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 ${G/G^n}$.

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 ${\leq 1/k}$.