TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-11

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Fundamental theorem of finite abelian groups: invariant-factor form

Statement

For every finite abelian group $G$ there is a unique list $1 < n_{1} ∣ \dots ∣ n_{r}$ such that $G ≅ C_{n_{1}} \times \dots \times C_{n_{r}}$ . Moreover $∣ G ∣ = n_{1} \dots n_{r}$ . The trivial group corresponds to the empty list and empty product.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Every finite abelian group is isomorphic to a finite direct product of cyclic groups of prime-power order. The multiset of their orders is uniquely determined by the group, up to permutation of the factors. (Fundamental theorem of finite abelian groups: elementary-divisor form).

[L2]

Every multiset of prime-power elementary divisors regroups in exactly one way into an invariant-factor list. (Elementary divisors regroup uniquely into invariant factors).

[L3]

An invariant-factor list for a finite abelian group $G$ is a finite list of integers $1 < n_{1} ∣ n_{2} ∣ \dots ∣ n_{r}$ together with an isomorphism $G ≅ C_{n_{1}} \times \dots \times C_{n_{r}}$ . The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. Unit factors are omitted. The trivial group has the empty list. (Invariant-factor data for a finite abelian group).

[L4]

If $G$ and $H$ are finite groups, then their external direct product is finite and has order $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ . (For finite groups $G$ and $H$ , $∣ G \times H ∣ = ∣ G ∣ ∣ H ∣$ ).

Proof

technique · direct

1.1givenL1L2L3L4

The elementary-divisor theorem supplies a unique multiset of prime powers, and the regrouping lemma converts it into a unique invariant-factor list.

2.1step 1.1∎

The order formula for finite direct products gives $∣ G ∣ = \prod_{i} n_{i}$ ; for the empty list this product is $1$ , the order of the trivial group.

Fundamental theorem of finite abelian groups: invariant-factor form

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources