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A Reidemeister-Schreier presentation for a surface subgroup

Example

Let

$G = ⟨ a, b ∣ b a b^{- 1} a ⟩,$

the Klein bottle group. The subgroup $H = ⟨ a, b^{2} ⟩$ has index $2$ and Reidemeister-Schreier gives the presentation

$H ≅ ⟨ x, y ∣ x y x^{- 1} y^{- 1} ⟩ ≅ Z^{2},$

so $H$ is the fundamental group of the torus.

Facts & Assumptions

Given: The Klein bottle presentation above and the subgroup $H = ⟨ a, b^{2} ⟩$ .

[L1]

Reidemeister-Schreier presents a subgroup by rewritten Schreier generators and conjugated relators (The Reidemeister-Schreier presentation theorem).

Verification

technique · direct

1.1givenconstruct

The quotient by $H$ has cosets $H$ and $H b$ , so $T = {1, b}$ is a right transversal. The nontrivial Schreier generators are $x = s (1, a) = a$ , $z = s (b, a) = b a b^{- 1}$ , and $y = s (b, b) = b^{2}$ .

2.1L1step 1.1algebra

The defining relator is $r = b a b^{- 1} a$ . Reidemeister-Schreier rewrites the two conjugates determined by $T$ as $τ (r) = z x$ and $τ (b r b^{- 1}) = y x y^{- 1} z$ . Therefore [L1] yields the presentation $⟨ x, y, z ∣ z x, y x y^{- 1} z ⟩$ .

3.1step 2.1algebra∎

The first relator gives $z = x^{- 1}$ . Substituting this into the second relator yields $y x y^{- 1} x^{- 1}$ . Hence the presentation from step 2.1 simplifies to $⟨ x, y ∣ y x y^{- 1} x^{- 1} ⟩$ , so the presented group is the free abelian group on two generators, that is, $Z^{2}$ . Thus $H$ is a surface subgroup of torus type.

Depends on

The Reidemeister-Schreier presentation theorem

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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