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

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Fundamental theorem of finite abelian groups: elementary-divisor form

Statement

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.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

If $G$ is finite abelian and $∣ G ∣ = \prod_{i < r} p_{i}^{a_{i}}$ is its prime factorisation, then the subgroups $G (p_{i})$ form an internal direct product of $G$ . Thus $G ≅ \prod_{i < r} G (p_{i}) .$ For the trivial group, this is the empty product. (A finite abelian group is the internal direct product of its primary components).

[L2]

Every finite abelian $p$ -group is isomorphic to a finite direct product of cyclic groups of prime-power order. The trivial $p$ -group is the empty product. (Every finite abelian p-group is a direct product of cyclic p-groups).

[L3]

An elementary-divisor decomposition of a finite abelian group $G$ is an isomorphism $G ≅ C_{q_{0}} \times \dots \times C_{q_{r - 1}},$ where every $q_{i} > 1$ is a prime power. The unordered multiset of the $q_{i}$ , counted with multiplicity, is the elementary-divisor data. The cyclic factors and product use thm-classification-of-cyclic-groups and def-external-direct-product-of-groups. The data records factor isomorphism types, not distinguished internal subgroups; the trivial group has empty data. (Elementary-divisor data for a finite abelian group).

[L4]

Suppose $G ≅ \prod_{j < r} C_{p^{e_{j}}}$ with $e_{j} \geq 1$ , and in additive notation write $p^{i} G = {p^{i} g : g \in G}$ . Define $d_{i}$ by $∣ p^{i} G / p^{i + 1} G ∣ = p^{d_{i}}$ . Then $d_{i} = ∣ {j : e_{j} \geq i + 1} ∣ .$ Consequently, for every $k \geq 1$ , the number of summands of order $p^{k}$ is $d_{k - 1} - d_{k}$ , so the elementary divisors are intrinsic. (The successive quotients p^iG/p^{i+1}G recover the cyclic summand multiplicities of a finite abelian p-group).

[L5]

If $G = ⟨ g ⟩$ is cyclic, then exactly one of the following applies: - if $g$ has infinite order, $G ≅ (Z, +)$ ; - if $g$ has finite order $n$ , necessarily $n \geq 1$ , then $G ≅ (Z / n, +)$ . (Every cyclic group is isomorphic to $(Z, +)$ or to $(Z / n, +)$ for its finite order $n \geq 1$ ).

Proof

technique · direct

1.1givenL1L2L3L4L5

Primary decomposition separates $G$ into its intrinsic $p$ -primary components, and cyclic decomposition expresses each component as a product of cyclic $p$ -groups. This proves existence.

2.1step 1.1

For a fixed prime $p$ , the successive quotients $p^{i} G (p) / p^{i + 1} G (p)$ recover the multiplicity of every cyclic order $p^{k}$ .

3.1step 2.1∎

Doing this independently for each prime proves uniqueness of the multiset of elementary divisors. The assertion concerns factor isomorphism types, not uniqueness of internal complements.

Fundamental theorem of finite abelian groups: elementary-divisor form

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources