Notes of Dominic Gruber’s Cambridge lecture 13-01-2017

Small cancellation theory over Burnside groups

Joint with Rémi Coulon.

1. A flexible tool for producing infinite periodic groups

Here are possible approaches.

1. Understand the proof that {B(S,n)} is infinite, and add a few relators to an infinite presentation. Olshanskii’s Tarski monster 1982. Hard.

2. Let {G} be hyperbolic and torsion free. Then {G} has infinite periodic quotients (Olshanskii 1991, Gromov-Delzant). Drawback: the exponent depends on {G}.

3. Our approach.

Theorem 1 (Coulon-Gruber) There exists {n_0} such that for all {n\geq n_0}, {n} odd, the following holds. Let {G=\langle S|R\rangle} be a {C'(\frac{1}{6})} presentation such that

  • {|S|\geq 2}, no relator of length {\leq 2}.
  • No relator is a proper power.
  • No third power is a subword of a relator.

Then {G/G^n} is infinite. Furthermore, no nonempty proper subword of a element of {R} represents the identity in {G/G^n}.

2. Examples of admissible data

In a group presentation, a piece is a common subword of two cyclic conjugates of relators. Say the presentation with cyclically reduced relators satisfies {C'(\lambda)} if whenever a piece {u} is a subword of a cyclic conjugate of a relator {r}, then {|u|<\lambda|r|}.

Examples.

  1. In the standard presentation of the genus 2 surface, all pieces have length 1.

  2. Let {S=\{a,b,t\}}. Let {x_i} be words in {a} and {b} of length {i}. Let {R} contain all words of the form {tx_{100N+1}tx^{100N+2}t\cdots tx^{100N+100}}. Then this satisfies {C'(\frac{1}{6})}.
  3. Let {S=\{a,b\}}. Let {x_\infty} be the Thue-Morse sequence. It contains no third power as subwords. Let {x_1=a}, {x_2=bb}, {x_3=aba}, … be the successive length {i} subwords. Apply previous construction, get a {C'(\frac{1}{6})} presentation. Our theorem provides an infinite periodic group.

    For {I\subset{\mathbb N}}, let {R(I)=\{r_i\,;\,i\in I\}}. This gives uncountably many different {n}-periodic groups.

3. Applications

1.Get {n}-periodic groups with coarsely embedded expander graphs.

2. Using the existence of {n}-periodic groups whose word problem is unsolvable, we get the following.

Theorem 2 Let {n\geq n_0} be not a prime. Let {(P)} be a property of groups such that

  • There exists a relatively finite presented {n}-periodic group {A} that has {(P)}.
  • There exists a relatively finite presented {n}-periodic group {B} such that any group containing {B} does not have {(P)}.

Then there does not exist an algorithm that takes a relatively finite presentation of an {n}-periodic group and decides wether it has {(P)} or not.

Here a relatively finite presentation means that one kills finitely many elements in {B(S,n)}.

Examples of suitable properties. Triviality, finiteness, being cyclic, abelian, nilpotent, solvable, amenable…

4. Periodic quotients of groups acting acylindrically on hyperbolic spaces

Let {X} be {\delta}-hyperbolic. Say {G} acts {(L,N,\epsilon)}-acylindrically on {X} if for all {x,y\in X} with {d(x,y)\geq L}, the number of elements of {G} moving each f {x} and {y} at most distance {\epsilon} away is at most {N}.

In 2013, Coulon showed that there exists {n_0} such that for all odd {n\geq n_0}, periodic quotients exist: let {G} act {(L,N,100\delta)}-acylindrically on a {\delta}-hyperbolic space {X} without elliptics. Then {G_n:=G/G^n} is infinite, and {G\mapsto G_n} is injective on a ball of radius 3 in the pseudo-metric induced by {X}.

The point is to understand the normal closure of high powers of all loxodromics simultaneously.

The method is small cancellation: {C'(\frac{1}{6})} implies that {Cay(G,S)} is hyperbolic and any cycle labelled by a relator is an isometrically embedded cycle graph. Small overlap implies a linear isometric inequality.

How does this extend to infinite presentations ?

Theorem 3 (Gruber-Sisto) Let {W\subset G} be a subset that contains all subwords of elements of {R}. Then {Cay(G,S\cup W)} is {\delta}-hyperbolic.

Apply this here. With our assumptions, {X} is nonelementary. Using the assumption that no {k+1}-st power is a subword of a relator, we show (Abbot and Hume did something similar) that the action of {G} on {X} is {(L,N,100\delta)}-acylindrical, for {N=N(k)}.

Why {k=2} ? If {\omega^k} is a subword of a relator, then the translation length of {\omega} on {X} is {\leq 1/k}.

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About metric2011

metric2011 is a program of Centre Emile Borel, an activity of Institut Henri Poincaré, 11 rue Pierre et Marie Curie, 75005 Paris, France. See http://www.math.ens.fr/metric2011/
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