CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$

Statement

Let $N ⊴ G$ . If $[G : N]$ is finite, then the quotient group $G / N$ is finite and

$∣ G / N ∣ = [G : N] .$

In particular, if $G$ is finite, then

$∣ G / N ∣ = \frac{∣ G ∣}{∣ N ∣} .$

Facts & Assumptions

Given: A group $G$ and a normal subgroup $N ⊴ G$ .

[F1]

The index $[G : N]$ is the finite cardinality of the left-coset set $G / N$ when that set is finite (The coset set $G / H$ and the index $[G : H]$ of a subgroup).

[L1]

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

[F2]

The order $∣ K ∣$ of a finite group $K$ is the cardinality of its underlying set (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

[L2]

The quotient group has the left cosets of $N$ as its underlying set (For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

Proof

technique · direct

1.1F1F2L2

If $[G : N]$ is finite, then by [F1] the coset set underlying $G / N$ is finite with cardinality $[G : N]$ ; hence [F2] and [L2] give $∣ G / N ∣ = [G : N]$ .

2.1step 1.1L1algebra

If $G$ is finite, then [L1] gives $∣ G ∣ = [G : N] ∣ N ∣$ . Since $N$ contains the identity, $∣ N ∣ \neq 0$ , and step 1.1 yields $∣ G / N ∣ = [G : N] = ∣ G ∣ / ∣ N ∣$ .

3.1step 1.1step 2.1∎

The two asserted formulas follow.

Depends on

Dependency tree · 2 levels

Sources

T. W. Judson, Abstract Algebra: Theory and Applications, Factor Groups and Normal Subgroups