ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

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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.

$\langle a,b\mid aba^{-1}b^{-1}\rangle\cong(\mathbb Z,+)\times(\mathbb Z,+)$

Example

Adjoining a single commutator relator to the free group on two generators presents the direct product of two copies of the additive integers:

$\langle a,b\mid aba^{-1}b^{-1}\rangle\cong(\mathbb Z,+)\times(\mathbb Z,+),$

where $a$ and $b$ correspond to $(1,0)$ and $(0,1)$ .

Facts & Assumptions

Given: The presentation $P=\langle a,b\mid aba^{-1}b^{-1}\rangle$ .

[L1]

A map of generators that sends every relator to the identity extends uniquely to a homomorphism from the presented group (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

[L2]

$(\mathbb Z,+,\cdot,0,1)$ is a commutative ring (The integers form a commutative ring).

[L3]

Integer powers satisfy $g^{m+n}=g^mg^n$ ; powers of commuting elements satisfy $(gh)^n=g^nh^n$ (Exponent laws in a group: $g^{m+n} = g^{m}g^{n}$ and $(g^{m})^{n} = g^{mn}$ for all $m, n \in \mathbb{Z}$ , and $(gh)^{n} = g^{n}h^{n}$ when $g$ and $h$ commute).

Verification

technique · constructive

1.1

Sending $a$ to $(1,0)$ and $b$ to $(0,1)$ kills the commutator in the abelian direct product, so [L1] constructs a homomorphism $\pi:P\to\mathbb Z\times\mathbb Z$ .

L1L2construct

1.2

The relator gives $ab=ba$ , so group algebra and [L3] move all powers of $a$ before all powers of $b$ and write every element of $P$ as $a^mb^n$ for integers $m,n$ .

L3given

2.1

The map $\pi$ sends $a^mb^n$ to $(m,n)$ by [L2], so it is surjective and step 1.2 shows that its kernel is trivial: an element mapping to $(0,0)$ has the form $a^0b^0=e$ .

L2step 1.1step 1.2

3.1

Therefore $\pi$ is the claimed isomorphism $P\cong\mathbb Z\times\mathbb Z$ .

step 2.1discharge-construct∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 74 results over 22 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

Nicholas Touikan, An Introduction to Combinatorial and Geometric Group Theory, Exercises §1.6 (standard reference, not scraped)
John McKernan, Presentations and Groups of Small Order, Lecture 12 (standard reference, not scraped)

⟨a,b∣aba−1b−1⟩≅(Z,+)×(Z,+)\langle a,b\mid aba^{-1}b^{-1}\rangle\cong(\mathbb Z,+)\times(\mathbb Z,+)⟨a,b∣aba−1b−1⟩≅(Z,+)×(Z,+)