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

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Orbit-stabiliser cardinality: $|G\cdot x|=[G:G_x]$ whenever either side is finite, and $|G|=|G_x|\,|G\cdot x|$ for finite $G$

Statement

For an action of $G$ on $X$ and $x\in X$ ,

$|G\cdot x|=[G:G_x]$

whenever either side is finite. In particular, if $G$ is finite, then

$|G|=|G_x|\,|G\cdot x|.$

Facts & Assumptions

Given: A left action of $G$ on $X$ and a point $x\in X$ .

[L1]

The map $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a bijection (Orbit-stabiliser: $G/G_x\to G\cdot x$ , $gG_x\mapsto g\cdot x$ , is a well-defined bijection).

[L2]

The index is the finite cardinality $[G:H]=|G/H|$ when the coset set is finite (The coset set $G/H$ and the index $[G:H]$ of a subgroup).

[L3]

Finite cardinality is preserved by a bijection (The cardinality $\lvert A\rvert$ of a finite set).

[L4]

If $G$ is finite and $H\le G$ , then $|G|=[G:H]|H|$ (Lagrange's theorem: $|G|=[G:H]|H|$ for every subgroup $H$ of a finite group $G$ ).

Proof

technique · direct

1.1

By [L1], the sets $G/G_x$ and $G\cdot x$ are bijective; [L2] and [L3] therefore give $|G\cdot x|=|G/G_x|=[G:G_x]$ whenever they are finite.

L1L2L3

2.1

If $G$ is finite, [L4] applied to $G_x\le G$ gives $|G|=[G:G_x]|G_x|=|G\cdot x|\,|G_x|$ .

step 1.1L4algebra∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 68 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

T. W. Judson, Abstract Algebra: Theory and Applications, 14.1, Theorem 14.11 (standard reference, not scraped)

Orbit-stabiliser cardinality: ∣G⋅x∣=[G:Gx]|G\cdot x|=[G:G_x]∣G⋅x∣=[G:Gx​] whenever either side is finite, and ∣G∣=∣Gx∣ ∣G⋅x∣|G|=|G_x|\,|G\cdot x|∣G∣=∣Gx​∣∣G⋅x∣ for finite GGG