CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13

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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.

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The order of a finite group is the product of the orders of its composition factors

Statement

If $G = G_{0} ▹ \dots ▹ G_{n} = 1$ is a composition series of a finite group, then $∣ G ∣ = \prod_{i = 0}^{n - 1} ∣ G_{i} / G_{i + 1} ∣ .$ For the trivial group, $n = 0$ and the empty product is $1$ .

Facts & Assumptions

Given: A composition series $G = G_{0} ▹ \dots ▹ G_{n} = 1$ of a finite group.

[F1]

The factors of the displayed composition series are $G_{i} / G_{i + 1}$ for $0 \leq i < n$ (Composition series, composition factors, and composition length).

[L1]

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

[L2]

If $N ⊴ K$ and $K$ is finite, then $∣ K / N ∣ = [K : N]$ (If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ ).

Proof

technique · direct

1.1

For every $0 \leq i < n$ , [L1] and [L2] give $∣ G_{i} ∣ = ∣ G_{i} / G_{i + 1} ∣ ∣ G_{i + 1} ∣$ .

givenL1L2

2.1

Multiplying the identities of step 1.1 and cancelling the intermediate positive integers gives $∣ G_{0} ∣ = (\prod_{i = 0}^{n - 1} ∣ G_{i} / G_{i + 1} ∣) ∣ G_{n} ∣$ .

step 1.1algebra

3.1

Since $G_{0} = G$ and $G_{n} = 1$ , one has $∣ G_{n} ∣ = 1$ , proving the formula. When $n = 0$ , step 2.1 reads $∣ 1 ∣ = 1$ by the empty-product convention.

givenstep 2.1F1∎

Depends on

Used by

Two composition series of C₁₂ have the same factors in different orders Example

Dependency tree · two levels

16 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. S. Milne, Group Theory, Chapter 6 (standard reference, not scraped)
K. Conrad, Subgroup Series I (standard reference, not scraped)
K. Igusa, Notes on Jordan-Hölder, section 5 (standard reference, not scraped)