CounterexampleConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-03 (gpt-5.6-sol-codex-subscription)

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$2\mathbb{Z}$ has index $2$ in $\mathbb{Z}$ and is nevertheless equinumerous with $\mathbb{Z}$

Statement refuted

If a proper subgroup $H<G$ has finite index, then $H$ cannot be equinumerous with $G$ .

Facts & Assumptions

Given: The additive group $\mathbb Z$ and its subgroup $2\mathbb Z$ .

[F1]

The set $2\mathbb Z$ is a subgroup of $(\mathbb Z,+)$ (Every subgroup of $(\mathbb{Z}, +)$ is $\langle n \rangle = n\mathbb{Z}$ for exactly one natural number $n$ , The integers form a commutative ring).

[F2]

The index is the number of cosets, additive cosets have the form $a+2\mathbb Z$ , and at modulus $2$ every congruence class has exactly one representative $r$ with $0\le r<2$ , hence representative $0$ or $1$ (The coset set $G/H$ and the index $[G:H]$ of a subgroup, Left and right cosets $gH$ and $Hg$ of a subgroup, For $n\ge 1$ , every class in $\mathbb{Z}/n$ has one representative $r$ with $0\le r<n$ , so $\lvert\mathbb{Z}/n\rvert=n$ ; while $\mathbb{Z}/0$ is in bijection with $\mathbb{Z}$ ).

[F3]

Two sets are equinumerous when a bijection between them exists (Equinumerous sets, $A \approx B$ and $A \preceq B$ , Injection, surjection, bijection).

[F4]

Multiplication by a nonzero integer can be cancelled: if $xz=yz$ and $z\ne0$ , then $x=y$ (The integers have no zero divisors; multiplicative cancellation).

Counterexample

technique · direct

1.1

The two cosets are $2\mathbb Z$ and $1+2\mathbb Z$ . Indeed [F2] writes every integer as $a=2q+r$ with $r\in\{0,1\}$ ; then $a+2\mathbb Z=r+2\mathbb Z$ , since $a+2k=r+2(q+k)$ and, conversely, $r+2t=a+2(t-q)$ . Thus every coset is one of the displayed two, and they are distinct because one contains $0$ while the other does not. Hence $[\mathbb Z:2\mathbb Z]=2$ .

F1F2

1.2

The map $f:\mathbb Z\to2\mathbb Z$ given by $f(k)=2k$ is surjective by the definition of $2\mathbb Z$ and injective because $2k=2\ell$ implies $k=\ell$ by cancellation at the nonzero factor $2$ . Hence it is a bijection.

F3F4construct

2.1

Thus the proper subgroup $2\mathbb Z$ has finite index and is equinumerous with $\mathbb Z$ , refuting the statement.

step 1.1step 1.2F1∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 78 results over 20 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)

2Z2\mathbb{Z}2Z has index 222 in Z\mathbb{Z}Z and is nevertheless equinumerous with Z\mathbb{Z}Z