TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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The two-strand braid group is infinite cyclic

Statement

The braid group $B_{2}$ is cyclic, generated by $σ_{1}$ , and $σ_{1}$ has infinite order. Hence $B_{2}$ is an infinite cyclic group.

Facts & Assumptions

Given: The Artin presentation of the braid groups.

[L1]

The group $B_{2}$ is presented by one generator $σ_{1}$ and no braid relations (The braid group by Artin presentation).

[L2]

Free groups are torsion-free (Free groups are torsion-free).

[L3]

A map of generators satisfying the relators extends uniquely from a presented group (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group).

Proof

technique · direct

1.1

By [L1], the presentation of $B_{2}$ has one generator and no defining relator, so every element of $B_{2}$ is a power of $σ_{1}$ . Thus $B_{2}$ is cyclic.

givenL1algebra

2.1

With no relators present, the presented group is the free group on one generator. By [L2], that generator has infinite order, so $σ_{1}^{m} \neq 1$ for every nonzero integer $m$ .

L1L2L3step 1.1algebra

3.1

Therefore the cyclic group $B_{2} = ⟨ σ_{1} ⟩$ is infinite, so it is infinite cyclic.

step 1.1step 2.1∎

Depends on

Used by

The braid category is braided but not symmetric Counterexample
The two-strand braiding in the braid category has infinite order Example
Braided coherence fails in the symmetric form Theorem

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

Michael Muger, Tensor Categories: A Selective Guided Tour, Section 4 (standard reference, not scraped)