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The short exact sequence to $S_{n}$ does not prove $B_{n}$ torsion-free

Statement refuted

Let $1 \to K \to G \to Q \to 1$ be a short exact sequence of groups with $K$ torsion-free and $Q$ finite. Then $G$ is torsion-free. In particular the configuration braid short exact sequence $1 \to P B_{n} \to B_{n}^{c o n f} \to S_{n} \to 1$ , whose kernel $P B_{n}$ is torsion-free and whose quotient $S_{n}$ is finite, would by itself show that $B_{n}^{c o n f}$ is torsion-free.

Facts & Assumptions

Given: the Axiom of Choice, an integer $n \geq 0$ , the groups $(Z, +, 0)$ and $(Z / 2, +)$ , and their external direct product $Z \times (Z / 2)$ .

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[F1]

Assume AC. Then $P B_{n}$ is torsion-free for every $n \geq 0$ : if $g \in P B_{n}$ and $g^{m} = e$ for some $m \geq 1$ , then $g = e$ (Pure braid groups are torsion-free). AC is used only for this citation (AC implies DC implies countable choice).

[F2]

For every $n \geq 0$ the sequence $1 \to P B_{n} \overset{p_{*}}{\to} B_{n}^{c o n f} \overset{π}{\to} S_{n} \to 1$ of the configuration groups is short exact: $p_{*}$ is injective, $π$ is surjective and $im p_{*} = \ker π$ (The configuration braid short exact sequence $1 \to P B_{n} \to B_{n}^{c o n f} \to S_{n} \to 1$ ); moreover $∣ S_{n} ∣ = n!$ , so $S_{n}$ is finite (The Lehmer code gives $∣ S_{n} ∣ = n!$ again).

[F3]

$(Z, +, \cdot, 0, 1)$ is a commutative ring and a totally ordered ring: the order is total and compatible with addition, so $x \leq y$ implies $x + z \leq y + z$ , and exactly one of $x < y$ , $x = y$ , $y < x$ holds for all integers (The integers form a commutative ring, The integers form a totally ordered ring).

[F4]

$(Z / 2, +, [0]_{2})$ is an abelian group with $[a]_{2} + [b]_{2} = [a + b]_{2}$ (For every natural $n$ , $(Z / n, +)$ is an abelian group, multiplication is a commutative monoid operation, and both distributive laws hold). Its exactly two classes are $[0]_{2}$ and $[1]_{2}$ (For $n \geq 1$ , every class in $Z / n$ has one representative $r$ with $0 \leq r < n$ , so $∣ Z / n ∣ = n$ ; while $Z / 0$ is in bijection with $Z$ , The congruence class $[a]_{n}$ and the quotient set $Z / n$ ), so $[1]_{2} \neq [0]_{2}$ and $[1]_{2} + [1]_{2} = [2]_{2} = [0]_{2}$ ; hence it has order $2$ and is generated by $[1]_{2}$ .

[F5]

For any groups $G, H$ the componentwise operation $(g, h) (g^{'}, h^{'}) = (g g^{'}, h h^{'})$ makes $G \times H$ a group with identity $(e_{G}, e_{H})$ , and the coordinate projections are group homomorphisms (The external direct product $G \times H$ with componentwise multiplication, $G \times H$ is a group with identity $(e_{G}, e_{H})$ , coordinatewise inverses, and homomorphic coordinate projections).

[F6]

In a group, $g^{0} = e$ and $g^{k + 1} = g^{k} \cdot g$ for natural powers, and $g$ has finite order $k \geq 1$ exactly when $k$ is the least positive integer with $g^{k} = e$ (Powers $g^{n}$ : natural exponents in a monoid and integer exponents in a group, with $g^{0} = e$ , The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity).

Counterexample

technique · direct

1.1F3F4

The two factors. By [F3] the additive group $(Z, +, 0)$ is a group, and by [F4] $(Z / 2, +)$ is an abelian group of order $2$ with $[1]_{2} \neq [0]_{2}$ and $[1]_{2} + [1]_{2} = [2]_{2} = [0]_{2}$ .

1.2F3F6

$Z$ is torsion-free. Let $a \in Z$ with $a \neq 0$ and let $m \geq 1$ . By trichotomy [F3] either $a > 0$ or $- a > 0$ ; assume $a > 0$ , the other case being symmetric with $m (- a) = - (m a)$ . Since $0 < a$ and the order is compatible with addition, $a < a + a = 2 a$ and inductively $k a < (k + 1) a$ for every $k \geq 1$ : if $k a > 0$ then $(k + 1) a = k a + a > k a > 0$ . Hence $m a > 0$ for every $m \geq 1$ , so $m a \neq 0$ ; an element $a$ of finite additive order would have $m a = 0$ for some $m \geq 1$ , so no nonzero element of $Z$ has finite order and $(Z, +, 0)$ is torsion-free.

2.1step 1.2F4F5

The sequence $1 \to Z \to Z \times (Z / 2) \to Z / 2 \to 1$ . By [F5] the set $G : = Z \times (Z / 2)$ with componentwise addition is a group with identity $e_{G} = (0, [0]_{2})$ , and the second-coordinate projection $π : G \to Z / 2$ is a group homomorphism; it is surjective because $π (a, [b]_{2}) = [b]_{2}$ for every $[b]_{2}$ . Its kernel is $K : = {(a, [0]_{2}) : a \in Z}$ , and the first-coordinate map $K \to Z$ , $(a, [0]_{2}) \mapsto a$ , is a group isomorphism; hence $K$ is torsion-free by step 1.2, and $G / K ≅ Z / 2$ has order $2$ by [F4], so the quotient is finite. Thus $1 \to K \to G \overset{π}{\to} Z / 2 \to 1$ is a short exact sequence with torsion-free kernel and finite quotient.

3.1step 2.1F4F5F6

The middle group has an element of order two. Put $g : = (0, [1]_{2}) \in G$ . Using the componentwise operation of [F5] and the addition of [F4], $g^{2} = (0 + 0, [1]_{2} + [1]_{2}) = (0, [2]_{2}) = (0, [0]_{2}) = e_{G}$ , while $g \neq e_{G}$ because $[1]_{2} \neq [0]_{2}$ . By [F6] the element $g$ has order $2$ , so $G$ is not torsion-free although $K$ is torsion-free and $G / K$ is finite. This refutes the general implication of the statement: a torsion-free kernel and a finite quotient do not force the middle group to be torsion-free.

4.1A1step 2.1step 3.1F1F2

The braid reading. Under the standing assumption [A1], which discharges the Axiom-of-Choice hypothesis that [F1] attaches to its own citation, the configuration braid sequence of [F2] has the two abstract features used above: its kernel $P B_{n}$ is torsion-free by [F1], and its quotient $S_{n}$ is finite by [F2]. Since the general implication fails already for the infinite cyclic kernel $Z$ and the two-element quotient $Z / 2$ of step 2.1, those two features alone cannot establish torsion-freeness of $B_{n}^{c o n f}$ ; only a finer argument about the specific groups can do that. This counterexample makes no claim that $B_{n}^{c o n f}$ itself has torsion.

∎

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