LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-01

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Compatible tuples form a subgroup of the product group

Statement

The compatible tuples in an inverse system form a subgroup of the full product group.

Facts & Assumptions

Given: An inverse system of groups indexed by a directed set $I$ .

[L1]

The inverse limit is the set of tuples satisfying $φ_{i j} (g_{j}) = g_{i}$ for every comparable pair $i \leq j$ (The inverse limit is the set of compatible tuples in the Cartesian product).

[F1]

A subset of a group is a subgroup exactly when it contains the identity and is closed under products and inverses (Subgroup, Group and abelian group).

Proof

technique · direct

1.1

The identity tuple $(e_{i})_{i \in I}$ is compatible, because every transition map is a homomorphism and therefore sends $e_{j}$ to $e_{i}$ . So the inverse limit is nonempty.

L1F1given

1.2

If $(g_{i})$ and $(h_{i})$ are compatible, then for every $i \leq j$ one has $φ_{i j} (g_{j} h_{j}) = φ_{i j} (g_{j}) φ_{i j} (h_{j}) = g_{i} h_{i} .$ Likewise $φ_{i j} (g_{j}^{- 1}) = g_{i}^{- 1}$ . Hence coordinatewise products and coordinatewise inverses remain compatible.

L1givenalgebra

2.1

By [F1], step 1.1 and step 1.2 prove that the compatible tuples form a subgroup of the product group.

F1step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

10 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

Brian Osserman, Math 6112 notes on inverse limits and profinite groups (standard reference, not scraped)
H. W. Lenstra, Profinite groups and Galois groups (standard reference, not scraped)