CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

$[G : H] = 1$ if and only if $H = G$

Statement

For any subgroup $H \leq G$ , finite or infinite,

$[G : H] = 1 ⟺ H = G .$

Facts & Assumptions

Given: A group $G$ and a subgroup $H \leq G$ .

[F1]

The coset set is $G / H = {g H : g \in G}$ . Its index is its finite cardinality when the coset set is finite and is the non-natural symbol $\infty$ otherwise; hence $[G : H] = 1$ says that $G / H$ is finite of cardinality $1$ (The coset set $G / H$ and the index $[G : H]$ of a subgroup, Left and right cosets $g H$ and $H g$ of a subgroup).

[L1]

The cosets partition $G$ , and $g H = H$ exactly when $g \in H$ (The left cosets of a subgroup partition the group, $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

[F2]

A finite set has cardinality $1$ exactly when it is a singleton: a bijection to $1 = {0}$ has one fibre and hence one element, while the unique map from a singleton to $1$ is a bijection (The cardinality $∣ A ∣$ of a finite set).

Proof

technique · direct

1.1

If $H = G$ , then every coset $g H$ equals $G$ , so $G / H = {G}$ and $[G : H] = 1$ .

givenF1F2

1.2

Conversely, if $[G : H] = 1$ , then $G / H$ is the singleton containing $H = e H$ . Thus $g H = H$ for every $g \in G$ , and [L1] gives $g \in H$ . Hence $G \subseteq H$ .

givenF1F2L1

2.1

Since always $H \subseteq G$ , step 1.2 gives $H = G$ ; together with step 1.1 this proves the equivalence.

step 1.1step 1.2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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

Thomas W. Judson, Abstract Algebra: Theory and Applications, Cosets and Lagrange's Theorem (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.1: Cosets (standard reference, not scraped)
Thomas W. Judson, Abstract Algebra: Theory and Applications, §6.2: Lagrange's Theorem (standard reference, not scraped)

[G:H]=1[G:H]=1 if and only if H=GH=G