One-relator quotients of Partially Commutative Groups

TL;DR

This paper extends Magnus's Freiheitssatz to partially commutative groups, providing conditions for subgroup embeddings, element orders, and decidability of the word problem in certain quotients, with applications to cycle graphs.

Contribution

It generalizes a key theorem from one-relator group theory to partially commutative groups, establishing subgroup embedding criteria and decidability results.

Findings

Magnus subgroups embed in quotients under specific conditions.

The order of elements in quotients matches their roots in the original group.

Decidability of the word and conjugacy problems is established for many cases.

Abstract

We generalise a key result of one-relator group theory, namely Magnus's Freiheitssatz, to partially commutative groups, under sufficiently strong conditions on the relator. The main theorem shows that under our conditions, on an element $r$ of a partially commutative group $\mathbb{G}$, certain Magnus subgroups embed in the quotient $G=\mathbb{G}/N(r)$; that if $r=s^n$ has root $s$ in $\mathbb{G}$ then the order of $s$ in $G$ is $n$, and under slightly stronger conditions that the word problem of $G$ is decidable. We also give conditions under which the question of which Magnus subgroups of $\mathbb{G}$ embed in $G$ reduces to the same question in the minimal parabolic subgroup of $\mathbb{G}$ containing $r$. In many cases this allows us to characterise Magnus subgroups which embed in $G$, via a condition on $r$ and the commutation graph of $\mathbb{G}$, and to find further examples of quotients $G$ where the word and conjugacy problems are decidable. We give evidence that situations in which our main theorem applies are not uncommon, by proving that for cycle graphs with a chord $\Gamma$, almost all cyclically reduced elements of the partially commutative group $\mathbb{G}(\Gamma)$ satisfy the conditions of the theorem.

Equations

$$R=\{[x_i,x_j]\in \mathbb{F}(A)\mid x_i,x_j\in A\text{ and there is an edge from }{x}_i\text{ to }{x}_j\text{ in }\Gamma \}.$$

$$x(a^{n}w){=}_{G}x{t}^{-m}=t^{-m}x{=}_G(a^{n}w)x.$$

$$\text{HNN}(t)$$

$$p=g_0{t}^{{\epsilon }_1}\cdots t^{{\epsilon }_n}{g}_n,$$

$$|p{|}_t=\sum _{i=1}^n|{\epsilon }_i|.$$

$$u=g_0{t}^{{\epsilon }_1}\cdots t^{{\epsilon }_m}{g}_m\text{ and }v=h_0{t}^{{\delta }_1}\cdots t^{{\delta }_n}{h}_n,$$

$$|u_1\cdots u_n{|}_t=\sum _{i=1}^n|u_i{|}_t$$

$$a_{l}d{u}^{\prime }u{b}_r^{-1}{d}^{-1}{v}^{\prime -1}=au{b}^{-1}=v,$$

$$\sigma (p)=\sigma (g_0)t^{{\epsilon }_1}\cdots t^{{\epsilon }_n}\sigma (g_n).$$

$$\overline{\varphi }([{\gamma }_{i-1+km},{\delta }_{i+km}])={\bar{g}}_{i-1},$$

$$\overline{\varphi }([{\gamma }_{km},{\gamma }_{(k+1)m}])={\overline{s}}_0={\bar{g}}_0{t}^{{\epsilon }_1}\cdots {\bar{g}}_{m-1}{t}^{{\epsilon }_m}\in F_D,$$

$$\mathrm{Sep}(\Delta )=\{{\alpha }_i,{\beta }_i:1\le i\le n\}.$$

$$\mathrm{Sep}(\mu )=\mathrm{Sep}(\Delta )\cap [\mu \backslash \{\beta \}].$$

$$l(\varphi (\beta ))\ge \{\begin{array}{l}\left(n-\frac{k+2}{2}\right)l(s)+2,\text{ if k is even}\\ \left(n-\frac{k+1}{2}\right)l(s)+1,\text{ if k is odd}\end{array}.$$

$$w=\prod _{i=1}^k{y}_i^{-1}{r}^{{\epsilon }_i}{y}_i.$$

$$\mathbb{G}={\mathbb{A}}_0{\ast }_U{\mathbb{A}}_1.$$

$$G=\mathbb{G}/N\cong (U\times K/M){\ast }_U{\mathbb{A}}_1.$$

$$g_1{u}_1{h}_1\cdots g_k{u}_k{h}_k{u}_{k+1}=g_1^{\prime }{u}_1^{\prime }{h}_1^{\prime }\cdots g_l^{\prime }{u}_l^{\prime }{h}_l^{\prime }{u}_{l+1}^{\prime }.$$

$$g_1{u}_1{h}_1\cdots h_{k-1}{u}_k{u}_{k+1}{g}_k=g_1^{\prime }{u}_1^{\prime }{h}_1^{\prime }\cdots h_{l-1}{u}_l^{\prime }{u}_{l+1}^{\prime }{g}_l^{\prime },$$

$$g_1{u}_1{h}_1\cdots g_k{u}_k{h}_k{u}_{k+1}=g_1^{\prime }{u}^{-1}{u}_1^{\prime }{h}_1^{\prime }\cdots g_l^{\prime }{u}_l^{\prime }{h}_l^{\prime }{u}_{l+1}^{\prime }u.$$

$$g_i=g_i^{\prime },h_i=h_i^{\prime },[h_i,u_{i+1}^{\prime }\cdots u_{k+1}^{\prime }u{u}_{k+1}^{-1}\cdots u_{i+1}^{-1}]=1,$$

$$u^{-1}{u}_1^{\prime }\cdots u_{k+1}^{\prime }u=u_1\cdots u_{k+1}.$$

$$x^{-1}(u_{i+1}^{\prime }\cdots u_{k+1}^{\prime }{)}^{-1}{h}_i{u}_{i+1}^{\prime }\cdots u_{k+1}^{\prime }x$$

$$x^{-1}{u}_1^{\prime }\cdots u_{k+1}^{\prime }x$$

$$a{u}_1{h}_1\cdots a{u}_k{h}_k{u}_{k+1}=u^{-1}a{u}_1^{\prime }{h}_1^{\prime }\cdots a{u}_l^{\prime }{h}_l^{\prime }{u}_{l+1}^{\prime }u,$$

$$\rho (S)=\underset{d,k\to \infty }{\lim }\frac{|S\cap L(d,k)|}{|L(d,k)|}.$$

$$(\prod _{j=1}^r{a}_{i+1}^{{\alpha }_j}{a}_{i-1}^{{\beta }_j})a_i^{\delta },$$

$$L_S^{\text{ref}it:composed2}(d,k,r,P)=\{u{g}_1{t}^{{\alpha }_1}\cdots g_r{t}^{{\alpha }_r}\in L_S^{\text{ref}it:composed2}(d,k):(|{\alpha }_1|,\dots ,|{\alpha }_r|)=P\}.$$

$$l_H(d)=1+8d{3}^{d-1},\text{ when }n=5.$$

$$2(n-1){\gamma }^{m-1}\le l_{H,S}(m)\le 2(n-1){\alpha }^{m-1},\text{ for all }m\ge 2.$$

$$1+\frac{n-1}{n-4}[{\gamma }^d-1]\le l_H(d)\le 1+\frac{n-1}{n-3}[{\alpha }^d-1],$$

Full text

††footnotetext: Keywords. One-relator group theory; partially commutative groups, right-angled Artin groups, HNN-extensions of groups.††footnotetext: Mathematics Subject Classification (2010): 20F05; 20F36, 20E06

1 Introduction

Partially commutative groups have been extensively studied in several different guises and are variously known as semi-free groups; right-angled Artin groups; trace groups; graph groups or even locally free groups. Let $\Gamma$ be a finite, undirected, simple graph. Let $A=V(\Gamma)$ be the set of vertices of $\Gamma$ and let $\mathbb{F}(A)$ be the free group on $A$. For elements $g,h$ of a group we denote the commutator $g^{-1}h^{-1}gh$ of $g$ and $h$ by $[g,h]$. Let

[TABLE]

We define the *free partially commutative group with (commutation) graph * $\Gamma$ to be the group $\mathbb{G}=\mathbb{G}(\Gamma)$ with free presentation $\left<A\mid R\right>.$ We shall refer to finitely generated free partially commutative groups as partially commutative groups. (Strictly speaking we have defined the class of finitely generated partially commutative groups.) The class of partially commutative groups contains finitely generated free, and free Abelian groups; has provided several crucial examples in the theory of finitely presented groups and has applications both in mathematics and computer science. For an introduction and survey of the literature see [7] or [12]. It emerges, from the work of Sageev, Haglund, Wise, Agol and others, that many well-known families of groups virtually embed into partially commutative groups: among these are Coxeter groups, certain one-relator groups with torsion, limit groups, and fundamental groups of closed $3$-manifold groups (see for example [25] for details and references). It is therefore natural to consider related classes, such as their one-relator quotients as we do here.

There are generalisations of one-relator group theory in several directions; for instance to two-relator groups, to one-relator quotients of surface groups [18, 2] and very successfully to one-relator quotients of free products of groups, see e.g. [6, 16, 17] and see [13] for details and fuller references. A key result of one-relator group theory is Magnus’s Freiheitssatz which states that if $\mathbb{F}$ is the free group of rank $n$, and $r$ is an element of $\mathbb{F}$, involving every generator of $\mathbb{F}$, then the subgroup generated by $n-1$ of the generators embeds in the one-relator group $\mathbb{F}/N(r)$. (We use $N(r)$ to denote the normal closure of $r$ in a given group.) The Freiheitssatz has several immediate consequences: for instance if $r=s^{m}$, where $s$ is not a proper power in $\mathbb{F}$ then $s$ has order precisely $m$ in $\mathbb{F}/N$; if $m=1$ then $\mathbb{F}/N$ is torsion-free; if $m>1$ then any element of finite order in $\mathbb{F}/N$ is conjugate to a power of $s$; and one-relator groups have solvable word problem (see for example [13]).

Turning to one-relator quotients of partially commutative groups, Antolin and Kar [1] proved that a Freiheitssatz holds for one-relator quotients of starred partially commutative groups: $\mathbb{G}(\Gamma)$ is starred if $\Gamma$ has no full subgraph isomorphic to the four cycle or the path graph on four vertices. More generally they prove versions of the Freiheitssatz for one-relator quotients of starred partially commutative products and show that the word problem is solvable, both for one-relator quotients of starred partially commutative groups and of starred partially commutative products of polycyclic groups.

Here, with appropriate restrictions on the relator, we show that a Freiheitssatz holds for one-relator quotients of any partially commutative group and that, with slightly stronger restrictions, the word problem is solvable. A summary of the situations in which a Freiheitssatz is known to hold, combining our results and those of [1], is given in the Examples and final summary of this section.

In order to state our main results we need some definitions. If $w$ is an element of the free monoid $(A\cup A^{-1})^{*}$ then we denote by $\operatorname{supp}(w)$ the set of elements $x\in A$ such that $x$ or $x^{-1}$ occurs in $w$. For an element $g\in\mathbb{G}$ we define $\operatorname{supp}(g)$ to be equal to $\operatorname{supp}(w)$, where $w$ is an element of $(A\cup A^{-1})^{*}$ of minimal length amongst all those elements representing $g$. For more detail (including the fact that $\operatorname{supp}(g)$ is well-defined) see Section 2. An element $g$ of $\mathbb{G}$ is said to be cyclically minimal if it is represented by a word $w\in(A\cup A^{-1})^{*}$ of length no greater than any other word representing an element of the conjugacy class of $g$. (See Section 2, Lemma 2.1.)

A subset $Y$ of $A$ is called a clique if the full subgraph $\Gamma_{Y}$ of $\Gamma$ generated by $Y$ is a complete graph and independent if $\Gamma_{Y}$ is a null graph. For $x\in A$ define the link of $x$ to be $\operatorname{lk}(x)=\{y\in A|d(x,y)=1\}$ (edges of the graph have length $1$) and the star of $x$ to be $\operatorname{st}(x)=\operatorname{lk}(x)\cup\{x\}$. The definitions of $t$-thick and $t$-root appear in Section 3.1, Definition 3.2 and Section 3.3, Definition 3.7, respectively.

To illustrate some of the possibilities we give some examples.

Example 1.1**.**

If $\Gamma$ is a complete or null graph then $\mathbb{G}=\mathbb{G}(\Gamma)$ is free Abelian or free, respectively. In this case, if $s\in\mathbb{G}$ is not a proper power, $t\in\operatorname{supp}(s)$ and $r=s^{n}$, for some positive integer $n$, then $\langle A\backslash\{t\}\rangle$ embeds in $G=\mathbb{G}/N(r)$ and $s$ has order $n$ in $G$; using standard results from the theory of finitely generated Abelian groups, or of one-relator groups, as appropriate. 2. 2.

Let $A=\{a,b,c\}$, $R=\{[a,b],[b,c]\}$ and $\mathbb{G}=\langle A|R\rangle\cong\mathbb{F}_{2}\times\mathbb{Z}$. Let $s=abc$, let $N$ be the normal closure of $s^{n}$ in $\mathbb{G}$, for some $n>0$, and let $G=\mathbb{G}/N$. Here $b\in\operatorname{supp}(s)$ and $r=(ac)^{n}b^{n}$, so $[(ac)^{n},a]=1$ in $G$, but $[(ac)^{n},a]\neq 1$ in $\mathbb{G}$ (as $C_{\mathbb{G}}(a)=\langle a,b\rangle$). Hence $\langle A\backslash\{b\}\rangle=\langle a,c\rangle$ does not embed in $G$.

Theorem 1.2**.**

Let $\mathbb{G}=\langle A|R\rangle$ and let $s\in\mathbb{G}$ be a cyclically minimal element such that, for some $t\in\operatorname{supp}(s)$,

$\operatorname{lk}(t)$* is a clique (or empty),* 2. 2.

$s$* is $t$-thick,* 3. 3.

$s\notin\langle\operatorname{st}(t)\rangle$* and* 4. 4.

$s$* is a $t$-root.*

Then, for $n\geq 3$,

1. (a)

$\langle A\backslash\{t\}\rangle$* embeds in $\mathbb{G}/N$, where $N$ denotes the normal closure of $s^{n}$ in $\mathbb{G}$, and* 2. (b)

the order of $sN$ in $\mathbb{G}/N$ is $n$.

Moreover if $n\geq 4$ then the word problem is solvable in $\mathbb{G}/N$.

Given a cyclically reduced element $r$ of the free group $\mathbb{F}(A)$ on $A$, a Magnus subgroup is a subgroup generated by a subset $B$ of $A$ such that $r\notin\langle B\rangle$. As given above, Magnus’s Freiheitssatz for one-relator groups applies when $r$ involves every generator of the free group in question so every proper subset of $A$ generates a Magnus subgroup; which embeds in the quotient by the normal closure of $r$. In the free group setting the general case reduces to this special case: indeed, if $r$ is an arbitrary element of $\mathbb{F}(A)$ and $Y=\operatorname{supp}(r)\subseteq A$ with $X=A\backslash Y$ non-empty then we have $\mathbb{F}(A)/N(r)\cong[\mathbb{F}(Y)/M(r)]*\mathbb{F}(X)$, where $N(r)$ and $M(r)$ are the normal closures of $r$ in $\mathbb{F}(A)$ and $\mathbb{F}(Y)$, respectively. From the Freiheitssatz as stated above, if $Y^{\prime}$ is a proper subset of $Y$ then the subgroup $\langle Y^{\prime}\rangle$ of $\mathbb{F}(Y)$ embeds in $\mathbb{F}(Y)/M(r)$, so $\langle X\cup Y^{\prime}\rangle$ embeds in $\mathbb{F}(A)/N(r)$. Thus the general Freiheitssatz for Magnus subgroups of $\mathbb{F}(A)$ reduces to the special case of Magnus subgroups for $\mathbb{F}(Y)$.

To obtain an analogous reduction for one-relator quotients of partially commutative groups we first note that it follows from work of B. Baumslag and S.J. Pride [4], together with the fact, proved by Diekert and Mushcoll [8], that equations are decidable over partially commutative groups; that if $\mathbb{G}=\mathbb{G}_{1}*\cdots*\mathbb{G}_{k}$ is a free product (so $\Gamma$ has $k$ connected components) and $\operatorname{supp}(r)$ contains vertices in at least two components of $\Gamma$, then each of the $\mathbb{G}_{i}$ embeds in $G$. In general, we consider the case where the group $\mathbb{G}$ decomposes as a free product with amalgamation, and one factor is generated by $\operatorname{supp}(r)$. More precisely, we make the following definition.

Definition 1.3**.**

For a subset $B$ of $A$ define $\operatorname{lk}(B)=\cap_{b\in B}\operatorname{lk}(b)$ and for $g\in\mathbb{G}$ define $\operatorname{lk}(g)=\operatorname{lk}(\operatorname{supp}(g))$. A subset $Y$ of $A$ is called synchronised if, for all vertices $v$ of $Y$, the star $\operatorname{st}(v)$ of $v$ is a subset of $Y\cup\operatorname{lk}(Y)$.

If $Y$ is synchronised then, writing $X=A\backslash\{Y\cup\operatorname{lk}(Y)\}$, it follows also that, for all vertices $v$ of $X$, the star $\operatorname{st}(v)$ is a subset of $X\cup\operatorname{lk}(Y)=A\backslash Y$. Hence, if $Y$ is synchronised, setting $\mathbb{A}_{0}=\langle Y\cup\operatorname{lk}(Y)\rangle$, $\mathbb{A}_{1}=\langle X\cup\operatorname{lk}(Y)\rangle$ and $U=\langle\operatorname{lk}(Y)\rangle$, it follows that $\mathbb{G}=\mathbb{A}_{0}*_{U}\mathbb{A}_{1}$.

Theorem 1.4**.**

Let $\mathbb{G}=\langle A|R\rangle$ and let $s\in\mathbb{G}$ be a cyclically minimal element such that $\operatorname{supp}(s)$ is synchronised. Let $\mathbb{A}_{0}=\langle\operatorname{supp}(s)\cup\operatorname{lk}(s)\rangle$, $\mathbb{A}_{1}=\langle A\backslash\operatorname{supp}(s)\rangle$ and $U=\langle\operatorname{lk}(s)\rangle$; so $\mathbb{G}=\mathbb{A}_{0}*_{U}\mathbb{A}_{1}$. Let $r=s^{n}$, for some $n\geq 1$, let $K=\langle\operatorname{supp}(s)\rangle$, denote by $N$ the normal closure of $r$ in $\mathbb{G}$ and by $M$ the normal closure of $r$ in $K$, and let $G=\mathbb{G}/N$. Then the following hold.

1. (a)

If $Y^{\prime}$ is a subset of $\operatorname{supp}(s)$ such that the subgroup $\langle Y^{\prime}\rangle$ of $K$ embeds in $K/M$ then $\langle Y^{\prime}\cup(A\backslash\operatorname{supp}(s))\rangle$ embeds in $G$; 2. (b)

if $s$ has order $n$ in $K/M$ then $s$ has order $n$ in $G$, 3. (c)

if the word problem is solvable in $K/M$ then the word problem is solvable in $G$ and 4. (d)

if the conjugacy problem is solvable in $K/M$ then the conjugacy problem is solvable in $G$.

In terms of Magnus subgroups of $\mathbb{G}$ (defined exactly as for free groups) with the notation, and under the hypotheses, of the theorem, a Magnus subgroup generated by a subset $B$ of $A$ embeds in $G$ if and only if the Magnus subgroup $\langle B\cap\operatorname{supp}(s)\rangle$ of $K$ embeds in $K/M$.

In cases where $\operatorname{supp}(s)$ is a clique we obtain a complete characterisation of the conditions under which all Magnus subgroups of $\mathbb{G}$ embed in $G$.

Corollary 1.5**.**

Let $\mathbb{G}=\langle A|R\rangle$ and let $s\in\mathbb{G}$ be a cyclically minimal element such that $\operatorname{supp}(s)$ is a clique. Let $r=s^{n}$, for some $n\geq 1$, denote by $N$ the normal closure of $r$ in $\mathbb{G}$ and let $G=\mathbb{G}/N$. Then the following are equivalent.

$\mathbb{G}_{t}=\langle A\backslash\{t\}\rangle$* embeds in $G$, for all $t\in\operatorname{supp}(s)$. * 2. 2.

$\operatorname{supp}(s)$* is synchronised.*

*Moreover, if $\operatorname{supp}(s)$ is synchronised then $s$ has order $n$ in $G$, and the word and conjugacy problem are decidable in $G$. *

Proof.

Suppose $\operatorname{supp}(s)$ is a clique and not synchronised. This means there exists $t\in\operatorname{supp}(s)$ and $x\in A\backslash(\operatorname{supp}(s)\cup\operatorname{lk}(s))$ such that $[x,t]=1$. Since $x\notin\operatorname{lk}(s)$, there is $a\in\operatorname{supp}(s)$ such that $[x,a]\neq 1$. As $K=\langle\operatorname{supp}(s)\rangle$ is Abelian we have $r=s^{n}=t^{m}a^{n}w$, for some $w$ such that $\operatorname{supp}(w)=\operatorname{supp}(s)\backslash\{a,t\}$. Then

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Hence $[x,a^{n}w]=_{G}1$ while, as $[x,a]\neq 1$, $[x,a^{n}w]\neq 1$ in $\mathbb{G}$. As $[x,a^{n}w]\in\langle A\backslash\{t\}\rangle$, this implies that $\langle A\backslash\{t\}\rangle$ does not embed in $G$.

The converse follows from Theorem 1.4, since when $K$ is Abelian $\langle\operatorname{supp}(s)\backslash\{t\}\rangle$ embeds in $K/M$, for all $t\in\operatorname{supp}(s)$, and $s$ has order $n$ in $K/M$; as in Example 1.1.1. As $K/M$ is a finitely generated Abelian group, it has solvable conjugacy problem, so the remaining statement follows from Theorem 1.4 (d). ∎

In the dual case, where $\operatorname{supp}(s)$ is independent, we obtain a sufficient, but not necessary, condition for all Magnus subgroups of $\mathbb{G}$ to embed in $G$.

Corollary 1.6**.**

Let $\mathbb{G}=\langle A|R\rangle$ and let $s\in\mathbb{G}$ be a cyclically minimal element such that $\operatorname{supp}(s)$ is independent. Let $r=s^{n}$, for some $n\geq 1$, denote by $N$ the normal closure of $r$ in $\mathbb{G}$ and let $G=\mathbb{G}/N$. If $\operatorname{supp}(s)$ is synchronised then $\mathbb{G}_{t}=\langle A\backslash\{t\}\rangle$ embeds in $G$, for all $t\in\operatorname{supp}(s)$, $s$ has order $n$ in $G$ and the word problem is solvable in $G$. Moreover, if $n\geq 2$ then the conjugacy problem is solvable in $G$.

Proof.

This follows from Theorem 1.4, since when $K$ is free $\langle\operatorname{supp}(s)\backslash\{t\}\rangle$ embeds in $K/M$, for all $t\in\operatorname{supp}(s)$ and $s$ has order $n$ in $K/M$; as in Example 1.1.1. As $K/M$ is a one-relator group it has solvable word problem, and if $n\geq 2$ then solvable conjugacy problem; so the final statement follows from Theorem 1.4 (c) and (d). ∎

Example 1.7**.**

Corollaries 1.5 and 1.6 generalise Example 1.1.1. In Example 1.1.2, $\operatorname{supp}(s)=A$ so, although it is synchronised, Theorem 1.4 gives no new information. In this example $\operatorname{lk}(b)$ is not a clique but $\operatorname{lk}(c)=\operatorname{lk}(a)=\{b\}$, a clique. However (taking $c=t$ in Theorem 1.2) we have $s=abc$ and $ab\notin\operatorname{Maln}(\langle b\rangle)$, so $s$ is not $c$-thick, and Theorem 1.2 does not apply. 2. 2.

Let $A=\{a,b,c,t\}$ and $R=\{[t,a],[a,b],[b,c]\}$ so $\mathbb{G}=\langle A\,|\,R\rangle$ has commutation graph the path graph $P_{4}$ of length $3$ (and $4$ vertices). If $s=ct$ then $\operatorname{lk}(t)$ is a clique and $s$ satisfies the hypothesis of Theorem 1.2. Setting $r=s^{3}$ it follows that $\langle a,b,c\rangle$ embeds in $\mathbb{G}/N(r)$. Note that, since $G=\mathbb{G}/N(s)$ is isomorphic to $\mathbb{Z}^{3}$, the subgroup $\langle a,c\rangle$ of $\mathbb{G}$ does not embed in $G$; showing that the condition $n\geq 3$ in Theorem 1.2 cannot be entirely removed. 3. 3.

Generalising the previous example, let $\Gamma$ be a tree, let $t$ be a leaf of $\Gamma$ and let $a$ be the vertex of $\Gamma$ to which $t$ is adjacent; so $\operatorname{lk}(t)=\{a\}$. Let $w$ be any word in $\langle A\backslash\operatorname{st}(a)\rangle$. Then $s=wt$ satisfies all the hypotheses of Theorem 1.2 so $\langle A\backslash\{t\}\rangle$ embeds in $\mathbb{G}/N(s^{n})$, when $n\geq 3$. Moreover, if $w_{i}\in\langle A\backslash\operatorname{st}(a)\rangle\cup\langle a\rangle$ and $\varepsilon_{i}\in\{\pm 1\}$ are chosen for $i=0,\ldots,m$, such that $w_{i}\notin\langle a\rangle$, for at least one $i$, and $s=w_{0}t^{\varepsilon_{1}}w_{1}\cdots w_{m-1}t^{e_{m}}w_{m}$ is cyclically minimal and not a proper power in $\mathbb{G}$, then again Theorem 1.2 implies that $\langle A\backslash\{t\}\rangle$ embeds in $\mathbb{G}/N(s^{n})$, for $n\geq 3$. 4. 4.

Consider the graph $C_{4}^{\prime}$ on the left of Figure 1. This is a starred graph, therefore Theorem A of [1] applies. Every subgraph of a connected starred graph contains a central vertex; that is a vertex incident to all other vertices. In this example the set of central vertices is $B=\{a,c\}$. From [1, Theorem A] it follows that if $r\notin\langle a,b,c\rangle$ then $\langle a,b,c\rangle$ embeds in $G=\mathbb{G}/N(r)$. Moreover (loc. cit.), if $r\notin\langle a,c,d\rangle$ then $\langle a,c,d\rangle$ embeds in $G$ and if $r\notin\langle a,c\rangle$ then $\langle a,c\rangle$ embeds in $G$. On the other hand, $\{a,c\}$ is a synchronised clique, so from our Corollary 1.5 it follows that if $r\in\langle a,c\rangle$ and $t\in\operatorname{supp}(r)$ then $\langle A\backslash\{t\}\rangle$ embeds in $G$. Again, $\{b,d\}$ is a synchronised and independent, so from Corollary 1.6, if $r\in\langle b,d\rangle$ and $t\in\operatorname{supp}(r)$ then $\langle A\backslash\{t\}\rangle$ embeds in $G$. The same applies when $\Gamma=C_{4}$, as shown on the right hand side of Figure 1. In this case, if $\operatorname{supp}(r)=\{b,d\}$ and $t=b$ or $d$, then Corollary 1.6 implies that $\langle A\backslash\{t\}\rangle$ embeds in $\mathbb{G}(C_{4})/N(r)$.

As the set of central vertices of a starred graph always forms a synchronised clique, for starred graphs Corollary 1.5 always complements [1, Theorem A], as in this example.

The previous examples of a four cycle and a four cycle with a chord may be extended to $n$-cycles where $n\geq 5$. Let $C_{n}^{\prime}$ be the $n$-cycle with a chord, $n\geq 5$, as shown on the left of Figure 2, let $\mathbb{G}^{\prime}=\mathbb{G}(C_{n}^{\prime})$ and let $s^{\prime}=a_{2}a_{n-2}t\in\mathbb{G}^{\prime}$. Then $\mathbb{G}^{\prime}$, $t$ and $s^{\prime}$ satisfy the conditions of Theorem 1.2, so setting $r^{\prime}=s^{\prime 3}$ the subgroup $\langle a_{1},a_{2},\ldots,a_{n-2},a_{n-1}\rangle$ of $\mathbb{G}^{\prime}$ embeds in $G^{\prime}=\mathbb{G}^{\prime}/N(r^{\prime})$.

Next let $C_{n}$ be $n$-cycle on the right of Figure 2, let $\mathbb{G}=\mathbb{G}(C_{n})$, let $s=a_{2}a_{n-2}t\in\mathbb{G}$ and $r=s^{3}$. In this case $\operatorname{lk}(t)$ is not a clique so Theorem 1.2 does not apply. There is a natural surjection $\pi^{\prime}$ from $\mathbb{G}$ to $\mathbb{G}^{\prime}$ and composing this with the canonical map $\rho^{\prime}$ from $\mathbb{G}^{\prime}$ to $G^{\prime}$ gives a surjection $\rho^{\prime}\pi^{\prime}$ of $\mathbb{G}$ onto $G^{\prime}$. Since $\pi^{\prime}$ maps the subgroup $H=\langle a_{2},\ldots,a_{n-2}\rangle$ of $\mathbb{G}$ isomorphically to the subgroup $H^{\prime}=\langle a_{2},\ldots,a_{n-2}\rangle$ of $\mathbb{G}^{\prime}$ and, from the above, $\rho^{\prime}$ embeds $H^{\prime}$ into $G^{\prime}$, it follows that $\rho^{\prime}\pi^{\prime}$ restricts to an embedding of $H$ into $G^{\prime}$. On the other hand, denoting the canonical map from $\mathbb{G}$ to $G=\mathbb{G}/N(r)$ by $\rho$, and the canonical map of $G$ to $G^{\prime}$ by $\pi$ we have $\pi\rho=\rho^{\prime}\pi^{\prime}$. Therefore $\rho$ restricts to an embedding of $H$ into $G$. This gives a restricted Freiheitssatz in the case where $\operatorname{lk}(t)$ is not a clique.

To put the conditions and conclusions of Theorem 1.2 in context we consider circumstances under which the Theorem fails. As in Example 1.7.5, we may salvage some form of Freiheitssatz in the case where $\operatorname{lk}(t)$ is not a clique; so we assume that, for some vertex $t$ of $\operatorname{supp}(s)$, the link of $t$ is a clique. To keep matters simple we also assume that $s=wt$, where $\operatorname{supp}(w)=A\backslash\{t\}$. In this case the conditions of Theorem 1.2 fail if either

1. (i)

$\operatorname{supp}(s)\subset\operatorname{st}(t)$, if and only if $\operatorname{supp}(w)\subset\operatorname{lk}(t)$; or 2. (ii)

$s$ is not a $t$-root; or 3. (iii)

$s$ is not $t$-thick.

Since we have assumed that $\operatorname{supp}(s)=A$ the first of these possibilities, together with the assumption that $\operatorname{lk}(t)$ is a clique, implies that $\mathbb{G}$ is a free Abelian group. In this case we defer to standard results for finitely generated free Abelian groups; and no longer need our Theorems. Moreover, the form $s=wt$ that we have chosen implies that $s$ is not a $t$-root. (Indeed, from the definitions in Section 3, this holds for every $s$ with prime $t$-length.) This leaves the case where $s$ is not $t$-thick. When $s=wt$ is not $t$-thick it follows from Lemma 3.3 that there is a element $b$ which belongs to both $Y=\{y\in A\,:\,\operatorname{st}(y)=A\}$ (the set of central vertices of $\Gamma$) and to $\operatorname{supp}(w)\backslash\operatorname{lk}(t)$. The centre $Z$ of $\mathbb{G}$ is equal to $\langle Y\rangle$ and so, setting $A_{0}=A\backslash Y$ and $\mathbb{G}_{0}=\langle A_{0}\rangle$, we have $\mathbb{G}=\mathbb{G}_{0}\times Z$ and $w=w_{0}w_{z}$, where $w_{0}\in\mathbb{G}_{0}$, $w_{z}\in Z$. As $s\notin\operatorname{st}(t)$, we have $t\in A_{0}$ and, setting $U_{0}=\langle\operatorname{lk}(t)\backslash Y\rangle$, Lemma 3.3 implies that $w_{0}$ is in $\operatorname{Maln}_{\mathbb{G}_{0}}(U_{0})$ so $s_{0}=w_{0}t$ is $t$-thick. Moreover, from the definition in Section 3.3, it follows that $s_{0}$ is not a $t$-root and certainly $s_{0}$ is not in the star of $t$ (in the full subgraph of $\Gamma$ generated by $A_{0}$). The hypotheses of Theorem 1.2 therefore hold for the element $s_{0}$ of $\mathbb{G}_{0}$ and the letter $t$ of $\operatorname{supp}(s_{0})$. Thus $\langle A_{0}\backslash\{t\}\rangle$ embeds in $G_{0}=\mathbb{G}_{0}/N$, where $N$ is the normal closure of $s_{0}^{n}$ in $\mathbb{G}_{0}$, for some $n\geq 3$. In $\mathbb{G}$ we have $r=s^{n}=w_{z}^{n}(w_{0}t)^{n}=w_{z}^{n}s_{0}^{n}$ and so $G=\mathbb{G}/N(r)=G_{0}\times Z/W$, where $W=\langle w_{z}\rangle$. It follows that the subgroup $\langle A_{0}\backslash\{t\}\rangle$ of $\mathbb{G}$ embeds in $G$. Also, using Theorem 1.2 (b), $sN(r)$ has order $n$ in $G$.

From the results of [1], the examples above and this analysis we conclude that a version of the Freiheitssatz holds for many one-relator quotients of partially commutative groups. A precise classification requires a more delicate analysis of the ways in which $s$ may fail to be $t$-thick or a $t$-root (or a proof which avoids some of these conditions). To add some credence to our claim however, we show in Section 5 that Theorem 1.2 applies for almost all cyclically reduced words in $\mathbb{G}(C_{n}^{\prime})$ (as defined in Example 1.7.5).

In Section 2 we review the parts of the theory of partially commutative groups necessary for the paper. In Section 3, we describe small cancellation theory over the HNN-presentation of $\mathbb{G}$, in situations where the there is an element $t$ in $\operatorname{supp}(s)$ such that $\operatorname{lk}(t)$ is a clique. In Section 4 we prove Theorem 1.2 and Theorem 1.4. Finally, in Section 5 we show that Theorem 1.2 is almost always applicable in the situation of Example 1.7.5.

2 Preliminaries

First we recall some of the notation and definitions of [3] and [12]. As above, if $w\in(A\cup A^{-1})^{\ast}$ then $\operatorname{supp}(w)$ is the set of elements $x\in A$ such that $x$ or $x^{-1}$ occurs in $w$. (In [12] $\alpha(x)$ is used instead of $\operatorname{supp}(x)$.) For $u,v\in(A\cup A^{-1})^{*}$ we use $u=v$ to mean $u$ and $v$ are equal as elements of $\mathbb{G}$. Equality of words $u,v$ in $(A\cup A^{-1})^{*}$ is denoted by $u\equiv v$. If $g\in\mathbb{G}$ and $w\in(A\cup A^{-1})^{*}$ is a word of minimal length representing $g$ then we say that $w$ is a minimal form of $g$ (or just that $w$ is minimal).

The Cancellation Lemma, [3, Lemma 4], asserts that if $w$ is a non-minimal word in $(A\cup A^{-1})^{*}$ then $w$ has a subword $xux^{-1}$, where $x\in A\cup A^{-1}$, $u\in(A\cup A^{-1})^{*}$ and $x$ commutes with every letter occurring in $u$. The Transformation Lemma, [3, Lemma 5.5.1] (see also [12, Lemma 2.3]) asserts that, if $u$ and $v$ are minimal and $u=v$ then we can transform the word $u$ into the word $v$ using only commutation relations from $R$ (that is, without insertion or deletion of any subwords of the form $x^{\varepsilon}x^{-\varepsilon},x\in A$). From the Transformation Lemma it follows that if $u$ and $v$ are minimal and $u=v$ then $\operatorname{supp}(u)=\operatorname{supp}(v)$ and $|u|=|v|$ (where $|w|$ denotes the length of the word $w$). Therefore, for any element $g\in\mathbb{G}$ we may define $\operatorname{supp}(g)=\operatorname{supp}(w)$, and the length $l(g)$ of $g$ as $l(g)=|w|$, where $w$ is a minimal form of $g$. For a subset $S$ of $\mathbb{G}$ we define $\operatorname{supp}(S)=\cup_{s\in S}\operatorname{supp}(s)$.

If $g,h\in\mathbb{G}$ such that $l(gh)=l(g)+l(h)$ then we write $gh=g\cdot h$. (Written $g\circ h$ in [12].) It follows that $gh=g\cdot h$ if and only if, for all minimal forms $u$ and $v$ of $g$ and $h$, respectively, $uv$ is a minimal form for $gh$. Clearly if $w\in(A\cup A^{-1})^{*}$ is minimal and $w\equiv uv$ then, in $\mathbb{G}$, $w=u\cdot v$. If $k=g\cdot h$ then we say that $g$ is a left divisor of $k$ (and $h$ is a right divisor of $k$).

We say that $h\in\mathbb{G}$ is cyclically minimal if $l(h)\leq l(h^{g})$, for all $g\in\mathbb{G}$. If $w\in(A\cup A^{-1})^{*}$ and $w\equiv uv$ then we call $vu\in(A\cup A^{-1})^{*}$ a cyclic permutation of $w$.

Lemma 2.1** ([11, Lemma 2.2]).**

Let $w\in(A\cup A^{-1})^{*}$ be a minimal form for an element $h$ of $\Gamma$. Then the following are equivalent.

1. (i)

$h$* is cyclically minimal.* 2. (ii)

If $y\in A\cup A^{-1}$ is a left divisor of $h$ then $y^{-1}$ is not a right divisor of $h$. 3. (iii)

All cyclic permutations of $w$ are minimal forms.

Moreover, if $w$ is cyclically minimal then $\operatorname{supp}(w^{g})\supseteq\operatorname{supp}(w)$, for all $g\in\mathbb{G}$.

If $w$ is a minimal form of a cyclically minimal element $g\in\mathbb{G}$ then we say that $w$ is a cyclically minimal form. From, for example, Proposition 3.9 of [9], if $g\in\mathbb{G}$ then there exist $u,w\in(A\cup A^{-1})^{*}$, with $w$ a cyclically minimal form, such that $g=u^{-1}\cdot w\cdot u$. Thus $g$ is cyclically minimal if and only if $u=1$. Observe that if $g$ is cyclically minimal then $l(g^{n})=nl(g)$. Therefore partially commutative groups are torsion free.

Lemma 2.2**.**

Let $U$ be a subgroup of $\mathbb{G}$ and let $D$ be a set of double coset representatives of $U$ in $\mathbb{G}$. The following are equivalent.

1. (i)

If $d\in D$ and $g\in UdU$ then $l(g)\geq l(d)$. 2. (ii)

If $d\in D$ then $d$ has no left or right divisor in $U$.

There exists a unique set $D$ of double coset representatives of $U$ in $\mathbb{G}$ such that i holds and this set satisfies

1. (iii)

$1\in D$* and* 2. (iv)

if $d\in D$ then $d^{-1}\in D$.

In particular, if $UdU=Ud^{-1}U$, for some $d\in D$, then $d=1$.

Proof.

For each $g\in\mathbb{G}$ we may choose a minimal form $d\in\mathbb{G}$ such that $l(d)\leq l(w)$, for all $w\in UgU$. Thus there exists a set $D$ of double coset representatives having property i. Suppose $D$ is such a set and $d\in D$. Then $d$ satisfies ii, since it has minimal length amongst elements of $UdU$. If $d\in D$, $g\in UdU$ and $l(g)=l(d)$ then $g=udv$, for some $u,v\in U$. In fact, as $d$ has no left or right divisor in $U$ we may choose such $u$ and $v$ so that $g=u\cdot d\cdot v$. Unless $u=v=1$ this implies that $l(g)>l(d)$, and if $u=v=1$ then we have $g=d$. Therefore $d$ is the unique element of minimal length in $UdU$; and $D$ is uniquely determined by condition i. On the other hand, if $D^{\prime}$ is a set of double coset representatives satisfying ii a similar argument shows that $D^{\prime}$ satisfies i.

Assume now that $D$ satisfies i. Then (iii), a special case of i, also holds. If $d\in D$ and $Ud^{-1}U$ contains an element $g$ such that $l(g)\leq l(d^{-1})$ then $g^{-1}\in UdU$ and $l(g^{-1})\leq l(d)$; so from the above $g=d^{-1}$. It follows that if $d\in D$ then so is $d^{-1}$, and so (iv) holds. Moreover, since $d\neq d^{-1}$, we have $UdU\cap Ud^{-1}U=\emptyset$, for all $d\in D\backslash\{1\}$. ∎

The elements of $A$ are termed the canonical generators of $\mathbb{G}(\Gamma)$ and a subgroup of $\mathbb{G}$ generated by a subset $Y$ of $\mathbb{G}$ is called a canonical parabolic subgroup. Let $Y\subseteq A$ and denote by $\Gamma_{Y}$ the full subgraph of $\Gamma$ with vertex set $Y$. Let $\mathbb{G}(\Gamma_{Y})$ be the partially commutative group with commutation graph $\Gamma_{Y}$. It follows from the Transformation Lemma, [3, Lemma 5.5.1], that $\mathbb{G}(\Gamma_{Y})=\langle Y\rangle$, the canonical parabolic subgroup of $\mathbb{G}$ generated by $Y$.

For cyclically minimal elements $g,h\in\mathbb{G}$ we define $g\sim_{0}h$ if $g=u\cdot v$ and $h=v\cdot u$, for some $u,v\in\mathbb{G}$. Then let $\sim$ be the transitive closure of $\sim_{0}$ and denote by $[g]$ the equivalence class of $g$ under the equivalence relation $\sim$ on cyclically minimal elements. The following appears in [11, Corollary 2.4] (where the set $[g]$ is incorrectly defined).

Corollary 2.3**.**

Let $w,g$ be (minimal forms of) elements of $\mathbb{G}$ and let $w=u^{-1}\cdot v\cdot u$, where $v$ is cyclically minimal. Then there exist minimal forms $a$, $b$, $c$, $d_{1}$, $d_{2}$ and $e$ such that $g=a\cdot b\cdot c\cdot d_{2}$, $u=d_{1}\cdot a^{-1}$, $d=d_{1}\cdot d_{2}$, $w^{g}=d^{-1}\cdot e\cdot d$, $[e]=[v]$, $e=v^{b}$, $\operatorname{supp}(b)\subseteq\operatorname{supp}(v)$ and $[\operatorname{supp}(b\cdot c),\operatorname{supp}(d_{1})]=[\operatorname{supp}(c),\operatorname{supp}(v)]=1$.

The next result is a direct consequence of this lemma.

Corollary 2.4**.**

Let $Y$ be a subset of $A$ and let $w,g$ be (minimal forms of) elements of $\mathbb{G}$ such that $w$ and $w^{g}$ belong to $\langle Y\rangle$ and $g$ has no right or left divisor in $\langle Y\rangle$. Then $[\operatorname{supp}(g),\operatorname{supp}(w)]=1$. (That is $[x,y]=1$, for all $x\in\operatorname{supp}(g)$ and $y\in\operatorname{supp}(w)$.)

Proof.

In the notation of Corollary 2.3 we have $u,v,d\in\langle Y\rangle$ since $w$ and $w^{g}$ are in $\langle Y\rangle$. Hence $a,d_{1},d_{2}$ and $b$ are in $\langle Y\rangle$. As $g$ has no left or right divisors in $\langle Y\rangle$ it follows that $g=c$ and $u=d_{1}$. To complete the proof we use the fact that $[\operatorname{supp}(b\cdot c),\operatorname{supp}(d_{1})]=[\operatorname{supp}(c),\operatorname{supp}(v)]=1$. ∎

Corollary 2.5**.**

Let $Y\subseteq A$ be a clique and let $w,g$ be (minimal forms of) elements of $\mathbb{G}$ such that $w$ and $w^{g}$ belong to $\langle Y\rangle$. Then $[\operatorname{supp}(g),\operatorname{supp}(w)]=1$.

Proof.

If $g\in\langle Y\rangle$, the result holds as $Y$ is a clique. Otherwise we may write $g=a\cdot b\cdot c$, where $a,c\in\langle Y\rangle$ and $b$ has no left or right divisor in $\langle Y\rangle$. Then $w^{g}=w^{b\cdot c}\in\langle Y\rangle$. Again, as $Y$ is a clique and $c\in\langle Y\rangle$, we have $w^{b\cdot c}=w^{b}$. From Corollary 2.4, $[\operatorname{supp}(b),\operatorname{supp}(w)]=1$, and as $Y$ is a clique, $[\operatorname{supp}(g),\operatorname{supp}(w)]=1$, as claimed. ∎

For a subset $U$ of $A$ denote by $\Gamma_{U}$ the full subgraph of $\Gamma$ generated by $U$ and for a cyclically minimal word $w$ over $\mathbb{G}$ set $\Gamma_{w}=\Gamma_{\operatorname{supp}(w)}$. For a subset $Y\subset A$ define $\operatorname{lk}(Y)=\cap_{y\in Y}\operatorname{lk}(y)$ and $\operatorname{st}(Y)=\cap_{y\in Y}\operatorname{st}(y)$. For an element $w\in\mathbb{G}$ define $\operatorname{lk}(w)=\operatorname{lk}(\operatorname{supp}(w))$ and $\operatorname{st}(w)=\operatorname{st}(\operatorname{supp}(w))$. From [3, Korollar 3], for $a\in A$ we have $C_{\mathbb{G}}(a)=\langle a\rangle\times\langle\operatorname{lk}(a)\rangle$.

As necessary we shall use a normal form for elements of the partially commutative group $\mathbb{G}(\Gamma)$, which we now define. Let $\Delta$ be the complement of $\Gamma$ ($\Delta$ has the same vertex set as $\Gamma$ and $\{u,v\}$ is an edge of $\Delta$ if and only if $\{u,v\}$ is not an edge of $\Gamma$). For any element $w$ of $\mathbb{G}$ define $\Delta(w)$ to be the full subgraph of $\Delta$ with vertices $\operatorname{supp}(w)$. If $\Delta(w)$ is connected we call $w$ a block. If $\Delta(w)$ has connected components $\Delta_{1},\ldots,\Delta_{k}$ then it follows that $w=w_{1}\cdot\,\cdots\,\cdot w_{k}$, where $\Delta(w_{i})=\Delta_{i}$ and $[\operatorname{supp}(w_{i}),\operatorname{supp}(w_{j})]=1$, for $i\neq j$. We call this factorisation of $w$ the block decomposition of $w$.

In the sequel, we shall use the notation and results of this section without further mention.

3 Partially commutative groups as HNN-extensions

In this section we review the special case of diagrams over HNN-extensions we need in the proof of Theorem 1.2. We refer the reader to [21, Pages 291–292] for basic results on diagrams over HNN-extensions and to [19] for a more general version of the definitions given here. As above, let $\mathbb{G}$ be a partially commutative group with commutation graph $\Gamma$ and canonical presentation $\langle A|R\rangle$. To realise $\mathbb{G}$ as an HNN-extension, given any $t\in A$, set $\mathbb{G}_{t}=\langle A\backslash\{t\}\rangle$ and define the HNN-extension

[TABLE]

By definition HNN$(t)$ is the group with presentation $\langle A\,|\,R_{t}\cup\{t^{-1}ut=u,u\in\langle\operatorname{lk}(t)\rangle\}\rangle$, where $R_{t}=\{[x,y]\in R\,|\,x\neq t\textrm{ and }y\neq t\}$. We may perform Tietze transformations on the latter to replace it with the presentation $\langle A\,|\,R_{t}\cup\{t^{-1}xt=x,x\in\operatorname{lk}(t)\}\rangle$. As $R_{t}\cup\{t^{-1}xt=x,x\in\operatorname{lk}(t)\}=R$, it follows that HNN$(t)=\mathbb{G}$. We call HNN$(t)$ the HNN-presentation of $\mathbb{G}$ with respect to $t$.

For notational simplicity, write

• •

$F$ for the group $\mathbb{G}=\operatorname{HNN}(t)$ with the HNN-presentation with respect to $t$,

• •

$H=\mathbb{G}_{t}$ and $U=\langle\operatorname{lk}(t)\rangle$, and in addition

• •

let $D$ be a set of double coset representatives for $U$ in $H$, satisfying the properties of Corollary 2.2, and let $\sigma$ be the function from $H$ to $D$ given by $\sigma(g)=d$, where $UgU=UdU$.

(We assume that all elements of $F$ are represented as reduced words of the free product $H*\langle t\rangle$, unless an explicit exception is made.) Given $p\in F$, the factorisation

[TABLE]

where $g_{i}\in H$ and $\varepsilon_{i}\in\{\pm 1\}$, is called reduced, over $F$, if either $n\leq 1$ or $n>1$ and, for $1\leq i\leq n-1$, if $g_{i}\in U$ then $\varepsilon_{i}\varepsilon_{i+1}=1$. In this case we say that $p$ has $t$-length

[TABLE]

Every element of $F$ has a reduced factorisation, and two reduced factorisations which represent the same element have the same $t$-length. We explicitly allow reduced words to contain sub-words of the form $t^{\varepsilon}ut^{\varepsilon}$, where $u\in U$ and $\varepsilon=\pm 1$.

If $u$ and $v$ are elements of $F$ such that $|uv|_{t}=|u|_{t}+|v|_{t}$, then we say the product $uv$ is reduced (if and only if, given reduced factorisations

[TABLE]

either $m+n\leq 1$ or $g_{m}h_{0}\notin U$ or $\varepsilon_{m}=\delta_{1}$). More generally, if $u_{1},\ldots,u_{n}$ are reduced factorisations of elements of $F$ such that

[TABLE]

we say that the product $u_{1}\cdots u_{n}$ is reduced. An element $p$ of $F$ is said to be cyclically reduced if the product $pp$ is reduced. (That is, given a reduced factorisation $p=g_{0}t^{\varepsilon_{1}}\cdots t^{\delta_{n}}g_{n}$, either $n\leq 1$ or $g_{n}g_{0}\notin U$ or $\varepsilon_{n}=\varepsilon_{1}$.) Again, for $u_{1},\ldots,u_{n}$ as above, the product $u_{1}\cdots u_{n}$ is said to be cyclically reduced if it is reduced and $u_{n}u_{1}$ is reduced.

The aim of the definitions in the remainder of this sub-section is to allow us to avoid products which split elements of $H$ in a non-trivial fashion. For example, if $a,b$ and $g$ are elements of $H\backslash U$ with $g=ab$, and we have $s=tgt$ then the factorisation $pq$, where $p=ta$ and $q=bt$, splits the factor $g=ab$. With this in mind, let the element $p\in F$ have reduced factorisation (3.2). If $g_{0}=1$ then $p$ is said to begin with a $t$-letter and if $g_{n}=1$ then $p$ is said to end with a $t$-letter. Now if $u_{1},\ldots,u_{k}$ are reduced factorisations of elements of $F$ then the product $u_{1}\cdots u_{k}$ is said to be right integral at $i$, where $1\leq i\leq k$, if it is cyclically reduced and $u_{i}$ ends with a $t$-letter and left integral at $i$ if $u_{i+1}$ begins with a $t$-letter (subscripts modulo $k$). The product $u_{1}\cdots u_{k}$ is integral at $i$ if it is either left or right integral at $i$. (Note that this is not quite the same as the definition in [19].) The product $u_{1}\cdots u_{k}$ is said to be (right) integral if it is (right) integral at $i$, for all $i\in\{1,\ldots,k\}$. We say that $u$ is a (right) integral subword of $w$ if $w=u\cdot v$, where $uv$ is a (right) integral factorisation. [Note that the integral property depends on the choice of reduced factorisation of factors. For example, $w=tuht$, where $u\in U$, $h\in H\backslash U$, can be written as the product $pq$, where $p=tu$ and $q=ht$. This product is not integral, but we may write $p=ut$ in $F$ and using this factorisation of $p$ the product $pq$ is integral.]

3.1 Thick subwords of HNN$(t)$

Let $G$ be a group and $K$ a non-trivial subgroup of $G$. Define $\operatorname{Maln}_{G}(K)=\{x\in G\,:\,x^{-1}Kx\cap K=\{1\}\}$. It follows that $x\in\operatorname{Maln}_{G}(K)$ if and only if $x^{-1}\in\operatorname{Maln}_{G}(K)$. Also, $x\in\operatorname{Maln}_{G}(K)$ if and only if $KxK\subseteq\operatorname{Maln}_{G}(K)$. Indeed, if $x\in\operatorname{Maln}_{G}(K)$ and $u,v\in K$ then $(uxv)^{-1}K(uxv)\cap K=v^{-1}x^{-1}Kxv\cap K=v^{-1}x^{-1}Kxv\cap v^{-1}Kv=v^{-1}(x^{-1}Kx\cap K)v=\{1\}$.

Lemma 3.1**.**

Let $V$ be a subgroup of the group $\mathbb{G}$ and let $D_{V}$ be a set of representatives of double cosets $VgV$ in $\mathbb{G}$, satisfying the conditions of Corollary 2.2. If $a,b\in\operatorname{Maln}_{\mathbb{G}}(V)$ and $u,v\in V$ are such that $aub^{-1}=v$ then $VaV=VbV$ and there exist $d\in D_{V}$, $u^{\prime},v^{\prime}\in V$ such that $a=(vv^{\prime})\cdot d\cdot u^{\prime}$ and $b=v^{\prime}\cdot d\cdot(u^{\prime}u)$.

Proof.

Note first that $x\in\operatorname{Maln}_{\mathbb{G}}(V)$ if and only if $VxV\subseteq\operatorname{Maln}_{\mathbb{G}}(V)$. As $aub^{-1}\in V$ we have $VaV=VbV$. Let $d$ the element of $D_{V}$ such that $VaV=VdV=VbV$. Then there exist $a_{l},b_{r},u^{\prime},v^{\prime}\in V$ such that $a=a_{l}\cdot d\cdot u^{\prime}$ and $b=v^{\prime}\cdot d\cdot b_{r}$. By assumption

[TABLE]

and since $a,b\in\operatorname{Maln}_{\mathbb{G}}(V)$ we have $d\in\operatorname{Maln}_{\mathbb{G}}(V)$, so $u^{\prime}ub_{r}^{-1}=a_{l}^{-1}vv^{\prime}=1$, from which the final statement follows. ∎

Definition 3.2**.**

Let $w=g_{0}t^{\varepsilon_{1}}\cdots g_{k-1}t^{\varepsilon_{k}}g_{k}$ be a reduced factorisation of an element of $HNN(t)$, where $g_{i}\in\mathbb{G}_{t}$ and $\varepsilon_{i}=\pm 1$, and let $U=\langle\operatorname{lk}(t)\rangle\leq\mathbb{G}_{t}$. We say $w$ is $t$-thick if $g_{i}\in U\cup\operatorname{Maln}_{\mathbb{G}_{t}}(U)$, for $i=0,\ldots,k$. We say $w$ is cyclically $t$-thick if $w$ is cyclically reduced, $t$-thick and $g_{k}g_{0}\in U\cup\operatorname{Maln}_{\mathbb{G}_{t}}(U)$.

The notion of $t$-thickness is well-defined since, if $w$ has another reduced factorisation $w=h_{0}t^{\delta_{1}}\cdots h_{m-1}t^{\delta_{m}}h_{m}$, where $\delta_{i}=\pm 1$, then $m=k$, $\varepsilon_{i}=\delta_{i}$ and there are elements $u_{0},\ldots,u_{m-1}\in U$ such that $g_{0}=h_{0}u_{0}^{-1}$, $g_{m}=u_{m-1}h_{m}$ and $g_{i}=u_{i-1}h_{i}u_{i}^{-1}$, for $1\leq i\leq m-1$. Hence, from the comment preceding Lemma 3.1, $h_{i}\in U\cup\operatorname{Maln}_{\mathbb{G}_{t}}(U)$ if and only if $g_{i}\in U\cup\operatorname{Maln}_{\mathbb{G}_{t}}(U)$.

Lemma 3.3**.**

Let $\mathbb{G}=\langle A\,|\,R\rangle$ be a partially commutative group, let $B$ be a subset of $A$ and let $K=\langle B\rangle$. If $B$ is a clique and $w$ is an element of $\mathbb{G}\backslash K$ then the following are equivalent.

1. (i)

$w$* belongs to $\operatorname{Maln}_{\mathbb{G}}(K)$.* 2. (ii)

For all $b\in B$, there exists $x\in\operatorname{supp}(w)\backslash B$ such that $[x,b]\neq 1$.

Proof.

To see that (i) implies (ii), assume $w\in\operatorname{Maln}_{\mathbb{G}}(K)$ and let $b\in B$. Then $w^{-1}bw\neq b$, so $w\notin C_{\mathbb{G}}(b)$ and it follows that there exists $x\in\operatorname{supp}(w)\backslash B$ such that $[x,b]\neq 1$.

To prove (ii) implies (i), consider $w\in\mathbb{G}\backslash K$ such that $w\notin\operatorname{Maln}_{\mathbb{G}}(K)$. Assume, in order to obtain a contradiction, that $w$ is of minimal length amongst all elements not in $K\cup\operatorname{Maln}_{\mathbb{G}}(K)$ which satisfy (ii). As $w\notin\operatorname{Maln}_{\mathbb{G}}(K)$, there exists $1\neq v\in K$ such that $w^{-1}vw\in K$. We may write $w=w_{0}\cdot w_{1}$, where $w_{0}\in K$ and $w_{1}$ is non-trivial and has no (non-trivial) left divisor in $K$. By minimality of $w$, we conclude that $w_{0}=1$, and so we may assume $w$ has no left divisor in $K$. As $w$ has no left divisor in $K$, (from the Cancellation Lemma) $w^{-1}vw=w^{-1}\cdot v\cdot w$, unless $w$ has a left divisor $x\in A^{\pm 1}$ such that $[x,v]=1$. In the latter case, $x$ must commute with all elements of $\operatorname{supp}(v)$. We may write $w=x\cdot w_{2}$ and, by the assumption on $w$, $\operatorname{supp}(w_{2})$ must contain an element which does not commute with an element of $\operatorname{supp}(v)$; so $w_{2}\notin K$. Moreover $w_{2}^{-1}vw_{2}=w^{-1}vw\in K$. It follows that $w_{2}$ is shorter than $w$ and satisfies all the same properties, contrary to the choice of $w$. We conclude that $w^{-1}vw=w^{-1}\cdot v\cdot w$, so that $w\in K$, a contradiction. Therefore, for all $w$ not in $K\cup\operatorname{Maln}_{\mathbb{G}}(K)$ property (ii) fails, as required. ∎

3.2 Diagrams over HNN extensions

In this section we follow [21, Chapter V, Section 11]. Assume that $\mathbb{G}$ is expressed as the HNN-extension $F=\operatorname{HNN}(t)$, and set $H=\mathbb{G}_{t}$ and $U=\langle\operatorname{lk}(t)\rangle$, as above. A set $\mathcal{R}$ of elements of $F$ is said to be symmetrised if every element of $\mathcal{R}$ is cyclically reduced and, for all $r\in\mathcal{R}$, all cyclically reduced conjugates of $r$ and $r^{-1}$ are in $\mathcal{R}$. The symmetrised closure $\widetilde{\mathcal{S}}$ of a set $\mathcal{S}$ of cyclically reduced elements of $F$ is the smallest symmetrised subset containing $\mathcal{S}$; and consists of all cyclically reduced conjugates of $s$, for all elements $s\in\mathcal{S}\cup\mathcal{S}^{-1}$. From Collins’ Lemma (see [21, Theorem 2.5, p. 185]) if $r\in F$ is cyclically reduced and ends in $t^{\pm 1}$ then a cyclically reduced element $s\in F$, ending in $t^{\pm 1}$, is a conjugate of $r$ if and only if it may be obtained by taking a cyclic permutation of $r$ and then conjugating by an element of $U$. This means that even though $\mathcal{R}$ may be finite its symmetrised closure $\widetilde{\mathcal{R}}$ is, in general, infinite.

Definition 3.4**.**

Let $\widetilde{\mathcal{R}}$ be a symmetrised subset of $F$. An element $p\in F$ is a piece (over $\widetilde{\mathcal{R}}$) if there exist distinct elements $r_{1}$, $r_{2}\in\widetilde{\mathcal{R}}$, and elements $u_{1},u_{2}$ of $F$, such that $r_{i}$ factors as a reduced product $r_{i}=_{F}pu_{i}$, for $i=1$ and $2$.

Definition 3.5**.**

Let $\widetilde{\mathcal{R}}$ be a symmetrised subset of $F$ and let $m$ be a positive integer. If $w\in\widetilde{\mathcal{R}}$ has a reduced factorisation $w=p_{1}\cdots p_{k}$, where $p_{i}$ is a piece over $\widetilde{\mathcal{R}}$, then $w$ is said to have a $k$-piece factorisation. If no element of $\widetilde{\mathcal{R}}$ has a $k$-piece factorisation, where $k<m$, then $\widetilde{\mathcal{R}}$ is said to satisfy small cancellation condition $C(m)$.

For details of disc diagrams over HNN-extensions we refer the reader to [21, pp. 291–294], and the references therein. We outline here only what we need below, in particular defining diagrams on a disk. Let $\widetilde{\mathcal{R}}$ be a symmetrised subset of $F$ and $w$ a reduced factorisation of an element of $F$. An $\widetilde{\mathcal{R}}$ diagram with boundary label $w$ over $F$ consists of the following. A finite $2$-complex $M$ with underlying space $\Sigma$ a compact, connected, simply connected, subset of the real plane; and a distinguished vertex $O$ of $M$ on $\partial\Sigma$. ([math]-cells, $1$-cells and $2$-cells of $M$ are called vertices, edges and regions, respectively.) A labelling function $\phi$ from oriented edges of $M$ to $H\cup\{t^{\pm 1}\}$. (Strictly speaking $\phi$ maps edges of $M$ to freely reduced words in $(A\backslash\{t\})^{\pm 1}$ or elements of $\{t^{\pm 1}\}$.) For an oriented edge $a$ we write $\bar{a}$ for the same edge given the opposite orientation (if $f:[0,1]\rightarrow a$ is a homeomorphism determining the oriented edge $a$ then $\bar{a}$ is the edge determined by the map $\bar{f}$ mapping $x\in[0,1]$ to $f(1-x)$.) A boundary cycle of a region $\Delta$ of $M$ is a closed path traversing the boundary $\partial\Delta$ of $\Delta$ exactly once (beginning and ending at a vertex $v$ of $M$). A boundary cycle of $M$ is a closed path traversing the boundary $\partial\Sigma$ of $\Sigma$ exactly once, beginning and ending at $O$. In addition the following conditions must be satisfied.

1. (1)

If $a$ is an oriented edge with label $w=\phi(a)$ then $\phi(\bar{a})=w^{-1}$. 2. (2)

If $M$ has a boundary cycle $p=a_{1},\ldots,a_{n}$, beginning and ending at $O$, then the product $\phi(a_{1})\cdots\phi(a_{n})$ is reduced and equal, in $F$, to $w$ or $w^{-1}$. 3. (3)

If $\Delta$ is a region of $M$ and $\Delta$ has a boundary cycle $p=a_{1},\ldots,a_{n}$, then the product $\phi(a_{1})\cdots\phi(a_{n})$ is reduced and equal in $F$ to an element $\widetilde{\mathcal{R}}$.

It follows (see for example [21, Theorem 11.5, p. 292]) that there exists a diagram $\mathcal{D}$ over $F$ with boundary label $w$ if and only if $w=1$ in $F/N$, where $N$ is the normal closure of $\mathcal{R}$ in $F$.

If the label $\phi(a)$ of an edge $a$ is in $H$ then $a$ is called an $H$-edge, and if $\phi(a)=t^{\pm 1}$ then $a$ is called a $t$-edge. If $v$ is a vertex in the boundary $\partial\Delta$ of a region $\Delta$ and $v$ is incident to a $t$-edge of $\partial\Delta$, then we call $v$ a primary vertex, with respect to $\Delta$. The products $\phi(a_{1})\cdots\phi(a_{n})$ appearing in 2 and 3 are called boundary labels of $M$ and $\Delta$ respectively. In our case $\widetilde{\mathcal{R}}$ is the symmetrised closure of the single element $r$, which is cyclically minimal as an element of the partially commutative group $\mathbb{G}$. Every element $w$ of $\widetilde{\mathcal{R}}$ is therefore equal in $F$ to $u^{-1}r^{\prime}u$, for some $u\in U$ and cyclic permutation $r^{\prime}$ of $r$ or $r^{-1}$. Let $\Delta$ be a region with boundary label equal in $F$ to $r^{\prime}$, when read from an appropriate vertex $o$ on its boundary. Attaching an edge $e$ labelled $u$ by identifying its initial vertex with $o$, we obtain a diagram with a single region $\Delta$ and boundary label $w$. As $r^{\prime}$ is cyclically minimal we may replace 3 with

1. (3’)

If $\Delta$ is a region of $M$ and $\Delta$ has a boundary cycle $p=a_{1},\ldots,a_{n}$, then the product $\phi(a_{1})\cdots\phi(a_{n})$ is reduced and equal in $F$ to a cyclic permutation of $r$ or $r^{-1}$.

As usual we shall restrict to diagrams which do not have pairs of redundant regions of the following sort. Let $\Delta_{1}$ and $\Delta_{2}$ be distinct regions of the diagram $\mathcal{D}$ with boundary cycles $\rho_{1}=\mu\nu_{1}$ and $\rho_{2}=\mu\nu_{2}$ (where $\mu,\nu_{i}$ are subpaths of $\rho_{i}$) meeting in the connected boundary component $\mu$. If $\phi(\nu_{1})\equiv\phi(\nu_{2})$ (as words in the free monoid $(A\cup A^{-1})^{*}$) then, as in Figure 3, we may remove the interior of $\Delta_{1}\cup\Delta_{2}$ and identify the sub-paths $\nu_{1}$ and $\nu_{2}$ of the boundary cycles of $\Delta_{1}$ and $\Delta_{2}$, via their (equal) labels $\phi(\nu_{1})$ and $\phi(\nu_{2})$, to leave a new diagram $\mathcal{D}^{\prime}$ with the same reduced boundary label as $\mathcal{D}$ but fewer regions. The modification of $\mathcal{D}$ to produce $\mathcal{D}^{\prime}$ is called a cancellation of regions. A diagram in which no cancellation of regions is possible is called reduced.

It turns out that if a pair of distinct regions $\Delta_{1}$ and $\Delta_{2}$ have boundary cycles $\rho_{1}=\mu\nu_{1}$ and $\rho_{2}=\mu\nu_{2}$, as above, but instead of satisfying $\phi(\nu_{1})\equiv\phi(\nu_{2})$ the labels satisfy only $\phi(\nu_{1})=_{F}\phi(\nu_{2})$ then, after some minor modifications to the diagram, which do not alter the label of its boundary, we may again cancel the regions $\Delta_{1}$ and $\Delta_{2}$. The general process of modification is described in [21, Page 292]. Here we describe modifications sufficient for our particular case. There are two types of these; the first of which results in a new diagram in which the product of edge labels around a boundary cycle may not be freely reduced. The second type may then be used to freely reduce labels on edges of boundary cycles if necessary.

Shuffling labels.

Suppose that a subpath $\mu$ of the boundary of a region $\Delta$ has label with subword $xy$, where $x,y\in A^{\pm 1}$ and $xy=yx$. Then we may modify the diagram $\mathcal{D}$ (without altering the element of $F$ represented by its boundary label) to obtain a new diagram in which this subword $xy$ is replaced by $yx$, as follows. If $xy$ occurs as a subword of the label $\phi(e)=axyb$ of a single edge $e$ of $\mu$ then we modify $\phi$ by setting $\phi(e)=ayxb$. Otherwise $\mu$ contains a subpath $e_{1}e_{2}$, where $e_{1}$ and $e_{2}$ are edges, such that $\phi(e_{1})=ax$ and $\phi(e_{2})=yb$. In this case $e_{1}\subseteq\Delta\cap\Delta^{\prime}$ and $e_{2}\subseteq\Delta\cap\Delta^{\prime\prime}$, for some regions $\Delta^{\prime}$ and $\Delta^{\prime\prime}$ and we modify $\mathcal{D}$ as shown in Figure 4. There is a choice here: we may label the new edge, in $\Delta^{\prime}\cap\Delta^{\prime\prime}$ with $yx^{-1}$, as shown, or with $x^{-1}y$. In both cases we refer to this modification as a shuffle of the label of $\Delta$.

Free reduction of labels.

Let $\mathcal{D}$ be an $\widetilde{\mathcal{R}}$-diagram over $F$ and let $\Delta$ be a region of $\mathcal{D}$. Suppose a subpath $\mu$ of a boundary cycle of $\Delta$ has label containing a subword $xx^{-1}$, where $x\in A^{\pm 1}$. If $xx^{-1}$ is a subword of the label $\phi(e)=axx^{-1}b$ of a single edge $e$, where $a,b\in H$, then we modify $\phi$ by setting $\phi(e)=ab$. (Such edges do not occur in $\mathcal{D}$ itself, but may arise after a shuffling of labels as above.) Otherwise $\mu$ contains a subpath $e_{1}e_{2}$ of two edges such that $\phi(e_{1})=ax$ and $\phi(e_{2})=x^{-1}b$, where $a,b\in H$ (and if $x=t^{\pm 1}$ then necessarily $a=b=1$, since $\phi(e_{i})$ is by definition either an element of $H$ or of $\{t^{\pm 1}\}$). In this case we modify $\mathcal{D}$ as shown in Figure 5. We call this free reduction of the label of $\Delta$. Note that this reduces the sum of lengths of boundary labels of regions of $\mathcal{D}$, where length here means length as a word over $A\cup A^{-1}$, so we may repeat free reductions of labels until no such subpaths occur in the boundary label of any region.

Note that the boundary label of $\mathcal{D}$ is unaffected by such modifications. Both free reduction and shuffling of labels of $\mathcal{D}$ result in a new diagram $\mathcal{D}^{\prime}$, which has regions in one to one correspondence with the regions of $\mathcal{D}$. If $\Delta_{1}$ and $\Delta_{2}$ are regions of $\mathcal{D}$ with a common boundary component $\mu$ and boundary cycles $\mu\nu_{1}$ and $\mu\nu_{2}$, respectively, such that $\phi(\nu_{1})=_{F}\phi(\nu_{2})$ then these modifications may be used alter the diagram so that $\phi(\nu_{1})\equiv\phi(\nu_{2})$; and $\Delta_{1}$ and $\Delta_{2}$ become cancelling regions. We say a diagram $\mathcal{D}$ is strongly reduced if it satisfies the condition that, whenever $\Delta_{1}$ and $\Delta_{2}$ are distinct regions with a common boundary component $\mu$, of positive length, the label $\phi(\mu)$ is a piece. From [21, Theorem 11.5] we obtain the following.

Proposition 3.6**.**

Let $\mathcal{R}$ be a subset of $F$ and $\widetilde{\mathcal{R}}$ its symmetric closure. There exists a strongly reduced $\widetilde{\mathcal{R}}$ diagram $\mathcal{D}$ over $F$ with boundary label $w$ if and only if $w=1$ in $F/N$, where $N$ is the normal closure of $\mathcal{R}$ in $F$.

3.3 Reduction to the free product

In this section we specialise the methods of [19] to the case in hand. As above, let $D$ be a set of double coset representatives of $U$ in $H$ satisfying the properties of Corollary 2.2, and let $\sigma:H\rightarrow D$ be the function mapping $h\in H$ to $d\in D$ such that $UhU=UdU$. Partition the set $D$ into disjoint subsets $D^{+}$, $D^{-}$ and $\{1_{H}\}$ such that $d\in D^{+}$ if and only if $d^{-1}\in D^{-}$. Now define the free product $F_{D}=\mathbb{F}(D^{+})\ast\langle t\rangle$, where $\mathbb{F}(D^{+})$ is the free group on $D^{+}$.

The canonical map from $D^{+}$ to $\mathbb{F}(D^{+})$ extends to an injective map from $D$ to $\mathbb{F}(D^{+})$ by mapping $d^{-1}$ in $D^{-}$ to $d^{-1}\in\mathbb{F}(D^{+})$, for all $d\in D^{+}$, and mapping $1_{H}$ to the empty word. Composing this map with the canonical injection from $\mathbb{F}(D^{+})$ into $F_{D}$ we have an injective map $\iota$ from $D$ to $F_{D}$. The composition $\iota\circ\sigma$ is then a map from $H$ to $F_{D}$, which we shall now also refer to as $\sigma$. We extend $\sigma$ to a function from $F$ to $F_{D}$ as follows. If $p\in F$ has reduced factorisation $p=g_{0}t^{\varepsilon_{1}}\cdots t^{\varepsilon_{n}}g_{n}$ then we define

[TABLE]

If $g,h\in H$ and $u\in U$ are such that $h=ug$ then $\sigma(g)=\sigma(h)$, so $\sigma$ is a well-defined map from $F$ to $F_{D}$. To simplify notation we may write $\bar{p}$ for $\sigma(p)$.

Definition 3.7**.**

If $p\in F$ is such that $\sigma(p)$ is a not a proper power in $F_{D}$ then we say that $p$ is a $t$-root.

Note that a cyclically reduced element of $F$ of $t$-length at least $1$, which ends in a $t$ letter and which is a $t$-root; cannot be a proper power in $F$. Indeed, suppose that $p$ is a cyclically reduced element of $F$ ending in a $t$-letter, with $|p|_{t}\geq 1$, and $p=q^{n}$, for some element $q\in F$ and positive integer $n$. It follows, from [21, Chapter IV, Section 2], that $q$ is also cyclically reduced, of positive $t$-length and ends in a $t$-letter. Therefore $\sigma(p)=\sigma(q^{n})=\sigma(q)^{n}$, so $p$ is not a $t$-root. (Note that $\sigma$ is not a group homomorphism.)

Let $n\geq 2$ be a positive integer and let $s$ be a cyclically reduced word of non-zero $t$-length, $s=h_{0}t^{\varepsilon_{1}}\cdots h_{m-1}t^{\varepsilon_{m}}$, where $h_{i}\in\mathbb{G}_{t}$ and $\varepsilon_{i}\in\{\pm 1\}$, for $i=1,\ldots,m-1$. Let $r=s^{n}$, let $N$ be the normal closure of $\mathcal{R}=\{r\}$ in $\mathbb{G}$ and let $G=\mathbb{G}/N$. If $r_{1}$ and $r_{2}$ are elements of the symmetrised closure $\widetilde{\mathcal{R}}$, of $\mathcal{R}$, such that $\sigma(r_{1})=\sigma(r_{2})$ we say that $r_{1}$ and $r_{2}$ are in periodic position.

Lemma 3.8**.**

Assume $s$ has $t$-length $m\geq 1$, ends in a $t$-letter, is cyclically $t$-thick and a cyclically reduced $t$-root. Let $r=s^{n}$, for some $n\geq 1$ and $\mathcal{R}=\{r\}$. If $r_{1}$ and $r_{2}$ are cyclic permutations of $r$ or $r^{-1}$ such that $r_{1}=pq_{1}$ and $r_{2}=pq_{2}$ and $r_{1}$ and $r_{2}$ are in periodic position, then $\phi(q_{1})=_{F}\phi(q_{2})$.

Proof.

We may assume that $r_{1}=r$ and $r_{2}$ is obtained from $r^{\varepsilon}$, where $\varepsilon=\pm 1$, by cyclic permutation. If $\varepsilon=-1$ then $\sigma(r_{2})$ is a cyclic permutation of $\sigma(r_{1}^{-1})=\sigma(r_{1})^{-1}$ and $\sigma(r_{1})=\sigma(r_{2})$ in $F_{D}$. This implies $\sigma(r_{1})=1$, so $r_{1}\in U$, a contradiction. Therefore $\varepsilon=1$ and as $s$ is a $t$-root it follows that $r_{2}$ is obtained from $r$ by a cyclic permutation of length $k|s|=2km$, for some positive integer $k$. Hence $r_{2}=_{F}r$ and so $\phi(q_{1})=_{F}\phi(q_{2})$. ∎

Corollary 3.9**.**

If $M$ is a strongly reduced diagram and $\mu$ is the common boundary component of regions $\Delta_{1}$ and $\Delta_{2}$, with boundary cycles $\rho_{1}=\mu\nu_{1}$ and $\rho_{2}=\mu\nu_{2}$, respectively, then $\phi(\rho_{1})$ and $\phi(\rho_{2})$ are not in periodic position. In particular, $\phi(\mu)$ is a piece.

Proof.

As $M$ is strongly reduced $\phi(\mu)$ is a piece, so from the previous lemma and condition (3’) for diagrams, $\phi(\rho_{1})$ and $\phi(\rho_{2})$ are not in periodic position. ∎

For the remainder of this section assume that $s$ and $r$ are elements of $F$ satisfying the hypotheses of Lemma 3.8. Let $\mathbb{F}(X)$ be the free group on a basis $X$ and let $w$ be an an element of $\mathbb{F}(X)$, written as a reduced word. We say a cyclic subword $a$ of $w$ is uniquely positioned if no other cyclic subword of $w$ or $w^{-1}$ is equal to $a$. As $\bar{s}$ is cyclically reduced and not a proper power in $F_{D}$, it follows from [10, Theorem 2.1] that $\bar{s}$ has a cyclic permutation with reduced factorisation equal to $\bar{a}\bar{b}$, where $\bar{a}$ and $\bar{b}$ are non-empty uniquely positioned subwords of $\bar{s}$. Therefore $s$ has a cyclic permutation $\tilde{s}$, with a reduced factorisation $\tilde{s}=ab$, such that $a$ and $b$ are integral cyclic subwords of $s$, $\sigma(a)=\bar{a}$ and $\sigma(b)=\bar{b}$. In this definition, the words $a$ and $b$ are not necessarily uniquely determined. Suppose $s$ has a cyclic permutation $\tilde{s}$ which factors as $\tilde{s}=w_{0}t^{\varepsilon}ut^{\varepsilon}w_{1}$, where $u\in U$, then $\sigma(\tilde{s})=\bar{w}_{0}t^{\varepsilon}t^{\varepsilon}\bar{w}_{1}$. If, in this expression, $\bar{a}$ ends with the first occurrence of $t^{\varepsilon}$ and $\bar{b}$ begins with the second, then $a$ may be chosen to end $t^{\varepsilon}u$ or $t^{\varepsilon}$, with $b$ beginning $t^{\varepsilon}$ or $ut^{\varepsilon}$, respectively. To avoid such ambiguity, in this situation, we always choose $a$ to be right integral, that is $a$ ends in $t^{\varepsilon}$. Similarly, if this situation arises with the roles of $a$ and $b$ interchanged, we choose $b$ to be right integral. With this convention, the words $a$ and $b$ are uniquely determined cyclic subwords of $s$.

Now let $\Delta$ be a region, of an $\widetilde{\mathcal{R}}$ diagram over $F$, with boundary cycle $\rho$, such that $\phi(\rho)=r_{0}\in\widetilde{\mathcal{R}}$. By changing the base point of $\rho$, if necessary, we may assume that $r_{0}$ has reduced factorisation $r_{0}=(g_{0}t^{\varepsilon_{1}}\cdots g_{m-1}t^{\varepsilon_{m}})^{n}$, where $s_{0}=g_{0}t^{\varepsilon_{1}}\cdots g_{m-1}t^{\varepsilon_{m}}$ is a cyclic permutation of $s$ or $s^{-1}$. The primary vertices of $\Delta$ form a subsequence $\delta_{1},\gamma_{1}\ldots,\delta_{mn},\gamma_{mn}$ of the vertex sequence of $\rho$, such that, setting $\gamma_{0}=\gamma_{mn}$, $\phi([\delta_{i+km},\gamma_{i+km}])=t^{\varepsilon_{i}}$ and $\phi([\gamma_{i-1+km},\delta_{i+km}])=g_{i-1}$, for $1\leq i\leq m$ and $0\leq k\leq n-1$ (where $[x,y]$ denotes the subpath of $\rho$ from vertex $x$ to vertex $y$). In this notation, if $g_{i}=1$ then the primary vertices $\gamma_{i+km}$ and $\delta_{i+1+km}$ are the same.

Let $\overline{\phi}=\sigma\circ\phi$, defined on subintervals of $\partial\Delta$ beginning and ending at primary vertices; so

[TABLE]

and

[TABLE]

for $0\leq k\leq n-1$. From the above, $s_{0}$ has a cyclic permutation which factorises as $ab$, where $a$ and $b$ are integral cyclic subwords of $s$, and $\overline{s}_{0}$ has a corresponding cyclic permutation which factorises as $\bar{a}\bar{b}$, where $\bar{a}$, $\bar{b}$ are uniquely positioned cyclic subwords of a cyclic permutation of $\bar{s}$. Consequently, $\bar{r}_{0}=(\bar{s}_{0})^{n}$ has a cyclic permutation which factorises as $(\bar{a}\bar{b})^{n}$. Therefore there is a subsequence $\alpha_{1},\beta_{1},\ldots\alpha_{n},\beta_{n}$ of the sequence $\delta_{1},\gamma_{1},\ldots,\delta_{mn},\gamma_{mn}$ of primary vertices of $\Delta$, such that $\phi([\alpha_{i},\beta_{i}])=a$ and $\phi([\beta_{i},\alpha_{i+1}])=b$, and $\bar{\phi}([\alpha_{i},\beta_{i}])=\bar{a}$ and $\bar{\phi}([\beta_{i},\alpha_{i+1}])=\bar{b}$, for $i=1,\ldots,n$; as illustrated in Example 3.10.

Example 3.10**.**

Figure 6 illustrates a possible distribution of the first four of the vertices $\alpha_{i}$, $\beta_{i}$ on $\partial\Delta$, assuming that $s_{0}=g_{0}tg_{1}t^{-2}$, $g_{0}=g^{\prime}_{0}g_{0}^{\prime\prime}$, $g_{1}=g_{1}^{\prime}g_{1}^{\prime\prime}$, $\bar{a}=t\bar{g}_{1}t^{-1}$ and $\bar{b}=t^{-1}\bar{g}_{0}$. In the diagram, primary vertices are those with the larger diameter.

We define $\operatorname{Sep}(\Delta)$ to be

[TABLE]

If $\mu$ is a simple cyclic sub-path of $\rho$ then we may assume that $\mu$ is a path from points $\alpha$ to $\beta$, on $\partial\Delta$ (when read with the same orientation as $\rho$). We define

[TABLE]

(The terminal point of $\mu$ never contributes to $\operatorname{Sep}(\mu)$.) Thus, if $\rho$ has a cyclic subpath $\xi$ with decomposition $\xi=\mu\nu$, where $\mu$ and $\nu$ have disjoint interiors, then $|\operatorname{Sep}(\xi)|=|\operatorname{Sep}(\mu)|+|\operatorname{Sep}(\nu)|$.

Lemma 3.11**.**

*Let $s$ and $r$ be as in Lemma 3.8. Assume $\Delta$ is a region of a strongly reduced $\widetilde{\mathcal{R}}$ diagram over $F$, which has boundary cycle $\rho$, with a sub-interval $\mu$ such that $\phi(\mu)=p$, where $p$ is a piece over $\widetilde{\mathcal{R}}$. Then $|\operatorname{Sep}(\mu)|\leq 1$. *

Proof.

We may assume that $p\notin H$ and $p\notin\{t^{\pm 1}\}$, since otherwise the result holds immediately. Let $p_{0}$ be the maximal integral subword of $p$ (which must be non-empty under this assumption). Then $p_{0}$ is a piece over $\widetilde{\mathcal{R}}$, and as the diagram is strongly reduced, $p_{0}$ is not in periodic position, so $\bar{p}_{0}$ is a piece over the symmetrised closure of $\sigma(\widetilde{\mathcal{R}})$. As $\bar{a}$ and $\bar{b}$ are uniquely positioned subwords of $\bar{s}$, neither can be a subword of $\bar{p}_{0}$. As $a$ and $b$ are integral subwords of the boundary label of $\Delta$, by definition of $p_{0}$, if $a$ or $b$ is a subword of $p$ then it is a subword of $p_{0}$; so $\bar{a}$ or $\bar{b}$ is a subword of $\bar{p}_{0}$, a contradiction. Thus neither $a$ or $b$ is a subword of $p$, and the Lemma follows. ∎

Following [22] we make the following definition.

Definition 3.12**.**

Let $\Delta$ be a region of a diagram $M$ over $\widetilde{\mathcal{R}}$ and suppose that $\Delta$ has a boundary cycle that factors as $\mu_{1}\cdots\mu_{k}\beta$, where $\mu_{i}$ is the common boundary component of $\Delta$ and a region $\Delta_{i}$, and $\beta$ is a component of a boundary cycle of $M$. Then $\Delta$ is called a $k$-shell, with boundary component $\beta$.

Proposition 3.13**.**

Let $s$, $r=s^{n}$ and $\mathcal{R}$ be as in Lemma 3.8.

1. (a)

If $\Delta$ is a $k$-shell of a strongly reduced diagram over $\widetilde{\mathcal{R}}$, with boundary component $\beta$ then

[TABLE] 2. (b)

$\widetilde{\mathcal{R}}$* satisfies small cancellation condition $C(2n)$.*

We remark that this theorem applies to diagrams over $\widetilde{\mathcal{R}}$, and not to diagrams over the free presentation for $G$. Indeed, $G$ may contain subgroups isomorphic to $\mathbb{F}_{2}\times\mathbb{F}_{2}$, in which case, as shown by Bigdely and Wise [5], no free presentation for $G$ satisfies $C(6)$.

Proof.

1. (a)

Let $\Delta$ have boundary cycle $\rho$ with decomposition $\rho=\mu_{1}\cdots\mu_{k}\beta$, as in the Definition of $k$-shell. From Corollary 3.9, $\phi(\mu_{i})=p_{i}$ is a piece over $\widetilde{\mathcal{R}}$, for $1\leq i\leq k$. Therefore Lemma 3.11 implies that $\operatorname{Sep}(\mu_{1}\cdots\mu_{k})\leq k$. As $\operatorname{Sep}(\Delta)=2n$, this gives $\operatorname{Sep}(\beta)\geq 2n-k$. We claim that if $\nu$ is a subpath of $\rho$ with $\operatorname{Sep}(\nu)\geq K$ then $l(\phi(\nu))\geq((K-2)/2)l(s)+2$, if $K$ is even, and that $l(\phi(\nu))\geq((K-1)/2)l(s)+1$, if $K$ is odd. This certainly holds if $K=0,1$ or $2$, since one of $a$, $b$ may have length $1$. An elementary induction then shows that the claim holds for all $K\geq 0$. From the claim, with $K=2n-k$, the first part of the Lemma follows. 2. (b)

Let $w$ be an element of $\widetilde{\mathcal{R}}$ with a $k$-piece factorisation $w=p_{1}\cdots p_{k}$. First we note that, if $p$ is a piece over $\widetilde{\mathcal{R}}$, and $v=pq\in\widetilde{\mathcal{R}}$, where $v=u^{-1}v_{0}u$, for some cyclically minimal element $v_{0}$ and $u\in U$, then $v_{0}=upu^{-1}(uqu^{-1})$ and it follows that $u^{-1}pu$ is also a piece over $\widetilde{\mathcal{R}}$. Hence we may assume that $w$ is cyclically minimal and we may form a diagram with a single cell $\Delta$ and boundary cycle $\rho$ such that $\rho=\mu_{1}\cdots\mu_{k}$, where $\phi(\mu_{i})=p_{i}$. Then $2n=\operatorname{Sep}(\Delta)=\sum_{i=1}^{k}\operatorname{Sep}(\mu_{i})\leq k$, proving the second part of the lemma.

∎

4 Proofs of main theorems

Proof of Theorem 1.2.

Let $r=s^{n}$, $\mathcal{R}=\{r\}$ and $\mathbb{G}_{t}=\langle A\backslash\{t\}\rangle$. If $\operatorname{lk}(t)$ is empty then $\mathbb{G}=\mathbb{G}_{t}*\langle t\rangle$ and Theorem 1.2 follows from well known results on one-relator products. We may therefore assume $\operatorname{lk}(t)$ is a non-empty clique.

1. (a)

Let $w\in((A\backslash\{t\})^{\pm 1})^{*}$ be a freely reduced word such that $w=1$ in $G$. From Proposition 3.6, there exists a strongly reduced $\widetilde{\mathcal{R}}$-diagram $M$ over $F$, with boundary label $w$. From Proposition 3.13, $\widetilde{\mathcal{R}}$ satisfies $C(2n)$, where $n\geq 3$. Therefore Greendlinger’s Lemma for $C(6)$ diagrams (see for example [22, Theorem 9.4]) implies that $M$ has a $k$-shell, for $k\leq 3$. Hence $\phi(\beta)$ has length at least $(n-2)l(s)\geq l(s)$, so is a word of positive $t$-length, contrary to the assumption that $w\in\mathbb{G}_{t}$. 2. (b)

Let $M$ be a strongly reduced $\widetilde{\mathcal{R}}$-diagram, with boundary label $s^{k}$, for some positive integer $k<n$. As $\partial M$ has length $kl(s)$, $M$ must contain at least $2$ regions and since $s^{k}$ is freely reduced, from Greendlinger’s Lemma for $C(6)$ diagrams again, $M$ must contain at least three $d$-shells, where $d\leq 3$, so $|\partial M|\geq 3(n-2)l(s)$. As $n\geq 3$, $3(n-2)\geq n$, so this implies $kl(s)=|\partial M|\geq nl(s)$, a contradiction.

For the final statement let $w\in\mathbb{F}(A)$. We may assume without loss of generality that $w$ is a minimal form for a non-trivial element of $F$. If $w=1$ in $F/N$ then there exist elements $y_{1},\ldots y_{k}\in\mathbb{F}(A)$ and $\varepsilon_{1},\ldots,e_{k}\in\{\pm 1\}$, such that

[TABLE]

From this expression we construct, in the usual way, a diagram over $\widetilde{\mathcal{R}}$, with boundary label $w$ and regions $\Delta_{i}$, $i=1,\ldots,k$. After deleting cancelling regions if necessary we obtain a reduced diagram $M$ with at most $k$ regions and boundary label $w$. From Proposition 3.6 we may assume $M$ is strongly reduced. As $n\geq 4$, from Proposition 3.13, $\widetilde{\mathcal{R}}$ satisfies $C(8)$ and so from standard small cancellation arguments (e.g. [15, page 246, Proposition 27]) the number of regions of $M$ is at most $8l(w)$. To decide whether or not $w=1$ in $F/N$ therefore amounts to deciding whether or not one of equations (4.1), in variables $y_{i}$, with $k\leq 8l(w)$ has solution. From [8] equations are decidable over partially commutative groups, so the latter problem is decidable.

∎

Proof of Theorem 1.4.

As above, we may decompose $\mathbb{G}$ as a free product with amalgamation,

[TABLE]

By definition of $\operatorname{lk}(s)$, we have $\mathbb{A}_{0}=U\times K$ so

[TABLE]

If $\langle Y^{\prime}\rangle$ embeds in $K/M$ it follows that $\langle\operatorname{lk}(s)\cup Y^{\prime}\rangle$ embeds in $U\times K/M$, so $\langle(A\backslash\operatorname{supp}(s))\cup Y^{\prime}\rangle$ embeds in $(U\times K/M)\ast_{U}\mathbb{A}_{1}$, as required. Moreover, if $s$ has order $n$ in $K/M$, it follows that $s$ also has order $n$ in $G$.

Given an element $w\in\mathbb{F}(A)$ we may write $w$ in the form $w=g_{1}u_{1}h_{1}\cdots g_{k}u_{k}h_{k}u_{k+1}=[\prod_{i=1}^{k}(g_{i}u_{i}h_{i})]u_{k+1}$, where $k\geq 0$, $u_{i}\in U$, $g_{i}\in K$ and $h_{i}\in\mathbb{A}_{1}$; using the Transformation Lemma (see Section 2), and the fact that $[K,U]=1$. Moreover, we may assume $w$ is a minimal length word in $(A\cup A^{-1})^{*}$ representing its class as an element of $\mathbb{G}$, and that $h_{i}$ has no left or right divisor in $U$, for all $i$. Then $w\in N$ if and only if $g_{i}\in M$, for $i=1,\ldots,k$, and $u_{1}h_{1}\cdots u_{k}h_{k}u_{k+1}=1$ in $\mathbb{A}_{1}$. If the word problem is solvable in $K/M$ we may decide whether or not $g_{i}\in M$; and so the word problem is decidable in $G$.

Given elements $v,w\in\mathbb{F}(A)$, to decide whether or not $v$ and $w$ are conjugate in $G$, we apply [23, Theorem 4.6, page 212]. We may first replace $v$ and $w$ by elements of $\mathbb{F}(A)$ representing cyclically minimal elements of $\mathbb{G}$. To simplify notation let $\mathbb{A}_{2}=U\times K/M$. If $v\in U$, and $v$ is conjugate to $w$, then $w$ represents an element of $\mathbb{A}_{1}\cup\mathbb{A}_{2}$ and there exists a sequence $v,v_{1},\ldots v_{l}=w$, where $v_{i}\in U$, for $i<l$ and consecutive terms are conjugate in $\mathbb{A}_{j}$, for $j=1$ or $2$. As $\mathbb{A}_{2}=U\times K/M$, this implies that $v$ is conjugate to $w$ in $\mathbb{A}_{1}$, and this may be effectively verified, in the partially commutative group $\mathbb{A}_{1}$. If $v\notin U$ but $v\in\mathbb{A}_{1}\cup\mathbb{A}_{2}$ then $v$ and $w$ are conjugate only if they belong to the same factor and are conjugate in that factor. If $v$ and $w$ belong to $\mathbb{A}_{1}$ we may verify if they are conjugate or not, as in the previous case. If both belong to $\mathbb{A}_{2}$ then we use the solvability of the conjugacy problem in $K/M$ to decide whether or not they are conjugate.

Otherwise $v$ and $w$ are cyclically minimal as elements of $\mathbb{A}_{2}*_{U}\mathbb{A}_{1}$ which we may write (after a cyclic permutation if necessary) $w=g_{1}u_{1}h_{1}\cdots g_{k}u_{k}h_{k}u_{k+1}$ and $v=g^{\prime}_{1}u^{\prime}_{1}h^{\prime}_{1}\cdots g^{\prime}_{l}u^{\prime}_{l}h^{\prime}_{l}u^{\prime}_{m+1}$, where $k>0$, $u_{i},u_{j}^{\prime}\in U$, $g_{i},g_{j}^{\prime}\in K$, $h_{i},h_{j}^{\prime}\in\mathbb{A}_{1}$ and $h_{i}$, $h_{j}^{\prime}$ have no left or right divisors in $U$, for all $i,j$. We may identify any $i$ such that $g_{i}=_{G}1$ and rewrite $w$, replacing $g_{i}$ by $1$, using the fact that the conjugacy (so word) problem is solvable in $K/M$. After repeating this process sufficiently often we may assume that $g_{i}\notin M$, for all $i$; without altering the conjugacy class of $w$ in $G$. Similarly we may assume no $g_{i}^{\prime}\in M$. Note that $w$ is cyclically reduced as an element of $(U\times K/M)\ast_{U}\mathbb{A}_{1}$ if and only the first and last letters come from distinct factors (that is $g_{1}\neq 1$ and $h_{k}\neq 1$) and we may assume $w$ begins with an element of $K/M$, by cyclically permuting if necessary. We may therefore decide whether or not $w$ is cyclically reduced and, if it is not, replace it by a cyclic permutation with fewer factors. Hence we may assume that $w$ and $v$ have been rewritten as representatives of cyclically reduced forms (in the sense of free products with amalgamation) of elements of $\mathbb{A}_{2}\ast_{U}\mathbb{A}_{1}$, both beginning with an element of $K$. Then $v$ is conjugate to $w$ if and only if a cyclic permutation of $v$, followed by conjugation by an element of $U$, results in a representative of $w$. We may assume that $v$ as written above is an appropriate cyclic permutation and it remains to decide if $u^{-1}vu=w$, for some $u\in U$.

First consider the case where $u=1$ and $v=w$. Then we have

[TABLE]

We claim that this holds if and only if $k=l$, $g_{i}=g_{i}^{\prime}$, $h_{i}=h_{i}^{\prime}$, $[h_{i},u^{\prime}_{i+1}\cdots u^{\prime}_{k+1}u^{-1}_{k+1}\cdots u_{i+1}^{-1}]=1$, for $i=1,\ldots,k$ and $u^{\prime}_{1}\cdots u^{\prime}_{k+1}=u_{1}\cdots u_{k+1}$. To see this, first observe that if (4.2) holds then we have $h_{l}^{\prime}u^{\prime}_{l+1}u_{k+1}^{-1}h_{k}^{-1}\in U$, and since $h_{k}$ and $h_{l}^{\prime}$ have no left or right divisors in $U$, this forces $h_{l}^{\prime}=h_{k}$ and $[h_{k},u^{\prime}_{l+1}u_{k+1}^{-1}]=1$. Therefore

[TABLE]

from which we have $g_{l}^{\prime}g_{k}^{-1}\in U$, so $g_{l}^{\prime}=g_{k}$. Continuing this way, it follows that $k=l$, $g_{i}=g_{i}^{\prime}$ and $[h_{i},u^{\prime}_{i+1}\cdots u^{\prime}_{k+1}u^{-1}_{k+1}\cdots u_{i+1}^{-1}]=1$, for $i=1,\ldots,k$. This in turn, together with (4.2), implies that $u^{\prime}_{1}\cdots u^{\prime}_{k+1}=u_{1}\cdots u_{k+1}$. Conversely, if all these conditions hold, then so does (4.2), by direct computation.

In the general case, $u^{-1}vu=w$ if and only if

[TABLE]

From the claim above this holds if and only if $k=l$,

[TABLE]

for $i=1,\ldots,k$ and

[TABLE]

That is, $u^{-1}vu=w$ if and only if $k=l$, $g_{i}=g_{i}^{\prime}$, $h_{i}=h_{i}^{\prime}$ and the system of $k+1$ equations

[TABLE]

in the variable $x$, where $1\leq i\leq k$, has a solution in the group $U$. In the terminology of [8], $U$ is a normalised rational subset of the partially commutative group $\mathbb{A}_{1}$. Thus (4.4) is a system of equations over $\mathbb{A}_{1}$ with normalised rational constraints. From [8, Corollary 1], the system (4.4) is decidable. Therefore we may decide whether or not $v$ is conjugate to $w$ by an element of $U$. As $v$ has only finitely many cyclic permutations, combining the above we have a solution to the conjugacy problem in $G$.

The final part of this proof is unsatisfactory, in that it uses the decidability of all systems of equations over partially commutative groups, to lift to decidability of the conjugacy problem in $G$. We therefore give an alternative argument, which depends only on decidability of the conjugacy problem.111The authors are grateful to Armin Weiss for pointing out this approach. First let $a\in A\cap K$, such that $a\notin M$. Then, replacing $g_{i}$ and $g_{i}^{\prime}$ by $a$, for all $i$ in (4.3), the argument above shows that

[TABLE]

for some $u\in U$, if and only if $k=l$, $h_{i}=h_{i}^{\prime}$ and the system of equations (4.4) has a solution $x\in U$. Since $a\notin U$ and $h_{i}$ has no left or right divisor in $U$ the subgraph of the non-commutation graph $\Delta$ of $\mathbb{G}$ with vertices $\{a\}\cup\bigcup_{i=1}^{k}\operatorname{supp}(h_{i})$ is connected. Therefore $\hat{w}=au_{1}h_{1}\cdots au_{k}h_{k}u_{k+1}$ has block decomposition $\hat{w}=b_{0}b_{1}\cdots b_{p}$, where the $b_{i}$ are blocks, $b_{0}=av_{1}h_{1}\cdots av_{k}h_{k}v_{k+1}$, for some $v_{i}\in U$, and $b_{i}\in U$, for $i\geq 1$. If (4.5) holds, then it follows from [12, Proposition 5.7] that its right hand side, $\hat{v}=au^{\prime}_{1}h_{1}\cdots au^{\prime}_{k}h_{k}u^{\prime}_{k+1}$, has block decomposition $\hat{v}=b^{\prime}_{0}b^{\prime}_{1}\cdots b^{\prime}_{p}$, where the $b^{\prime}_{i}$ are blocks, $b^{\prime}_{0}=av^{\prime}_{1}h_{1}\cdots av^{\prime}_{k}h_{k}v^{\prime}_{k+1}$, for some $v^{\prime}_{i}\in U$ and $b^{\prime}_{i}\in U$, for $i\geq 1$. Moreover, in this case, after reordering the $b_{i}^{\prime}s$, $i\geq 1$, if necessary, the blocks $b_{i}$ and $b_{i}^{\prime}$ are cyclically minimal and conjugate. From the Transformation Lemma, $b_{0}$ is conjugate to $b_{0}^{\prime}$ if and only if there is an element $z_{0}\in U\cap\langle\operatorname{supp}(b_{0})\rangle$ such that $z_{0}^{-1}b_{0}^{\prime}z_{0}=b_{0}$, and for $i\geq 1$, $b_{i}$ is conjugate to $b_{i}^{\prime}$ if and only if there exist $z_{i}$ such that $z_{i}\in\langle\operatorname{supp}(b_{i})\rangle$, and $z_{i}^{-1}b_{i}z_{i}=b_{i}^{\prime}$, for $i=1,\ldots,p$. As $[\operatorname{supp}(b_{i}),\operatorname{supp}(b_{j})]=1$, for all $i\neq j$, it follows that there exists $u\in U$ such that $u^{-1}\hat{w}u=\hat{v}$ if and only if there exist such $z_{i}$. For $i\geq 1$ the question of whether or not such a $z_{i}$ exists is the conjugacy problem in the partially commutative group $\langle\operatorname{supp}(b_{i})\rangle$, so is decidable. For $i=0$, the question of existence of such a $z_{0}$ is decidable by [12, Proposition 5.8]. Therefore, we may decide whether or not $\hat{w}$ is conjugate to $\hat{v}$. This means that we may also decide whether or not (4.4) has a solution $x\in U$; and applying this argument to all cyclic permutations of $v$, the result follows. ∎

5 Cycle graphs with a chord

In this section we argue that Theorem 1.2 almost always applies, and a Freiheitssatz holds, in the situation of Example 1.7.5. We conjecture that a similar statement holds for arbitrary graphs. Here though we use only naive counting arguments; a study of the conjecture over arbitrary graphs is likely to require a more systematic approach.

As in Example 1.7.5, $\mathbb{G}^{\prime}=\mathbb{G}(C^{\prime}_{n})$, where $C_{n}^{\prime}$ is the graph on the left of Figure 2, and $n\geq 5$. The subgroup $H$ of $\mathbb{G}^{\prime}$ generated by $A_{0}=\{a_{1},\ldots,a_{n-1}\}$ is isomorphic to $\mathbb{G}(C_{n-1})$, $\operatorname{lk}(t)=\{a_{1},a_{n-1}\}$, $U=\langle\operatorname{lk}(t)\rangle$ and HNN$(t)=\langle H,t\,|\,t^{-1}at=a,\forall a\in\operatorname{lk}(t)\rangle$. As input to the question “for which elements of $\mathbb{G}^{\prime}$ does Theorem 1.2 hold?” we choose a particular set $L$ (see Section 5.1) of words over $(A\cup A^{-1})^{*}$ with the property that every cyclically minimal element of $H$ is represented by a unique element of $L$, every cyclically minimal element of $\mathbb{G}^{\prime}$ ending in a $t$-letter is represented by a unique element of $L$ and every element of $L$ represents a cyclically minimal element of $\mathbb{G}^{\prime}$. Every element of $L$ is reduced with respect to HNN$(t)$ and is written as $w=g_{1}t^{\alpha_{1}}g_{2}\cdots g_{m}t^{\alpha_{m}}$, where $m\geq 0$, $g_{i}\in H$, $\varepsilon_{i}=\pm 1$, and $\alpha_{i}\in\mathbb{Z}\backslash\{0\}$. The set of elements $w\in L$ for which $|\alpha_{1}|+\cdots+|\alpha_{m}|\leq k$ and $l(g_{i})\leq d$, for all $i$, is denoted by $L(d,k)$. We show that almost all words of $L$ satisfy the hypotheses of Theorem 1.2, in a sense which we now make precise. For a subset $S$ of $L$ we define the asymptotic density of $S$ to be

[TABLE]

(In this limit, both $d$ and $k$ must become, simultaneously, sufficiently large. The asymptotic density of $S$ is undefined if the limit does not exist.) For general discussion of asymptotic density and it properties we refer, for example, to [24, 20, 14]: the 2-parameter “bidimensional” asymptotic density defined in [14] plays a role equivalent to the limits as both $d$ and $k$ approach infinity, defined here.

As subset $S\subseteq L$ is said to be generic if $\rho(S)=1$ and negligible if $\rho(S)=0$.

Proposition 5.1**.**

In the above terminology, let $L_{Y}$ be the subset of $L$ consisting of words which satisfy the hypotheses of Theorem 1.2. Then $L_{Y}$ is generic.

In the remainder of this section we define the set $L$ and prove this proposition.

5.1 Normal forms

The definition of $L$ depends on a choice of normal forms for elements of $H$. The subgroup $H$ is $\mathbb{G}(C_{n-1})$ where $C_{n-1}$ is the cycle graph with vertices $A_{0}$. We may define a set $L_{H}$ of unique normal forms for all $n\geq 5$, as we do below, but observe that the case $n=5$ is special, as in this case $H=\mathbb{F}(a_{1},a_{3})\times\mathbb{F}(a_{2},a_{4})$, and an alternative normal form may be chosen using this decomposition. It turns out that some of the bounds we derive for our general normal form involve division by $n-5$, so would need special treatment when $n=5$. Moreover, the alternative normal form for the case $n=5$ allows relatively straightforward computation of the precise sizes of sets we’re interested in. This being the case, we make separate definitions of normal forms for elements of $H$ in the case $n=5$ and the case $n\geq 6$. These definitions result in different choices of sets $L$ in the two cases.

In the case $n=5$ we say a word $w$ over $(A_{0}\cup A_{0}^{-1})^{*}$ is a square normal form if $w=w_{1}w_{2}$, where $w_{1}$ is a reduced word in $\mathbb{F}(a_{2},a_{4})$ and $w_{2}$ is a reduced word in $\mathbb{F}(a_{1},a_{3})$. (The graph $C_{4}$ is a “square”.) Then every square normal form is a minimal form and every element of $H$ is represented by a unique square normal form.

For $n\geq 5$, we say a word $w$ over $(A_{0}\cup A_{0}^{-1})^{*}$ is a normal form if

1. (i)

$w$ is freely reduced and 2. (ii)

contains no subword of the form $a_{i+1}^{\varepsilon}a_{i-1}^{\beta}a_{i}^{\delta}$, (subscripts modulo $n-1$) where $1\leq i\leq n-1$, $\varepsilon,\delta\in\{\pm 1\}$ and $\beta\in\mathbb{Z}$ ($\beta$ may be zero).

Note that if $w$ is a word in normal form then it is minimal, because the centraliser of $a_{i}$ in $H$ is $C_{H}(a_{i})=\langle a_{i-1},a_{i},a_{i+1}\rangle$, and $w$ can contain no subword of the form $a_{i}^{-\delta}ua_{i}^{\delta}$, with $u\in\langle a_{i-1},a_{i+1}\rangle$. We claim that every element of $H$ is equal to a unique word in normal form. To see this, define a word to be prohibited if it has the form

[TABLE]

where $r\geq 1$, $\delta\in\{\pm 1\}$, $\alpha_{j}\neq 0$, for all $j$ and $\beta_{j}\neq 0$, for $j<r$ (where subscripts are written modulo $n-1$). Let $v$ be a minimal length word representing an element of $H$ and suppose that $v\equiv v_{0}v_{1}v_{2}$, where $v_{1}$ is prohibited and is the leftmost prohibited subword of $v$: that is, if $v\equiv u_{0}u_{1}u_{2}$, where $u_{1}$ is prohibited, then $|v_{0}|\leq|u_{0}|$. If $v_{1}=(\prod_{j=1}^{r}a_{i+1}^{\alpha_{j}}a_{i-1}^{\beta_{j}})a_{i}^{\delta}$ then let $v_{1}^{\prime}=a_{i}^{\delta}(\prod_{j=1}^{r}a_{i+1}^{\alpha_{j}}a_{i-1}^{\beta_{j}})$, so $v=_{H}v_{0}v_{1}^{\prime}v_{2}$ and if $v_{0}v_{1}^{\prime}v_{2}\equiv u_{0}u_{1}u_{2}$, where $u_{1}$ is prohibited, then $|u_{0}|>|v_{0}|$ (as the first letter $a_{i}^{\delta}$ of $v_{1}^{\prime}$ cannot be part of a prohibited subword). Continuing this way we may eventually write $v$ to a normal form $u$ with $v=_{H}u$. Hence every element of $H$ is represented by a normal form.

It remains to show that normal forms are unique, for which we shall use the fact that, if $w$ is a normal form and $w\equiv w_{0}a_{k}^{\gamma}w_{1}$, where $[a_{k},\operatorname{supp}(w_{1})]=1$, then $w_{0}w_{1}$ is also a normal form. Indeed, if $w_{0}w_{1}$ contains a subword $u=a_{i+1}^{\varepsilon}a_{i-1}^{\beta}a_{i}^{\delta}$, with $\varepsilon,\delta\in\{\pm 1\}$, then the first letter $a_{i+1}^{\varepsilon}$ of $u$ occurs in $w_{0}$ and the last letter $a_{i}^{\delta}$ occurs in $w_{1}$. Thus $w_{0}=w_{0}^{\prime}a_{i+1}^{\varepsilon}a_{i-1}^{\beta_{0}}$ and $w_{1}=a_{i-1}^{\beta_{1}}a_{i}^{\delta}w_{1}^{\prime}$, where $\beta_{0}+\beta_{1}=\beta$ and $w_{0}^{\prime},w_{1}^{\prime}$ are initial and terminal subwords of $w_{0}$ and $w_{1}$, respectively. As $[a_{k},\operatorname{supp}(w_{1})]=1$, we have $[a_{k},a_{i}]=1$ so $k\in\{i-1,i,i+1\}$. If $k=i-1$ then $w$ has a subword $a_{i+1}^{\varepsilon}a_{i-1}^{\beta+\gamma}a_{i}^{\delta}$. If $k=i$ then $w$ has a subword $a_{i+1}^{\varepsilon}a_{i-1}^{\beta_{0}}a_{i}^{\gamma}$ and if $k=i+1$ then $w$ has a subword $a_{i+1}^{\gamma}a_{i-1}^{\beta_{1}}a_{i}^{\delta}$. None of these are possible as $w$ is a normal form, so $w_{0}w_{1}$ is a normal form, as claimed.

We now use induction on length to show each element of $H$ has exactly one normal form. This is certainly true for the element of length [math], and we assume that it is true for all elements of length at most $m$, for some $m\geq 0$. Next suppose $w$ and $w^{\prime}$ are normal forms with $w=_{H}w^{\prime}$, $l(w)=m+1$ (so $l(w^{\prime})=m+1$) and $w\nequiv w^{\prime}$. Let $w\equiv w_{0}a_{i}^{\varepsilon}$, where $\varepsilon\in\{\pm 1\}$. Then $w^{\prime}\equiv w_{0}^{\prime}a_{j}^{\delta}$, for some $\delta\in\{\pm 1\}$, where $j\neq i$; otherwise $w_{0}$ and $w_{0}^{\prime}$ are distinct normal forms of length $m$ for the same element of $H$. Thus $w$ has right divisors $a_{i}^{\varepsilon}$ and $a_{j}^{\delta}$, so $[a_{i},a_{j}]=1$ and we have $w_{0}\equiv w_{1}a_{j}^{\delta}w_{2}$ and $w_{0}^{\prime}\equiv w_{1}^{\prime}a_{i}^{\varepsilon}w_{2}^{\prime}$, where $[a_{j},\operatorname{supp}(w_{2})]=[a_{i},\operatorname{supp}(w_{2}^{\prime})]=1$, $a_{j}\notin\operatorname{supp}(w_{2})$, $a_{i}\notin\operatorname{supp}(w_{2}^{\prime})$ and $j=i\pm 1$. Therefore, $w\equiv w_{1}a_{j}^{\delta}w_{2}a_{i}^{\varepsilon}$ and $w^{\prime}\equiv w_{1}^{\prime}a_{i}^{\varepsilon}w_{2}^{\prime}a_{j}^{\delta}$ and, using the fact above, both $w_{1}w_{2}a_{i}^{\varepsilon}$ and $w_{1}^{\prime}a_{i}^{\varepsilon}w_{2}^{\prime}$ are normal forms. As $w_{1}w_{2}a_{i}^{\varepsilon}=_{H}w_{1}^{\prime}a_{i}^{\varepsilon}w_{2}^{\prime}$, the inductive assumption implies $w_{1}w_{2}a_{i}^{\varepsilon}\equiv w_{1}^{\prime}a_{i}^{\varepsilon}w_{2}^{\prime}$ and, as $a_{i}\notin\operatorname{supp}(w_{2}^{\prime})$ it follows that $w_{2}^{\prime}\equiv 1$. Similarly $w_{2}\equiv 1$, and now we have $w_{1}\equiv w_{1}^{\prime}$. Therefore $w\equiv w_{1}a_{j}^{\delta}a_{i}^{\varepsilon}$ and $w^{\prime}\equiv w_{1}a_{i}^{\varepsilon}a_{j}^{\delta}$, with $j=i\pm 1$. This is a contradiction, since $w$ and $w^{\prime}$ cannot both be normal forms, and the result follows.

Therefore we have unique normal forms for elements of $H$, for $n\geq 5$. Next we define subsets of the set of normal forms needed for analysis of Theorem 1.2.

• •

For $n=5$, let $L_{H}$ be the set of square normal forms for elements of $H$, and for $n\geq 6$ let $L_{H}$ be the set of normal forms for elements of $H$. In both cases let $L_{H}(d)$ be the set of elements of $L_{H}$ of length at most $d$, and set $l_{H}(d)=|L_{H}(d)|$.

• •

Let $L_{H}^{U}\subseteq L_{H}$ be the set of normal forms for elements of $H$ which have no non-trivial left divisor in $U$. Let $L_{H}^{U}(d)=L_{H}^{U}\cap L_{H}(d)$ and let $l_{H}^{U}(d)=|L_{H}^{U}(d)|$.

• •

Let $L_{U}$ be the subset $\{a_{n-1}^{\alpha}a_{1}^{\beta}:\alpha,\beta\in\mathbb{Z}\}$ of $L_{H}$. Then $L_{U}$ is the subset of $L_{H}$ consisting of normal forms for elements of $U$ (for all $n\geq 5$). Let $L_{U}(d)$ be the set of elements of $L_{U}$ of length at most $d$ and set $l_{U}(d)=|L_{U}(d)|$.

• •

Let $L_{H,S}(d)$, $L_{H,S}^{U}(d)$ and $L_{U,S}(d)$ be the subsets $L_{H}$, $L_{H}^{U}$ and $L_{U}$, respectively, of elements of length exactly $d$. Set $l_{H,S}(d)=|L_{H,S}(d)|$, $l_{H,S}^{U}(d)=|L_{H,S}^{U}(d)|$ and $l_{U,S}(d)=|L_{U,S}(d)|$.

• •

Let $L_{t}(k)$ be the set of cyclically reduced words of $(A\cup A^{-1})^{*}$ representing elements of $\mathbb{G}^{\prime}$ which end in a $t$-letter and have the form $ug_{1}t^{\varepsilon_{1}}\cdots g_{k}t^{\varepsilon_{k}}$, where $g_{i}\in L_{H}^{U}$ and $u\in L_{U}$.

As the unique element of $L_{U}\cap L_{H}^{U}$ is $1$, the condition that elements of $L_{t}(k)$ are cyclically reduced amounts to the condition that, for all $i$; either $g_{i}\neq 1$ or $\varepsilon_{i-1}=\varepsilon_{i}$ (subscripts modulo $n$).

Fix $k\geq 1$ and let $w$ be an element of $L_{t}(k)$. We say that $w$ is composed if either

1. (i)

$w=ut^{k}$ or $w=ut^{-k}$, for some $u\in U$, or 2. (ii)

$w=ug_{1}t^{\alpha_{1}}\cdots g_{r}t^{\alpha_{r}}$, where $g_{i}\in L_{H}^{U}\backslash\{1\}$, $u\in L_{U}$, $|\alpha_{i}|\geq 1$ and $(|\alpha_{1}|,\ldots,|\alpha_{r}|)$ is a composition of $k$ into $r$ parts; for some $r$ such that $1\leq r\leq k$.

• •

Let $L(k)$ the set of composed elements of $L_{t}(k)$ and let $L^{\ref{it:composed1}}(k)$ and $L^{\ref{it:composed2}}(k)$ be the sets of elements of $L(k)$ of types (i) and (ii), respectively.

• •

For $d\geq 0$ denote by $L(d,0)$ the set of normal forms in $L_{H}$ representing cyclically minimal elements of $H$ and set $l(d,0)=|L(d,0)|$.

• •

For $d\geq 0$ and $k\geq 1$ denote by $L^{\ref{it:composed1}}_{S}(d,k)$ the set of elements of $L^{\ref{it:composed1}}(k)$ such that $u\in L_{U}(d)$. Set $l^{\ref{it:composed1}}_{S}(d,k)=|L^{\ref{it:composed1}}_{S}(d,k)|$.

• •

For $d\geq 0$ and $k\geq 1$ denote by $L^{\ref{it:composed2}}_{S}(d,k)$ the set of elements of $L^{\ref{it:composed2}}(k)$ such that $g_{i}\in L_{H}^{U}(d)$, for all $i$, $u\in L_{U}$ and $l(ug_{1})\leq d$. Set $l^{\ref{it:composed2}}_{S}(d,k)=|L^{\ref{it:composed2}}_{S}(d,k)|$.

• •

For $d\geq 0$ and $k\geq 1$ set $L_{S}(d,k)=L^{\ref{it:composed1}}_{S}(d,k)\cup L^{\ref{it:composed2}}_{S}(d,k)$ and let $l_{S}(d,k)=|L_{S}(d,k)|$.

• •

For $d\geq 0$ and $k\geq 1$ set $L^{\ref{it:composed1}}(d,k)=\cup_{l=1}^{k}L^{\ref{it:composed1}}_{S}(d,l)$, $L^{\ref{it:composed2}}(d,k)=\cup_{l=1}^{k}L^{\ref{it:composed2}}_{S}(d,l)$, $L(d,k)=L(d,0)\cup L^{\ref{it:composed1}}(d,k)\cup L^{\ref{it:composed2}}(d,k)$ and $l^{\ref{it:composed1}}(d,k)=|L^{\ref{it:composed1}}(d,k)|$, $l^{\ref{it:composed2}}(d,k)=|L^{\ref{it:composed2}}(d,k)|$ and $l(d,k)=|L(d,k)|$.

• •

For each integer $r$ such that $1\leq r\leq k$ and each composition $P$ of $k$ into $r$ parts, set

[TABLE]

Set $l^{\ref{it:composed2}}_{S}(d,k,r,P)=|L^{\ref{it:composed2}}_{S}(d,k,r,P)|$.

• •

Let $L=\bigcup_{d\geq 0,k\geq 0}L(d,k)$.

5.2 Counting normal forms

To find bounds on the size of $L(d,k)$ we first compute bounds on the sizes of $L_{H}(d)$, $L_{U}(d)$ and $L_{H}^{U}(d)$. We consider cases $n=5$ and $n\geq 6$ separately, though the bounds we find for $n\geq 6$ also apply in the case $n=5$, except in cases where division by $n-5$ is involved. First suppose that $n=5$. All words of length [math] and $1$ are in $L_{H}$, so $l_{H}(0)=1$ and $l_{H,S}(1)=2(n-1)=8$. For $k\geq 2$, the number of normal forms $w_{1}w_{2}$ such that $l(w_{1})=p$ and $l(w_{2})=k-p$, where $1\leq p\leq k-1$, is $4^{2}3^{k-2}$, and there are $8\cdot 3^{k-1}$ normal forms $w_{1}\cdot w_{2}$ of length $k$ with either $w_{1}$ or $w_{2}$ trivial. Hence $l_{H,S}(k)=8\cdot 3^{k-2}(2k+1)$, for $k\geq 1$. Summing over $k$ from [math] to $d$ gives

[TABLE]

Now consider the case $n\geq 6$. Again all words of length [math] and $1$ are in $L_{H}$, so $l_{H}(0)=1$ and $l_{H,S}(1)=2(n-1)$. A word of length $2$ is in $L_{H}$ if and only if it is of the form $a_{i}^{\varepsilon}a_{j}^{\delta}$, where $\varepsilon,\delta\in\{\pm 1\}$, and satisfies the conditions that $j\neq i-1$ and that if $i=j$ then $\varepsilon=\delta$. Therefore $l_{H,S}(2)=2(n-1)(2n-5)$. We claim that, setting $\alpha=2n-5$ and $\gamma=2n-7$,

[TABLE]

To verify this, assume that it holds for some $m\geq 2$. If $w$ is a normal form of length $m+1$ then $w=ua_{j}^{\delta}$, for some $u$ of length $m$, which is necessarily in $L_{H}$, $a_{j}\in A_{0}$ and $\delta=\pm 1$. From the argument used to establish the value of $l_{H,S}(2)$; for each such $u$, there are at most $2n-5$ choices for $a_{j}^{\delta}$. Hence $l_{H,S}(m+1)\leq\alpha l_{H,S}(m)\leq 2(n-1)\alpha^{m}$. Some of these choices do not give rise to normal forms, as they result in prohibited subwords of $w$. On the other hand if $u=va_{i}^{\varepsilon}$, and $i\notin\{j-1,j+1\}$ then (as $u\in L_{H}$) $ua_{j}^{\delta}$ contains no prohibited subword, so is in $L_{H}$ (as long as it is reduced). Hence there are at least $2n-7$ choices of $a_{i}^{\varepsilon}$ which do result in $ua_{i}^{\varepsilon}$ being a normal form. Therefore there are at least $\gamma l_{H,S}(m)$ possibilities for $w$. That is, $l_{H,S}(m+1)\geq\gamma l_{H,S}(m)\geq 2(n-1)\alpha\gamma^{m-1}$. Hence, as $\alpha\geq\gamma$, (5.2) follows by induction on $m$. Summing over $m$ from [math] to $d$ we obtain

[TABLE]

for all $d\geq 1$, in the case $n\geq 6$.

Normal forms of elements of $U$ all have the form $a_{n-1}^{a}a_{1}^{b}$, for some integers $a,b$, and for all $n\geq 5$. Therefore the number of elements of $U$ of length exactly $d$ is $4d$, and

[TABLE]

Next we consider $L_{H}^{U}$, the set of normal forms for elements of $H$ which have no non-trivial left divisor in $U$. When $n=5$ a normal form $w_{1}w_{2}$ belongs to $L_{H}^{U}$ if and only if $w_{1}$ does not begin with $a_{4}^{\pm 1}$ and $w_{2}$ does not begin with $a_{1}^{\pm 1}$. There are $2\cdot 3^{l-1}$ elements of $F(a_{2},a_{4})$ of length $l\geq 1$, beginning with $a_{4}^{\pm 1}$; and similarly for $F(a_{1},a_{3})$ and $a_{3}^{\pm 1}$. It follows that the number of elements of $L_{H}^{U}$ of length exactly $k$ is $4\cdot 3^{k-2}(2+k)$. Therefore

[TABLE]

In the case $n\geq 6$ we find bounds on the size of $L_{H}^{U}(d)$. We partition $L_{H}^{U}$ into three subsets: (a) normal forms beginning with elements $a_{i}^{\pm 1}$, where $3\leq i\leq n-3$; (b) normal forms beginning $a_{2}^{\pm 1}$ and (c) normal forms beginning $a_{n-2}^{\pm 1}$.

1. (a)

Normal forms beginning $a_{i}^{\pm 1}$, where $3\leq i\leq n-3$, have no left divisor in $U$. There are $2(n-5)$ choices for the first letter $a_{i}^{\varepsilon}$, where $3\leq i\leq n-3$, of such a normal form. As in the derivation of bounds in (5.2) above, there are $2n-5$ possibilities for the second letter, and between $2n-7$ and $2n-5$ possibilities for each subsequent letter, of such a normal form. Thus, for $d\geq 2$, $L_{H}^{U}$ contains at least $2(n-5)\alpha\gamma^{d-2}\geq 2(n-5)\gamma^{d-1}$ and at most $2(n-5)\alpha^{d-1}$ elements of type (a), with length exactly $d$. Therefore, the number $a(d)$ of elements of type (a) in $L_{H}^{U}(d)$ satisfies

[TABLE] 2. (b)

To bound the number $b(d)$ of normal forms of length at most $d$ which begin with $a_{2}^{\pm 1}$ and have no left divisor in $U$, we begin by finding bounds on the number $b_{S}(d)$ of such normal forms of length exactly $d$. We have $b_{S}(0)=0$ and $b_{S}(1)=2$, as the possibilities are $a_{2}^{\pm 1}$. To estimate $b_{S}(d)$, for $d>1$, it is convenient to consider normal forms beginning with elements of the set $M$ of reduced words in $\{a_{2}^{\pm 1}\}(\{a_{2}^{\pm 1},a_{n-1}^{\pm 1}\})^{*}$. To this end suppose $w$ is a normal form of type (b) and $w\equiv gh$, where $g$ is a maximal prefix of $w$ belonging to $M$ and $g$ has length $t\geq 1$. There are $2$ possibilities for the first letter of $g$ and $3$ for each subsequent letter; so $2.3^{t-1}$ such $g$ altogether. To find those $gh$ with no left divisor in $U$ we distinguish between those $g$ ending $a_{2}^{\pm 1}$ and those ending $a_{n-1}^{\pm 1}$. Let $N$ be the set of all elements of $M$ ending in $a_{2}^{\pm 1}$ and let $P=M\backslash N$. Let $n(t)$ and $p(t)$ be the number of elements of $N$ and $P$, respectively, of length $t\geq 1$. It follows (by induction) that

[TABLE]

for all $t\geq 1$.

If $w\equiv gh$, where $g\in P$ so $g$ ends in $a_{n-1}^{\pm 1}$, then $h$ cannot begin with a letter in $\{a_{2}^{\pm 1},a_{n-1}^{\pm 1}\}$, by maximality of $g$, cannot begin with $a_{n-2}^{\pm 1}$, by definition of normal form, and cannot begin with $a_{1}^{\pm 1}$ as $w$ has no left divisor in $U$. Thus the first letter of $h$ belongs to $\{a_{3}^{\pm 1},\ldots,a_{n-3}^{\pm 1}\}$. As neither $a_{1}$ or $a_{n-1}$ commutes with any element of this set, each subsequent letter of $h$ may be any member of $A_{0}^{\pm 1}$ which results in a normal form. As before, the the number of possibilities for such an $h$ of length $s$ is at most $2(n-5)\alpha^{s-1}$ and at least $2(n-5)\gamma^{s-1}$.

On the other hand, if $w\equiv gh$, where $g\in N$ so $g$ ends in $a_{2}^{\pm 1}$, then $h$ cannot begin with a letter in $\{a_{2}^{\pm 1},a_{n-1}^{\pm 1}\}$ and cannot begin with $a_{1}^{\pm 1}$. As $a_{n-1}$ does not commute with $a_{2}$ and $a_{1}$ does not commute with the first letter of $h$, each subsequent letter of $h$ may again be any element of $A_{0}^{\pm 1}$ which results in a normal form. The number of possibilities for $h$ of length $s$ in this case is then at most $2(n-4)\alpha^{s-1}$and at least $2(n-4)\gamma^{s-1}$.

Now, for $k\geq 2$, let $b(k,t)$ be the number of normal forms $w\in L_{H}^{U}$ of length $k$, of type (b), where $w\equiv gh$, as above, and $g$ has length $t\geq 1$; so $h$ is of length $s=k-t$. When $t<k$ (so $s\geq 1$)

[TABLE]

Similarly,

[TABLE]

where $s=k-t$. For the cases where $t=k$ and $h$ is trivial, $b(k,k)=n(k)+p(k)=2\cdot 3^{k-1}$. We leave consideration of $b(d)$ at this point and turn to normal forms of type (c), as this makes later computation simpler. 3. (c)

Again, to bound the number $c(d)$ of normal forms of length at most $d$, which begin with $a_{n-2}^{\pm 1}$ and have no non-trivial left divisor in $U$, we shall first bound the number $c_{S}(d)$ of such normal forms of length exactly $d$. As before, $c_{S}(0)=0$ and $c_{S}(1)=2$. To find bounds on $c_{S}(k)$ for $k\geq 2$ we consider normal forms $w$ which factor as $gh$, where $g$ is a maximal prefix in the set of reduced words in $\{a_{n-2}^{\pm 1}\}(\{a_{n-2}^{\pm 1},a_{1}^{\pm 1}\})^{*}$. Let the number of such $g$ of length $t\geq 1$ which end in $a_{n-2}$ be $n^{\prime}(t)$ and the number ending in $a_{1}$ be $p^{\prime}(t)$. Then $n^{\prime}(t)=n(t)$ and $p^{\prime}(t)=p(t)$, where $n(t)$ and $p(t)$ are as in case (b) above. Following through the argument of case (b), we find that the number $c(k,t)$ of normal forms $w\in L_{H}^{U}$ of length $k$, of type (c), where $w\equiv gh$, as above, and $g$ has length $t\geq 1$, satisfies

[TABLE]

where $k\geq 2$, $s=k-t$ and $\alpha$ and$\gamma$ are as above.

As

[TABLE]

combining (5.7), (5.8), (5.9) and (5.10) and setting $\beta=2n-9$, gives

[TABLE]

when $k\geq 2$ and $t\geq 1$. Also, $c(k,k)=b(k,k)=2\cdot 3^{k-1}$, when $k\geq 2$.

Therefore, for $n\geq 6$ and $k\geq 2$,

[TABLE]

Similarly, for $n\geq 6$ and $k\geq 2$,

[TABLE]

Note that in fact (5.11) and (5.12) hold for all $k\geq 1$. Then

[TABLE]

Similarly,

[TABLE]

As $l_{H}^{U}(d)=a(d)+b(d)+c(d)$, from (5.6), (5.13) and (5.14),

[TABLE]

In what follows we shall need to know how many elements of $L_{H}^{U}(d)$ do not belong to $U\cup\operatorname{Maln}_{H}(U)$. The elements of $H$ which are not in $\operatorname{Maln}_{H}(U)$ or $U$ are those which belong either to

• •

$\langle a_{n-1}\rangle\langle a_{n-2},a_{1}\rangle$ and have support containing $a_{n-2}$, or to

• •

$\langle a_{1}\rangle\langle a_{n-1},a_{2}\rangle$ and have support containing $a_{2}$.

As elements of $L_{H}^{U}$ have no non-trivial left divisor in $U$, elements of $L_{H}^{U}$ which do not belong to $U\cup\operatorname{Maln}_{H}(U)$ are those in $\langle a_{n-2},a_{1}\rangle$ which begin with $a_{n-2}^{\pm 1}$, or those in $\langle a_{2},a_{n-1}\rangle$ which begin with $a_{2}^{\pm 1}$, all of which are in normal form, as long as they are freely reduced. There are $4\cdot 3^{p-1}$ elements of length $p$ of these forms, so the number $e(d)$ of elements of $L_{H}^{U}(d)$ which do not belong to $U\cup\operatorname{Maln}_{H}(U)$ is

[TABLE]

To bound the size of $L(d,0)$ we simply note that

[TABLE]

Bounds on $l(d,k)$, for $k>0$, are found by considering $L^{\ref{it:composed1}}(d,k)$ and $L^{\ref{it:composed2}}(d,k)$ separately. From (5.4), there are

[TABLE]

elements $ut^{\pm k}$ of type (i) in $L^{\ref{it:composed1}}_{S}(d,k)$. Therefore

[TABLE]

for all $n\geq 5$.

Elements of $L^{\ref{it:composed2}}_{S}(d,k,r,P)$ have the form $ug_{1}t^{\alpha_{1}}\cdots g_{r}t^{\alpha_{r}}$ subject to the conditions of the definition on page • ‣ 5.1. Here $ug_{1}\in H$ and $l(ug_{1})\leq d$ and, as every element of $H$ may be uniquely expressed as $vg$, where $v\in U$ and $g$ has no left divisor in $U$, there are $l_{H}(d)$ possibilities for $ug_{1}$. For $i>1$, $g_{i}\in L_{H}^{U}$, so there are $l_{H}^{U}(d)^{r-1}$ possibilities for $(g_{2},\ldots,g_{r})$. Given the fixed composition $P=(a_{1},\ldots,a_{r})$ of $k$, there are $2^{r}$ choices for $(t^{\alpha_{1}},\ldots,t^{\alpha_{r}})$, with $(|\alpha_{1}|,\ldots,|\alpha_{r}|)=(a_{1},\ldots,a_{r})$. Therefore

[TABLE]

There are $\binom{k-1}{r-1}$ compositions of $k$ into $r$ parts so the number $l^{\ref{it:composed2}}_{S}(d,k,r)$ of elements of $L^{\ref{it:composed2}}_{S}(d,k)$ which involve a partition of $k$ into $r$ parts is

[TABLE]

As $l^{\ref{it:composed2}}_{S}(d,k)=\sum_{r=1}^{k}l^{\ref{it:composed2}}_{S}(d,k,r)$, (5.19) gives

[TABLE]

Now

[TABLE]

We are now ready to calculate the asymptotic density of subsets of $L$ which satisfy conditions 1 – 4 of Theorem 1.2.

5.3 Proof of Proposition 5.1

For each i, from 1 to 4, let $Z_{i}$ be the subset of $L$ which satisfies condition i of Theorem 1.2. We shall show that $\rho(Z_{i})=1$, for $1\leq i\leq 4$. As a finite intersection of generic sets is generic, and $L_{Y}=\cap_{i=1}^{4}Z_{i}$, this proves the proposition.

Theorem 1.2, condition 1.

An element $s\in L$ satisfies this condition if and only if it does not belong to $H$. Thus the intersection of $L(d,k)$ with the set $Z_{1}^{c}$ of these failing elements is $L(d,0)$. From (5.17), the asymptotic density of the set $Z_{1}^{c}$ is

[TABLE]

and from (5.1), (5.3), (5.20) and the definition of $L(d,k)$, it follows that $\rho(Z_{1}^{c})=0$. Therefore, $\rho(Z_{1})=1$; that is the set of elements of $L$ satisfying condition 1 has asymptotic density $1$.

Theorem 1.2, condition 2.

There are no $t$-thick elements in $L(d,0)$ and elements of $L^{\ref{it:composed1}}$, are by definition, all $t$-thick. Thus the number of $t$-thick elements in $L(d,k)$ is $z_{2}(d,k)=l^{\ref{it:composed1}}(d,k)+z_{2}^{(ii)}(d,k)$, where $z_{2}^{(ii)}(d,k)$ is the number of $t$-thick elements in $L^{\ref{it:composed2}}(d,k)$. To find the number of $t$-thick elements in $L^{\ref{it:composed2}}(d,k)$, begin by considering the $t$-thick elements of $L^{\ref{it:composed2}}_{S}(d,k,r,P)$. An element $w=ug_{1}t^{\alpha_{1}}\cdots g_{r}t^{\alpha_{r}}\in L^{\ref{it:composed2}}_{S}(d,k,r,P)$ fails to be $t$-thick only if at least one of the $g_{i}$ is not in $\operatorname{Maln}_{H}(U)$. For $i\geq 2$, $g_{i}$ is an element of $L_{H}^{U}(d)$ and, from (5.16), there are $e(d)=2(3^{d}-1)$ elements of $L_{H}^{U}(d)$ which are not in $\operatorname{Maln}_{H}(U)$. Therefore, there are $t_{H}^{U}(d)=l_{H}^{U}(d)-2(3^{d}-1)$ possible choices for each $g_{i}$ in the $t$-thick element $w$.

Now consider $ug_{1}$. By definition, $l(ug_{1})\leq d$ and supposing that $u$ has length $s$, where $1\leq s\leq d-1$, from (5.16) and the fact that there are $4s$ elements of $U$ of length $s$, there are $4s\cdot e(d-s)$ possible $ug_{1}$, with $g_{1}\notin\operatorname{Maln}_{H}(U)$. Summing over $s$, such that $0\leq s\leq d-1$, the total possible number of such $ug_{1}$ is $e^{\prime}(d)$ where

[TABLE]

The element $ug_{1}$ is an arbitrary element of $L_{H}(d)$, so there are $t_{H}(d)=l_{H}(d)-e^{\prime}(d)$ possible choices for $ug_{1}$ in the $t$-thick element $w$ of $L^{\ref{it:composed2}}_{S}(d,k,r,P)$. Combining these facts, as in the calculation of $l^{\ref{it:composed2}}(d,k)$, we see that the number $z_{2}^{(ii)}(d,k)$ of $t$-thick elements in $l^{\ref{it:composed2}}(d,k)$ is

[TABLE]

As the number of $t$-thick elements in $L(d,k)$ is $z_{2}(d,k)=l^{\ref{it:composed1}}(d,k)+z^{(ii)}_{2}(d,k)$,

[TABLE]

First note that

[TABLE]

If $n=5$ then, from (5.5) and (5.16),

[TABLE]

which has limit [math] as $d,k\rightarrow\infty$, so

[TABLE]

If $n\geq 6$ then $\gamma^{d}-1\leq l_{H}^{U}(d)$, and $\gamma\geq 5$, from (5.15), so $e(d)/(l_{H}^{U}(d)-e(d))$ has limit [math] as $d,k\rightarrow\infty$, and (5.21) holds again.

Next,

[TABLE]

for all $n\geq 5$. Finally, as

[TABLE]

[TABLE]

We have

[TABLE]

If $n=5$,

[TABLE]

which has limit [math] as $d,k\rightarrow\infty$.

If $n\geq 6$, then from (5.3), $l_{H}(d)\geq(n-1)(\gamma^{d}-1)/(n-4)$ so again $e^{\prime}(d)/l_{H}(d)$ has limit [math]. Therefore in all cases

[TABLE]

Also, it follows from (5.21) that

[TABLE]

Combining all the above we see that $\rho(Z_{2})=1$: that is the set of elements of $L$ which satisfy Theorem 1.2.2 has asymptotic density $1$.

Theorem 1.2, condition 3.

A word in $L$ represents an element of $\langle\operatorname{st}(t)\rangle$ if and only if it belongs to $L_{U}$ or $L^{\ref{it:composed1}}$. Therefore, denoting by $Z_{3}^{c}$ the set of elements of $L$ which do not satisfy Theorem 1.2, condition 3, $\rho(Z_{3}^{c})$ is the limit of $[l_{U}(d)+l^{\ref{it:composed1}}(d,k)]/l(d,k)$ as $d,k\rightarrow\infty$. As this limit is [math], it follows that $\rho(Z_{3})=1$.

Theorem 1.2, condition 4

If $w$ in $\mathbb{G}^{\prime}$ is not a $t$-root we say it is a $t$-power. An element $g\in L(d,0)$ has $\sigma(g)\in D$, so $\sigma(g)$ is a generator, or the inverse of a generator, of the factor $\mathbb{F}(D^{+})$ of $F_{D}$; so $\sigma(g)$ is not a proper power. Consequently there are no $t$-powers in $L(d,0)$. We shall count the number of elements of $L(d,k)$ which are $t$-powers and are in $L^{\ref{it:composed1}}$ and in $L^{\ref{it:composed2}}$ separately, and show that both have asymptotic density zero, from which it follows that the set of elements $Z_{4}^{c}$ of $L$ which do not satisfy condition 4 has asymptotic density zero. Elements $ut^{\pm k}$ of $L^{\ref{it:composed1}}$ are all $t$-powers. There are $2[1+d(d+1)]$ of these in $L(d,k)$, from (5.18), and it follows from (5.20) and the definition of $L(d,k)$, that the set of $t$-powers that are in $L^{\ref{it:composed1}}$ has asymptotic density zero.

Let $p^{(ii)}(d,k)$ denote the number of $t$-powers in $L^{\ref{it:composed2}}(d,k)$. If $w=ug_{1}t^{\alpha_{r}}\cdots g_{r}t^{\alpha_{r}}\in L^{\ref{it:composed2}}_{S}(d,k,r,P)$ is a $t$-power then there are elements $d_{i}\in D$ and $u_{i}\in U$ and an integer $p$, such that $1\leq p\leq\lfloor r/2\rfloor$, $r=pq$, $w=_{F}ud_{1}u_{1}t^{\alpha_{1}}\cdots d_{r}u_{r}t^{\alpha_{r}}$ and $\sigma(w)=d_{1}t^{\alpha_{1}}\cdots d_{r}t^{\alpha_{r}}=(d_{1}t^{\alpha_{1}}\cdots d_{p}t^{\alpha_{p}})^{q}$. (The $d_{i}$ are fixed generators of the free group $\mathbb{F}(D^{+})$ (or their inverses).) In this case it follows that $P=(|\alpha_{1}|,\ldots,|\alpha_{r}|)=P_{p}^{q}$, where $P_{p}=(|\alpha_{1}|,\ldots,|\alpha_{p}|)$ is a composition of $q=r/p$ into $p$ parts. Conversely, if there exist positive integers $p$, $q$ such that $p=r/q$, $2\leq q\leq r$ and $P=P_{p}^{q}$ for some composition $P_{p}$ of $q$ into $p$ parts, then for all

• •

elements $d_{i}\in D$ and $u,u_{i,j}\in U$ such that $1\leq i\leq p$ and $1\leq j\leq q$, with

• •

$l(ud_{1}u_{1,1})\leq d$, $l(d_{i}u_{i,j})\leq d$ and $[u_{i,j},d_{i}]\neq 1$, and

• •

all $(\alpha_{1},\ldots,\alpha_{p})$ such that $P_{p}=(|\alpha_{1}|,\ldots,|\alpha_{p}|)$;

[TABLE]

is a $t$-power in $L(d,k,r,P)$. Here $w$ has initial subword $w_{p}=ud_{1}u_{1,1}t^{\alpha_{1}}d_{2}u_{1,2}t^{\alpha_{2}}\cdots d_{p}u_{1,p}t^{\alpha_{p}}$ with normal form in $L^{\ref{it:composed2}}_{S}(d,q,p,P_{p})$ and $\sigma(w_{p})$ is the $q$th root of $\sigma(w)$. Therefore the $t$-power $w$ is determined by the element $w_{p}\in L^{\ref{it:composed2}}_{S}(d,q,p,P_{p})$ and the sequence $(u_{2,1},\ldots,u_{2,p},\ldots u_{q,1},\ldots,u_{q,p})$ of length $p(q-1)=k-p$, of elements of $U$ (each of length at most $d$). There are $(4d)^{k-p}$ sequences of length $k-p$, of such elements of $U$, so writing $p^{(ii)}(d,k,r,P)$ for the number of $t$-powers in $L^{\ref{it:composed2}}_{S}(d,k,r,P)$, and assuming an appropriate composition $P_{p}$ of $r/p$, exists for each $p$ such that $1\leq p\leq\lfloor r/2\rfloor$, we have

[TABLE]

Summing over all compositions of $k$ into $r$ parts and over $r$ from $1$ to $k$, and setting $a=l_{H}^{U}(d)$, $b=l_{H}(d)$ and $c=2d$, the number $p^{(ii)}_{S}(d,k)$ of $t$-powers in $L^{\ref{it:composed2}}_{S}(d,k)$ satisfies

[TABLE]

Therefore the number $p^{(ii)}(d,k)$ of $t$-powers in $L^{\ref{it:composed2}}(d,k)$, satisfies

[TABLE]

In current notation, from (5.20), $l(d,k)\geq l^{\ref{it:composed2}}(d,k)=b[(2a+1)^{k}-1]/a$, so

[TABLE]

For $n=5$ and $n\geq 6$, $c/a^{1/2}$ tends to [math] as $d\rightarrow\infty$ and so, in both cases, it can be seen that $\lim_{d,k\rightarrow\infty}\frac{p^{(ii)}(d,k)}{l(d,k)}=0$. Therefore the set of $t$-powers in $L^{\ref{it:composed2}}$ has asymptotic density [math] and we conclude that $\rho(Z_{4})=1$.

The conclusion is that the set of elements for which all the hypotheses of Theorem 1.2 hold has asymptotic density $1$. That is Theorem 1.2 holds on a generic subset of $L$.