assumptions

Given: A group $G$ and an element $x\in G$ .

[L1]

Orbit-stabiliser gives a bijection from the cosets of a point stabilizer to its orbit (Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a well-defined bijection).

[L2]

The finite cardinality of an orbit is the index of its stabilizer (Orbit-stabiliser cardinality: $|G\cdot x|=[G:G_x]$ whenever either side is finite, and $|G|=|G_x|\,|G\cdot x|$ for finite $G$ ).

[L3]

The conjugacy class is $\operatorname{Cl}_G(x)=\{gxg^{-1}:g\in G\}$ and the centralizer is $C_G(x)=\{g:gxg^{-1}=x\}$ (The conjugacy class $\operatorname{Cl}_G(x)$ and centralizer $C_G(x)$ of an element).

[L4]

The centralizer $C_G(x)$ is a subgroup of $G$ ( $C_G(x)$ and $N_G(H)$ are subgroups of $G$ ).

[L5]

The maps $x\mapsto gxg^{-1}$ form the conjugation homomorphism (The map $g\mapsto(x\mapsto gxg^{-1})$ is a homomorphism $G\to\operatorname{Aut}(G)$ with kernel $Z(G)$ and image $\operatorname{Inn}(G)$ ).

[L6]

A homomorphism into a symmetric group defines a group action (Actions of $G$ on $X$ correspond exactly to homomorphisms $G\to\operatorname{Sym}(X)$ ).

Proof

technique · direct

1.1

By [L5] and [L6], $G$ acts on itself by conjugation. By [L3], the orbit of $x$ is $\operatorname{Cl}_G(x)$ and its stabilizer is $C_G(x)$ , which is a subgroup by [L4].

L3L4L5L6

2.1

Applying [L1] to this action gives the displayed well-defined bijection $gC_G(x)\mapsto gxg^{-1}$ .

step 1.1L1

3.1

Applying [L2] to the same orbit gives $|\operatorname{Cl}_G(x)|=[G:C_G(x)]$ whenever finite.

step 1.1step 2.1L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 62 results over 15 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

P. Brosnan, Undergraduate Algebra Notes, 3.14: G-Sets, Corollary 3.109 (standard reference, not scraped)
T. W. Judson, Abstract Algebra: Theory and Applications, 14.2 (standard reference, not scraped)

G/CG(x)→Cl⁡G(x)G/C_G(x)\to\operatorname{Cl}_G(x)G/CG​(x)→ClG​(x) is a bijection, so ∣Cl⁡G(x)∣=[G:CG(x)]|\operatorname{Cl}_G(x)|=[G:C_G(x)]∣ClG​(x)∣=[G:CG​(x)] whenever these cardinalities are finite