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TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11
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Every nontrivial finite abelian group is an internal direct product of indecomposable subgroups

Statement

Every nontrivial finite abelian group is an internal direct product of finitely many indecomposable subgroups.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

A nontrivial finite abelian group is indecomposable if it is not an internal direct product of two nontrivial subgroups in the sense of def-internal-direct-product-of-subgroups. It is decomposable if such a product exists. The trivial group is assigned neither label. (Indecomposable and decomposable nontrivial finite abelian groups).

[L2]

Let PP be a property of naturals such that for every nNn \in \mathbb{N}, if P(m)P(m) holds for all m<nm < n then P(n)P(n). Then P(n)P(n) holds for all nNn \in \mathbb{N}. (At n=0n = 0 the hypothesis is vacuous, so P(0)P(0) is forced.) (Strong (complete) induction).

[L3]

Let N0,,Nr1GN_0,\ldots,N_{r-1}\trianglelefteq G. The following are equivalent: the NiN_i form an internal direct product of GG; every gGg\in G has a unique expression g=n0nr1g=n_0\cdots n_{r-1} with niNin_i\in N_i; and the multiplication map μ:i<rNiG\mu:\prod_{i<r}N_i\to G is an isomorphism. These statements include the empty family and the one-factor case. (Internal direct products are external direct products, equivalently every element has a unique factorisation).

[L4]

Let GG be a finite group and HGH\le G. Then G=[G:H]H.|G|=[G:H]\,|H|. Consequently, under the canonical embedding ι:NZ\iota:\mathbb N\to\mathbb Z, H|H| divides G|G|. (Lagrange's theorem: G=[G:H]H|G|=[G:H]|H| for every subgroup HH of a finite group GG).

Proof

technique · induction
1.1

For strong induction on G|G|, the order-one case is vacuous because the only group of that order is trivial.

basegivenL1L2L3L4
2.1

Fix a nontrivial GG and assume the result for every nontrivial finite abelian group of smaller order. If GG is indecomposable, the one-factor product is the required decomposition.

ihstep 1.1
3.1

If GG is decomposable, write G=BCG=B\oplus C with BB and CC nontrivial. Lagrange gives B,C<G|B|,|C|<|G|, so the induction hypothesis decomposes each into indecomposable factors.

step 2.1
4.1

Unique factorisation in BCB\oplus C and in the two inductive products combines to unique factorisation by all the smaller factors; internal-product recognition completes the induction.

step 3.1discharge-induction

Depends on

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Sources