Proof that the Braid Group is Torsion-Free

The braid group $B_n$ has no elements of finite order, a remarkable property distinguishing it from the symmetric group $S_n$ (which is finite). This result has deep consequences for the algebraic structure of braid groups and their faithful representations.

Statement

Theorem5.3Braid Groups are Torsion-Free

The braid group $B_n$ is torsion-free: if $\beta \in B_n$ and $\beta^k = 1$ for some positive integer $k$ , then $\beta = 1$ .

Proof

We present the proof via the orderability of the braid group, following Dehornoy's approach.

Step 1: Left-orderability. A group $G$ is left-orderable if there exists a strict total order $<$ on $G$ such that $\alpha < \beta$ implies $\gamma\alpha < \gamma\beta$ for all $\gamma \in G$ . We will show $B_n$ is left-orderable.

Step 2: The Dehornoy order. Define a braid $\beta \in B_n$ to be $\sigma_i$ -positive if it can be written as a word in $\sigma_1^{\pm 1},\ldots,\sigma_n^{\pm 1}$ such that the generator $\sigma_i$ appears but $\sigma_i^{-1}$ does not (though $\sigma_j^{\pm 1}$ for $j \neq i$ may appear freely). A braid is Dehornoy-positive ( $\beta > 1$ ) if it is $\sigma_i$ -positive for some $i$ .

The key properties (whose proofs require substantial combinatorial arguments using the braid group's Garside structure or its action on a free group):

(Trichotomy) For every $\beta \in B_n$ , exactly one holds: $\beta$ is Dehornoy-positive, $\beta = 1$ , or $\beta^{-1}$ is Dehornoy-positive.
(Closure under multiplication) If $\alpha$ and $\beta$ are both Dehornoy-positive, then $\alpha\beta$ is Dehornoy-positive.

Step 3: Left-invariant order. Define $\alpha < \beta$ if $\alpha^{-1}\beta$ is Dehornoy-positive. Trichotomy ensures this is a total order. Closure under multiplication ensures left-invariance: if $\alpha < \beta$ , then $\gamma\alpha < \gamma\beta$ (since $(\gamma\alpha)^{-1}(\gamma\beta) = \alpha^{-1}\beta$ is Dehornoy-positive).

Step 4: Left-orderable implies torsion-free. Suppose $\beta^k = 1$ for some $\beta \neq 1$ and $k \geq 2$ . By trichotomy, either $\beta > 1$ or $\beta < 1$ .

Case 1: $\beta > 1$ . By left-invariance: $\beta^2 = \beta\cdot\beta > \beta\cdot 1 = \beta > 1$ . By induction: $1 < \beta < \beta^2 < \cdots < \beta^k$ . But $\beta^k = 1$ , giving $1 < 1$ , a contradiction.

Case 2: $\beta < 1$ . Similarly: $\beta^k < \cdots < \beta < 1$ , giving $1 < 1$ , a contradiction.

In both cases we reach a contradiction, so $\beta = 1$ . $\square$

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ExampleExamples Illustrating Torsion-Freeness

In $B_2$ : $\sigma_1^k \neq 1$ for any $k \neq 0$ (the closure $\hat{\sigma}_1^k$ is a $(2,k)$ -torus link, which is non-trivial for $k \neq 0$ ). In $B_3$ : the element $\Delta^2 = (\sigma_1\sigma_2\sigma_1)^2$ is the full twist, which generates the center $Z(B_3) \cong \mathbb{Z}$ . Despite $\Delta^2$ acting as $2\pi$ rotation on physical braids, it is not the identity in $B_3$ . The quotient $B_n/Z(B_n)$ is also torsion-free (for $n \geq 3$ ). Compare with the symmetric group: $S_n$ has abundant torsion (every transposition has order 2).

RemarkOther Proofs and Generalizations

Alternative proof via configuration spaces: $B_n$ is the fundamental group of the configuration space $\mathrm{Conf}_n(\mathbb{C}) = \{(z_1,\ldots,z_n) \in \mathbb{C}^n : z_i \neq z_j\}/S_n$ . This space is a $K(\pi,1)$ (Eilenberg-MacLane space), meaning $\pi_k = 0$ for $k \geq 2$ . The universal cover is contractible, so $B_n$ is torsion-free by a standard result in algebraic topology (a group with a finite-dimensional contractible classifying space is torsion-free). Bi-orderability: $B_n$ is left-orderable but not bi-orderable for $n \geq 3$ . Generalizations: Artin groups of finite type are torsion-free (Charney-Davis), generalizing the braid group result to other Coxeter-type groups.