On the finite index subgroups of Houghton’s groups

Abstract

Houghton’s groups $H_2, H_3, \ldots$ are certain infinite permutation groups acting on a countably infinite set; they have been studied, among other things, for their finiteness properties. In this note we describe all of the finite index subgroups of each Houghton group, and their isomorphism types. Using the standard notation that d(G) denotes the minimal size of a generating set for G, we then show, for each $n\in \{2, 3,\ldots \}$ and U of finite index in $H_n$, that $d(U)\in \{d(H_n), d(H_n)+1\}$ and characterise when each of these cases occurs.

1 Introduction

Introduced in [11] by Houghton, the Houghton groups have since attracted attention for their finiteness properties [2, 13], their growth [3, 10], their many interesting combinatorial features [1, 9, 14, 15] as well as other properties.

Definition 1 For $X\ne \emptyset$, let ${{\,\mathrm{Sym}\,}}(X)$ denote the group of all bijections on X. For $g\in {{\,\mathrm{Sym}\,}}(X)$, let ${{\,\mathrm{supp}\,}}(g):=\{x\in X\;:\;(x)g\ne x\}$, called the support of g. Then ${{\,\mathrm{FSym}\,}}(X)=\{g\in {{\,\mathrm{Sym}\,}}(X)\;:\;|{{\,\mathrm{supp}\,}}(g)|<\infty \}$ and ${{\,\mathrm{Alt}\,}}(X)\le {{\,\mathrm{FSym}\,}}(X)$ consists of only the even permutations, meaning $[{{\,\mathrm{FSym}\,}}(X)\;:\;{{\,\mathrm{Alt}\,}}(X)]=2$.

We give a brief overview of these groups for our purposes; more detailed introductions can be found, for example, in [1, 4]. We will use right actions throughout. We define $\mathbb {N}:=\{1, 2,\ldots \}$, let $G\le _f H$ denote that G is a finite index subgroup of H, and for a group G and $g, h\in G$, let $[g, h]:=g^{-1}h^{-1}gh$ and $g^h:=h^{-1}gh$.

Definition 2 Let $n\in \{3, 4, \ldots \}$. Then the nth Houghton group, denoted $H_n$, is generated by $g_2, \ldots , g_n\in {{\,\mathrm{Sym}\,}}(X_n)$ where $X_n=\{1, \ldots , n\}\times \mathbb {N}$ and for $k\in \{2, \ldots , n\}$,

$$\begin{aligned} (i,m)g_k=\left\{ \begin{array}{ll}(1, m+1) &{} \mathrm {if}\ i=1\ \mathrm {and}\ m\in \mathbb {N},\\ (1,1) &{} \mathrm {if}\ i=k\ \mathrm {and}\ m=1,\\ (k, m-1) &{} \mathrm {if}\ i=k\ \mathrm {and}\ m\in \{2, 3, \ldots \},\\ (i, m) &{} \mathrm {otherwise}. \end{array}\right. \end{aligned}$$

(1)

For each $n\in \{3, 4, \ldots \}$, we have that ${{\,\mathrm{FSym}\,}}(X_n)\le H_n$. One way to see this is to first compute that $[g_2, g_3]=((1, 1), (1, 2))$, and then observe that any 2-cycle with support in $X_n$ can be conjugated, using the elements $g_2, \ldots , g_n$, to ((1, 1), (1, 2)). Furthermore, as observed in [15], we have a short exact sequence of groups

$$\begin{aligned} 1\longrightarrow {{\,\mathrm{FSym}\,}}(X_n){\mathop {\longrightarrow }\limits ^{}} H_n{\mathop {\longrightarrow }\limits ^{\pi }} \mathbb {Z}^{n-1}\longrightarrow 1. \end{aligned}$$

Here $\pi$ is induced by defining $\pi (g_i):=e_{i-1}$ for $i=2, \ldots , n$, where $e_i$ denotes the vector in $\mathbb {Z}^{n-1}$ with ith entry 1 and other entries 0.

Definition 3 The second Houghton group is generated by the two cycle ((1, 1), (1, 2)) together with the element $g_2$, defined analogously to (1) above. This is isomorphic to ${{\,\mathrm{FSym}\,}}(\mathbb {Z})\rtimes \langle t\rangle$ where $t\in {{\,\mathrm{Sym}\,}}(\mathbb {Z})$ sends each $z\in \mathbb {Z}$ to $z+1$.

There have been many papers with questions and results relating to the finite index subgroups of this family of groups, e.g. the questions on invariable generation in [15] and subsequent answers in [6], showing they all have solvable conjugacy problem in [4], and also all have the $R_\infty$ property [5, 8]. Some of these use the partial description of the finite index subgroups from [3]. We start by giving the first full description of them.

Theorem 1

Let $n\in \{3, 4, \ldots \}$ and $U\le _f H_n$. Then there exist $c_2, \ldots , c_n\in \mathbb {N}$ such that $\pi (U)=\langle c_2e_1, \ldots , c_ne_{n-1}\rangle$ and either U is:

• (i) equal to $\langle {{\,\mathrm{FSym}\,}}(X_n), g_i^{c_i}\;:\;i=2, \ldots , n\rangle$; or

• (ii) isomorphic to $\langle {{\,\mathrm{Alt}\,}}(X_n), g_i^{c_i}\;:\;i=2, \ldots , n\rangle$.

If $U\le _fH_2$, then there exists $c_2\in \mathbb {N}$ such that $\pi (U)=\langle c_2\rangle \le \mathbb {Z}$ and either (i) or (ii) above occurs or U is equal to $\langle {{\,\mathrm{Alt}\,}}(X_2), ((1, 1), (1, 2)g_2)\rangle$.

Given $U, U'\le _fH_n$ such that $\pi (U)=\langle c_2e_1, \ldots , c_ne_{n-1}\rangle$ for $c_2, \ldots , c_n\in \mathbb {N}$ and $\pi (U')=\langle d_2e_1, \ldots , d_ne_{n-1}\rangle$ for $d_2, \ldots , d_n\in \mathbb {N}$, one might wonder when $U\cong U'$. Clearly any permutation of the constants $c_2, \ldots , c_n$ produces an isomorphism. By considering ${{\,\mathrm{Aut}\,}}(U)$ and ${{\,\mathrm{Aut}\,}}(U')$, it seems that this is the only way for the groups to be isomorphic. Our methods do allow us to obtain the following.

Corollary 2

Let $n \in \{2, 3, \ldots \}$ and $c_2, \ldots , c_n\in \mathbb {N}$. If $U\le _fH_n$ and $\pi (U)=\langle c_2e_1, \ldots , c_ne_{n-1}\rangle$, then either

• at least two of $c_2, \ldots , c_n$ are odd and $U=\langle g_2^{c_2}, \ldots , g_n^{c_n}, {{\,\mathrm{FSym}\,}}(X_n)\rangle$; or

• U is one of exactly $2^{n-1}+1$ specific subgroups of $H_n$.

We then extend the work in [7], where the groups $\langle {{\,\mathrm{Alt}\,}}(X_2), g_2^c\rangle$ were shown to be 2-generated for each $c\in \mathbb {N}$, by investigating the generation properties of each of these groups.

Notation 1 For a finitely generated group G, let $d(G):=\min \{|S|\;:\;\langle S\rangle =G\}$.

Every finite index subgroup of a finitely generated group is itself finitely generated. In the case of $\mathbb {Z}^n$, where $n\in \mathbb {N}$, we have that each $A\le _f \mathbb {Z}^n$ satisfies $d(A)=d(\mathbb {Z}^n)$. There has been much work related to this notion, e.g. [12] and papers verifying the Nielsen–Schreier formula for families of groups other than free groups. Our second theorem states that there is a close connection between the minimal number of generators of a Houghton group and its finite index subgroups. Note that $d(H_2)=2$ and $d(H_n)=n-1$ for $n\in \{3, 4, \ldots \}$.

Theorem 3

If $U\le _f H_2$, then $d(U)=d(H_2)$. For $n\in \{3, 4 \ldots \}$ and $U\le _fH_n$, we have that $d(U)\in \{d(H_n), d(H_n)+1\}$. Furthermore, let $\pi (U)=\langle c_2e_1, \ldots , c_ne_{n-1}\rangle$. Then $d(U)=d(H_n)+1$ occurs exactly when both of the following conditions are met:

• i) that ${{\,\mathrm{FSym}\,}}(X_n)\le U$; and

• ii) either one or zero elements in $\{c_2, \ldots , c_n\}$ are odd.

Our proof involves providing a generating set. Theorem 1 allows us to replace U with either $\langle {{\,\mathrm{FSym}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$ or $\langle {{\,\mathrm{Alt}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$. Proposition states that, in the second case, $d(U)=n-1$. Lemma 3.5 and Lemma 3.6 combine to tell us that $\langle {{\,\mathrm{FSym}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle =\langle {{\,\mathrm{Alt}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$ when there are distinct $i, j \in \{2, \ldots , n\}$ that are both odd. Thus the only remaining possibility is that $U=\langle {{\,\mathrm{FSym}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle \ne \langle {{\,\mathrm{Alt}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$. Lemma 3.9 shows that in this case $d(U)=d(H_n)+1$, by determining that the abelianization of U, with these conditions, is $C_2\times \mathbb {Z}^{n-1}$. This complete categorisation provides us with subgroups of the Houghton groups with constant minimal number of generators on finite index subgroups.

Corollary 4

Let $n\in \{3, 4, \ldots \}$ and define $G_\mathbf{2}:=\langle {{\,\mathrm{Alt}\,}}(X_n), g_2^2, \ldots , g_n^2\rangle \le _f H_n$. If $U\le _f G_\mathbf{2}$, then $d(U)=d(G_\mathbf{2})=d(H_n)$.

2 The structure of finite index subgroups of $H_n$

In this section, we prove Theorem 1. The following is well known.

Lemma 2.1

Given $U\le _fG$, there exists $N\le _fU$ which is normal in G.

Some structure regarding finite index subgroups of $H_n$ is known; the following is particularly useful to us.

Lemma 2.2

([5, Prop. 2.5]). Let X be a non-empty set and ${{\,\mathrm{Alt}\,}}(X)\le G\le {{\,\mathrm{Sym}\,}}(X)$. Then G has ${{\,\mathrm{Alt}\,}}(X)$ as a unique minimal normal subgroup.

Let $n\in \{2, 3, \ldots \}$ and $U\le _f H_n$. By Lemma 2.1, U contains a normal subgroup of $H_n$ and so, by Lemma 2.2, ${{\,\mathrm{Alt}\,}}(X_n)\le U$. Furthermore, $U\le _f H_n$ and so $\pi (U)\le _f\pi (H_n)$ where $\pi : H_n\rightarrow \mathbb {Z}^{n-1}$ sends $g_i$ to $e_{i-1}$ for $i=2, \ldots , n$. Hence there exist minimal $k_2, \ldots , k_n\in \mathbb {N}$ such that $\pi (g_i^{k_i})\in \pi (U)$ for each $i\in \{2, \ldots , n\}$. But $\pi (g_i^{k_i})=k_ie_{i-1}$, and so the preimage in $H_n$ of $\pi (g_i^{k_i})$ is $\{\sigma g_i^{k_i}\;:\;\sigma \in {{\,\mathrm{FSym}\,}}(X_n)\}$. If ${{\,\mathrm{FSym}\,}}(X_n)\le U$, then $g_2^{k_2}, \ldots , g_n^{k_n} \in U$. Otherwise ${{\,\mathrm{FSym}\,}}(X_n)\cap U={{\,\mathrm{Alt}\,}}(X_n)$ and, for each $i\in \{2, \ldots , n\}$, either $g_i^{k_i}\in U$ or there exists $\omega _i\in {{\,\mathrm{FSym}\,}}(X_n)\setminus {{\,\mathrm{Alt}\,}}(X_n)$ such that $\omega _i g_i^{k_i}\in U$; since ${{\,\mathrm{Alt}\,}}(X_n)\le U$, we can specify that $\omega _i=((1, 1)\;(1, 2))$. Then, using the minimality of $k_2, \ldots , k_n$,

$$\begin{aligned} U=\langle g_2^{k_2}, \ldots , g_n^{k_n}, {{\,\mathrm{FSym}\,}}(X_n)\rangle \text { or }U=\langle \epsilon _2g_2^{k_2}, \ldots , \epsilon _ng_n^{k_n}, {{\,\mathrm{Alt}\,}}(X_n)\rangle \end{aligned}$$

(2)

where each $\epsilon _i$ is either trivial or $((1, 1)\;(1, 2))$. We now describe the isomorphism type for these subgroups. To do this, we introduce two families of finite index subgroups in $H_n$ where $n\in \{2, 3, \ldots \}$.

Notation 2 For any given $n\in \{2, 3, \ldots \}$ and any $\mathbf{c}=(c_2, \ldots , c_n)\in \mathbb {N}^{n-1}$, let $F_\mathbf{c}:=\langle {{\,\mathrm{FSym}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$ and $G_\mathbf{c}:=\langle {{\,\mathrm{Alt}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$.

There are some $\mathbf{c}\in \mathbb {N}^n$ such that $G_\mathbf{c}=F_\mathbf{c}$, e.g. if $n\ne 2$ and $c_2=\cdots =c_n=1$.

Notation 3 Let $I_n:=\{\mathbf{c}=(c_2, \ldots , c_n)\in \mathbb {N}^n\;:\;G_\mathbf{c}\ne F_\mathbf{c}\}$.

Lemma 3.5 and Lemma 3.6 together describe $I_n$. This description, together with (2), yields Corollary 2.

Notation 4 For $n\in \{2, 3, \ldots \}$ and $i\in \{1, \ldots , n\}$, let $R_i:=\{(i,m)\;:\;m\in \mathbb {N}\}\subset X_n$.

Proposition 2.3

Let $n\in \{3, 4, \ldots \}$, $\mathbf{c}=(c_2, \ldots , c_n)\in I_n$, each $\epsilon _2, \ldots , \epsilon _n$ be either trivial or $((i, 1)\;(i, 2))$, and $U=\langle {{\,\mathrm{Alt}\,}}(X_n), \epsilon _ig_i^{c_i}\;:\; i=2, \ldots , n, \rangle \le _f H_n$. Then U is isomorphic to $G_\mathbf{c}$.

Proof

If there are $i\ne j$ such that $c_i=c_j=1$, then ${{\,\mathrm{FSym}\,}}(X_n)\le G_\mathbf{c}$ and so $G_\mathbf{c}= F_\mathbf{c}$, i.e. $\mathbf{c}\not \in I_n$.

Now consider if $c_i=1$ for some i and that $\epsilon _i =((i, 1)\;(i, 2))$. Then $\epsilon _ig_i$ fixes (i, 1). Moreover, we have $(i, m)\epsilon _ig_i=(i,m-1)$ for all $m\ge 3$, $(i,2)\epsilon _ig_i=(1, 1)$, $(1, m)\epsilon _i=(1,m+1)$ for all $m\ge 1$, and that $\epsilon _ig_i$ fixes all other points in $X_n$. This allows us to relabel $X_n$ so that the point (i, 1) becomes part of another ray $j\in \{2, \ldots , n\}$.

If $c_i\ne 1$ and $\epsilon _i=((i,1)\;(i,2))$, then we can relabel $R_i$ as $R_i'$ by swapping the labels on the points (i, 1) and (i, 2) so that $\epsilon _ig_i^{c_i}$ acts on $R_i'\cup R_1$ in the same way as $g_i^{c_i}$ acts on $R_i\cup R_1$. $\square$

Lemma 2.4

Let $k\in \{2, 3, \ldots \}$, $\epsilon =((1, 1)\;(1, 2))$, and $U=\langle {{\,\mathrm{Alt}\,}}(X_n), \epsilon g_2^k \rangle \le _f H_2$. Then U is isomorphic to $\langle {{\,\mathrm{Alt}\,}}(X_2), g_2^k\rangle$.

Proof

In a similar way to the preceding proof, we can relabel $R_2$ as $R_2'$ (by swapping the labels on the points (2, 1) and (2, 2)) so that $\epsilon g_2^k$ acts on $R_1\cup R_2'$ in the same way that $g_2^k$ acts on $R_1\cup R_2$. $\square$

Remark 2.5

The preceding lemma may hold for $k=1$. Such an isomorphism would not be induced by a permutation of $X_2$, however: any element g in $\langle {{\,\mathrm{Alt}\,}}(X_2), ((1, 1)\;(1, 2)) t\rangle$ with $\pi (g)=1$ cannot have an infinite orbit equal to $X_2$.

3 Generation of the groups $F_\mathbf{c}$ and $G_\mathbf{c}$

We start with the case of $H_2$, first dealing with the ‘exceptional’ case.

Lemma 3.1

The group $\langle {{\,\mathrm{Alt}\,}}(X_2), ((1, 1)\;(1, 2))g_2\rangle$ is 2-generated.

Proof

Note that $\langle {{\,\mathrm{Alt}\,}}(X_2), ((1, 1)\;(1, 2))g_2\rangle \cong \langle {{\,\mathrm{Alt}\,}}(\mathbb {Z}), (1\;2)t\rangle \le {{\,\mathrm{FSym}\,}}(\mathbb {Z})\rtimes \langle t\rangle$ where $t: z\rightarrow z+1$ for all $z\in \mathbb {Z}$. Our aim will be to show that $S=\{(0\;1\;2), (0\;1)t\}$ generates $\langle {{\,\mathrm{Alt}\,}}(\mathbb {Z}), (1\;2)t\rangle$. To do this, we will use $\langle S\rangle$ to construct all 3-cycles $(0\;a\;b)$ with $0<a<b$. With these elements we can then also produce every 3-cycle of the form $(a\;b\;c)$ where $0<a<b<c$, by conjugating $(0\;a\;b)$ by $(0\;c\;2c)$. Then every 3-cycle in ${{\,\mathrm{Alt}\,}}(\mathbb {N}\cup \{0\})$, or its inverse, is accounted for by such elements. Conjugation by some power of $(0\;1)t$ yields every 3-cycle in ${{\,\mathrm{Alt}\,}}(\mathbb {Z})$, and we recall that ${{\,\mathrm{Alt}\,}}(\mathbb {Z})$ is generated by the set of all 3-cycles with support in $\mathbb {Z}$.

In addition to our first simplification, note that it is sufficient to show that $(0, 1, k+1)\in \langle (0\;1\;2), (0\;1)t\rangle$ for every $k\in \mathbb {N}$. This is because any 3-cycle $(0\;a\;b)$ with $0<a<b$ will be conjugate, by $(0\;1)t$, to an element of the form $(0, 1, k+1)$. We start with $\sigma _1:=(0, k, k+1)$. If $k=1$, then we are done. Otherwise, conjugate $\sigma _1^{-1}$ by $(0, k-1, k)$ to obtain $\sigma _2=(0, k-1, k+1)$. If $k>2$, conjugate $\sigma _2^{-1}$ by $(0, k-2, k-1)$ to obtain $\sigma _3$. Continuing in this way yields the result. $\square$

The following results conclude the $H_2$ case. We include the proof here as we will adapt it for $F_\mathbf{c}$ and $G_\mathbf{c}$. The notation $\Omega _{k}:=\{1, \ldots , k\}$ is helpful for these results.

Lemma 3.2

([5, Lem. 3.6]). Let $k\in \{3, 4, \ldots \}$ and $t: z\rightarrow z+1$ for all $z\in \mathbb {Z}$. Then $\langle {{\,\mathrm{Alt}\,}}(\mathbb {Z}), t^k\rangle$ is generated by ${{\,\mathrm{Alt}\,}}(\Omega _{2k}) \cup \{t^k\}$.

Lemma 3.3

([5, Lem. 3.7]). Let $k\in \mathbb {N}$ and $t: z\rightarrow z+1$ for all $z\in \mathbb {Z}$. Then $G_k:=\langle {{\,\mathrm{Alt}\,}}(\mathbb {Z}), t^k\rangle$ and $F_k:=\langle {{\,\mathrm{FSym}\,}}(\mathbb {Z}), t^k\rangle$ are 2-generated.

Proof

We start with $G_k$. We will show that we can find, for each $k\in \{3, 4, \ldots \}$, an $\alpha _k\in {{\,\mathrm{Alt}\,}}(\mathbb {Z})$ such that $\langle t^k, \alpha _k\rangle$ contains ${{\,\mathrm{Alt}\,}}(\mathbb {Z})$. Then, for $d\in \{1, 2, 3\}$, $\langle t^d, \alpha _6\rangle$ contains $\langle t^6, \alpha _6\rangle$ and so contains ${{\,\mathrm{Alt}\,}}(\mathbb {Z})$. We can therefore fix some $k\in \{3, 4, \ldots \}$, meaning all 3-cycles in ${{\,\mathrm{Alt}\,}}(\Omega _{2k})$ are conjugate.

Let $r=\left( {\begin{array}{c}2k\\ 3\end{array}}\right)$ and let $\omega _1, \ldots , \omega _r$ be a choice of distinct 3-cycles in ${{\,\mathrm{Alt}\,}}(\Omega _{2k})$ with $\omega _i\ne \omega _j^{-1}$ for every $i,j \in \{1, \ldots , r\}$. Thus $\langle \omega _1, \ldots , \omega _r\rangle ={{\,\mathrm{Alt}\,}}(\Omega _{2k})$. Set $\sigma _0=(1\;3)$ and $\sigma _{r+1}=(2\;3)$ and note that $(\sigma _{r+1}^{-1}\sigma _0^{-1}\sigma _{r+1})\sigma _0=(1\;2\;3)$. Now choose $\sigma _1, \ldots , \sigma _{r}\in {{\,\mathrm{Alt}\,}}(\Omega _{2k})$ so that for each $m\in \{1,\ldots , r\}$, we have $\sigma _m^{-1}(1\;2\;3)\sigma _m=\omega _m$.

Let $\alpha _k:=\prod _{i=0}^{r+1}t^{-2ik}\sigma _it^{2ik}$ and, for $m\in \{1,\ldots , r+1\}$, let $\beta _m:=t^{2mk}\alpha _k t^{-2mk}$. Then $\beta _{r+1}^{-1}\alpha _k^{-1}\beta _{r+1}\alpha _k=\sigma _{r+1}^{-1}\sigma _0^{-1}\sigma _{r+1}\sigma _0=(1\;2\;3)$ and, for $m\in \{1, \ldots , r\}$, we have that $\beta _m^{-1}(1\;2\;3)\beta _m=\omega _m$. Hence $\langle \alpha _k, t^k\rangle =G_k$.

We can now adapt the element $\alpha _k$ to an element $\gamma _k$ so, for each $k\in \{3, 4, \ldots \}$, we have that $\langle \gamma _k, t^k\rangle =F_k$. One way to do so is to set $\gamma _k:=(\sigma _0')(\prod _{i=1}^{r+1}t^{-2ik}\sigma _it^{2ik})$ where $\sigma _0'\!=\!(1\;3)(4\;5)$. This means that $[\sigma _{r+1}, \sigma _0']$ still equals $(1\;2\;3)$, and so ${{\,\mathrm{Alt}\,}}(\mathbb {Z})\le \langle \gamma _k, t^k\rangle$, but that ${{\,\mathrm{Alt}\,}}(\mathbb {Z})\cup \gamma _k({{\,\mathrm{Alt}\,}}(\mathbb {Z}))={{\,\mathrm{FSym}\,}}(\mathbb {Z})\le \langle \gamma _k, t^k\rangle$ as well. $\square$

We now work with a fixed $n\in \{3, 4, \ldots \}$. The following well-known commutator identities will be helpful for a few of our remaining proofs.

$$\begin{aligned}{}[a, bc]=a^{-1}(bc)^{-1}a(bc)=a^{-1}c^{-1}acc^{-1}a^{-1}b^{-1}abc=[a, c][a, b]^c \end{aligned}$$

(3)

$$\begin{aligned}{}[ab, c]=(ab)^{-1}c^{-1}(ab)c=b^{-1}(a^{-1}c^{-1}ac)bb^{-1}c^{-1}bc=[a, c]^b[b, c] \end{aligned}$$

(4)

Remark 3.4

Let $a, b, c \in H_n$ and $\mathrm {sgn}$ denote the sign function on ${{\,\mathrm{FSym}\,}}(X_n)$. From the identities (3) and (4) above, we have that $\mathrm {sgn}([a, bc])=\mathrm {sgn}([a, b])\mathrm {sgn}([a, c])$ and $\mathrm {sgn}([ab, c])=\mathrm {sgn}([a, c])\mathrm {sgn}([b, c])$.

Recall that $\mathbf{c}\in \mathbb {N}^{n-1}$ is in $I_n$ if and only if $G_\mathbf{c}\ne F_\mathbf{c}$.

Lemma 3.5

Let $\mathbf{c}=(c_2, \ldots , c_n)\in \mathbb {N}^n$. We have that $\mathbf{c}\in I_n$ if and only if $[g_i^{c_i}, g_j^{c_j}]\in {{\,\mathrm{Alt}\,}}(X_n)$ for all $i, j \in \{2, \ldots , n\}$.

Proof

If there exist $i, j\in \{2, \ldots , n\}$ such that $[g_i^{c_i}, g_j^{c_j}]\not \in {{\,\mathrm{Alt}\,}}(X_n)$, then $G_\mathbf{c}$ contains an odd permutation and so $G_\mathbf{c}= F_\mathbf{c}$. Recall that $G_\mathbf{c}=\langle {{\,\mathrm{Alt}\,}}(X_n), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$ and that we have a homomorphism $\pi : H_n\rightarrow \mathbb {Z}^{n-1}$ induced by sending $g_i$ to $e_{i-1}$ for $i=2, \ldots , n$. Then $\pi (G_\mathbf{c})\le _f\pi (H_n)=\mathbb {Z}^{n-1}$, and so $\pi (G_\mathbf{c})$ is free abelian of rank $n-1$. Define $S:=\{g_2^{c_2}, \ldots , g_n^{c_n}\}$ so that $\langle S\cup {{\,\mathrm{Alt}\,}}(X_n)\rangle = G_\mathbf{c}$ and $\pi (S)$ is a linearly independent set. Let $t_i:=\pi (g_i^{c_i})$ for $i=2, \ldots , n$. Thus

$$\begin{aligned} \pi (\langle S\rangle )=\langle t_2, \ldots , t_n\mid R\rangle \text { where }R=\{[t_i, t_j]\;:\;i\ne j\}. \end{aligned}$$

(5)

Now let $\alpha \in \langle S\rangle \cap {{\,\mathrm{FSym}\,}}(X_n)$. We will show that $\alpha \in {{\,\mathrm{Alt}\,}}(X_n)$, meaning that $\mathbf{c}\in I_n$. First, express $\alpha$ as a word w in $S^{\pm 1}$. Thus $\alpha =w(g_2^{c_2}, \ldots , g_n^{c_n})$ and $\pi (\alpha )=w(t_2, \ldots , t_n)$ for the same word w. But also $\pi (\alpha )=\underline{0}$ which, from (5), means that $w(t_2, \ldots , t_n)$ is in the normal closure of R. Thus $w(t_2, \ldots , t_n)$ is a product of conjugates of powers of commutators in terms of $t_2, \ldots , t_n$. Then $\alpha =w(g_2^{c_2}, \ldots , g_n^{c_n})$ must also be a product of conjugates of powers of commutators in terms of $g_2^{c_2}, \ldots , g_n^{c_n}$. Our assumption that $[g_i^{c_i}, g_j^{c_j}]\in {{\,\mathrm{Alt}\,}}(X_n)$ for all $i, j \in \{2, \ldots , n\}$ together with Remark 3.4 means that $\alpha \in {{\,\mathrm{Alt}\,}}(X_n)$. $\square$

The following gives a clearer description of the set $I_n$.

Lemma 3.6

Let $i, j\in \{2, \ldots , n\}$ and $c_i, c_j\in \mathbb {N}$. Then $[g_i^{c_i}, g_j^{c_j}]\not \in {{\,\mathrm{Alt}\,}}(X_n)$ if and only if $c_i$ and $c_j$ are both odd.

Proof

Note that $[g_i, g_j]\in {{\,\mathrm{FSym}\,}}(X_n)\setminus {{\,\mathrm{Alt}\,}}(X_n)$ whereas $[g_i^2, g_j^2], [g_i, g_j^2], [g_i^2, g_j] \in {{\,\mathrm{Alt}\,}}(X_n)$. Now, by repeatedly applying Remark 3.4, we can reduce $c_i$ and $c_j$ to be either 1 or 2, depending on whether they are odd or even respectively. $\square$

Combining Lemma 3.5 and Lemma 3.6, we see that $\mathbf{c}=(c_2, \ldots , c_n)\in \mathbb {N}^{n-1}$ is in $I_n$ if and only if at least two of $c_2, \ldots , c_n$ are odd.

Proposition 3.7

Given $\mathbf{c}=(c_2, \ldots , c_n)\in \mathbb {N}^{n-1}$, we have that $d(G_\mathbf{c})=d(H_n)$.

Proof

We start with the case that $\mathbf{c}\in I_n$. By possibly relabelling the branches of $X_n$, set $c_3:=\max \{c_2, \ldots , c_n\}$. Thus $c_3\ne 1$, as otherwise $c_2=c_3=1$ and $\mathbf{c}\not \in I_n$. Let $k:=2c_3$ and $\Omega ^*:=\{(1, 1), \ldots , (1, 2k)\}$. Thus $2k\ge 8$ and all 3-cycles in ${{\,\mathrm{Alt}\,}}(\Omega ^*)$ are conjugate in ${{\,\mathrm{Alt}\,}}(\Omega ^*)$. We claim that ${{\,\mathrm{Alt}\,}}(X_n)\le \langle {{\,\mathrm{Alt}\,}}(\Omega ^*), g_2^{c_2}, \ldots , g_n^{c_n}\rangle$. By Lemma 3.2, $\langle {{\,\mathrm{Alt}\,}}(\Omega ^*), g_3^{c_3}\rangle$ is a subgroup of $H_n$ which generates $\langle {{\,\mathrm{Alt}\,}}(R_1\cup R_3), g_3^{c_3}\rangle \cong \langle {{\,\mathrm{Alt}\,}}(\mathbb {Z}), t^{c_3}\rangle$. Most importantly, $\langle {{\,\mathrm{Alt}\,}}(\Omega ^*), g_3^{c_3}\rangle$ contains ${{\,\mathrm{Alt}\,}}(R_1)$. Now, given any $\sigma \in {{\,\mathrm{FSym}\,}}(X_n)$, there exists a word of the form $g_2^{d_2}\ldots g_n^{d_n}$ for some $d_i\in c_i\mathbb {N}$ which conjugates $\sigma$ to $\sigma '$, where ${{\,\mathrm{supp}\,}}(\sigma ')\subset R_1$. Hence $\langle {{\,\mathrm{Alt}\,}}(\Omega ^*), g_2^{c_2}, \ldots , g_n^{c_n}\rangle =G_\mathbf{c}$. Our next aim is to adapt the proof of Lemma 3.3. In particular, we will show that there exists a $\sigma \in {{\,\mathrm{Alt}\,}}(X_n)$ such that $\langle \sigma g_2^{c_2}, g_3^{c_3}, \ldots , g_n^{c_n}\rangle =G_\mathbf{c}$. For any $\omega \in {{\,\mathrm{FSym}\,}}(X_n)$, let $\sigma _\omega :=[\omega g_2^{c_2}, g_3^{2k}]$, which we can calculate using (4) as

$$\begin{aligned}{}[\omega , g_3^{2k}]^{g_2^{c_2}}[g_2^{c_2}, g_3^{2k}]=g_2^{-c_2}(\omega ^{-1}g_3^{-2k}\omega g_3^{2k})g_2^{c_2}[g_2^{c_2}, g_3^{2k}]. \end{aligned}$$

(6)

Observe that ${{\,\mathrm{supp}\,}}([g_2^{c_2}, g_3^{2k}])\subseteq \{(1, 1), \ldots , (1, c_2+2k)\}$ and, since $k=2c_3>2c_2$, that $(1, 4k+1), (1, 4k+3)\not \in {{\,\mathrm{supp}\,}}([g_2^{c_2}, g_3^{k}])$. Let $\delta _0:=((1, 2k-c_2+1), (1, 2k-c_2+3))$ and note, using (6), that ${{\,\mathrm{supp}\,}}(\sigma _{\delta _0})\cap \{(1, m)\;:\;m=4k+2 \text { or }m\ge 4k+4\}=\emptyset$ and $\sigma _{\delta _0}$ swaps $(1, 4k+1)$ and $(1, 4k+3)$. We now restrict to those $\omega \in {{\,\mathrm{Alt}\,}}(X_n)$ with

$${{\,\mathrm{supp}\,}}(\omega \delta _0)\subset \bigcup _{j\in \mathbb {N}}(\Omega ^*)g_3^{-4jk}.$$

The following implications of this restriction are all important in relation to (6):

• ${{\,\mathrm{supp}\,}}(\sigma _\omega )\subset R_1\cup R_3$;

• ${{\,\mathrm{supp}\,}}(\omega ^{-1})\cap {{\,\mathrm{supp}\,}}(g_3^{-2k}\omega g_3^{2k})=\emptyset$;

• $(x)g_2^{-c_2}\omega ^{-1}g_2^{c_2}=(x)\omega ^{-1}$ for all $x\in R_3$; and

• $(x)g_2^{-c_2}(g_3^{-2k}\omega g_3^{2k})g_2^{c_2}=(x)g_3^{-2k}\omega g_3^{2k}$ for all $x\in R_3$.

Now, by mimicking the proof of Lemma 3.3, we can choose one such $\omega$, which we will denote by $\alpha$, so that ${{\,\mathrm{Alt}\,}}(R_1)\le \langle g_3^k, \sigma _{\alpha }\rangle$. Let $r=\left( {\begin{array}{c}2k\\ 3\end{array}}\right)$ and $\omega _1, \ldots , \omega _r$ be a set of 3-cycles such that $\langle \omega _1, \ldots , \omega _r\rangle ={{\,\mathrm{Alt}\,}}(\Omega ^*)$. Choose $\sigma _1, \ldots , \sigma _r\in {{\,\mathrm{Alt}\,}}(\Omega ^*)$ such that $\sigma _i^{-1}((1, 1)\,(1, 2)\,(1, 3))\sigma _i=\omega _i$ for $i=1, \ldots , r$. Also set $\sigma _{r+1}:=((1, 2)\,(1, 3))$ and $\sigma _0:=((1, 1)\,(1, 3))$. Define $\delta _i:=g_3^{4ki}\sigma _ig_3^{-4ki}$ for $i=1, \ldots , r+1$ and $\alpha :=\delta _0\ldots \delta _{r+1}$. Define $\beta _m:=g_3^{-4mk}\sigma _\alpha g_3^{4mk}$ for $m=-1, \ldots , r+1$. Then, as in Lemma 3.3, we have that $[\beta _{r+1}, \beta _{-1}]=[\sigma _{d+1}, \sigma _0]$ and $\beta _i[\sigma _{d+1}, \sigma _0]\beta _i^{-1}=\omega _i$ for $i=1, \ldots , r$.

We end our proof with the case where $\mathbf{c}\not \in I_n$. From Lemma 3.5 and Lemma 3.6, there then exist $i, j \in \{2, \ldots , n\}$ such that $[g_i^{c_i}, g_j^{c_j}]\in {{\,\mathrm{FSym}\,}}(X_n)\setminus {{\,\mathrm{Alt}\,}}(X_n)$. Using the element $\sigma _\alpha \in {{\,\mathrm{Alt}\,}}(X_n)$ defined above, we claim that $\langle \sigma _\alpha g_2^{c_2}, g_3^{c_3}, \ldots , g_n^{c_n}\rangle = F_\mathbf{c}$. We note that if $[g_i^{c_i}, g_j^{c_j}]\in {{\,\mathrm{FSym}\,}}(X_n)\setminus {{\,\mathrm{Alt}\,}}(X_n)$, then, by Remark 3.4, so is $[\sigma g_i^{c_i}, g_j^{c_j}]$ for any $\sigma \in {{\,\mathrm{FSym}\,}}(X_n)$. Hence $\langle \sigma _\alpha g_2^{c_2}, g_3^{c_3}, \ldots , g_n^{c_n}\rangle$ contains both ${{\,\mathrm{Alt}\,}}(X_n)$ and an odd permutation, and so $d(G_\mathbf{c})=d(H_n)$. $\square$

We now look at the final case of determining $d(F_\mathbf{c})$ where $\mathbf{c}\in I_n$.

Lemma 3.8

Let $\mathbf{c}\in I_n$. Then the commutator subgroup of $F_\mathbf{c}$ equals ${{\,\mathrm{Alt}\,}}(X_n)$.

Proof

Note that $[{{\,\mathrm{Alt}\,}}(X_n), {{\,\mathrm{Alt}\,}}(X_n)]={{\,\mathrm{Alt}\,}}(X_n)$. Let $\sigma \prod _{i\in I} g_i^{p_i}$ and $\omega \prod _{j\in J} g_j^{q_j}$ be in $F_\mathbf{c}$, where $I, J\subseteq \{2, \ldots , n\}$, $\{p_i\;:\;i\in I\}\subset \mathbb {Z}$, $\{q_j\;:\;j\in J\}\subset \mathbb {Z}$, and $\sigma , \omega \in {{\,\mathrm{FSym}\,}}(X_n)$. Then their commutator, using (3), will be a product of conjugates of elements of the form $[\sigma \prod _{i\in I} g_i^{p_i}, \omega ]$ or $[\sigma \prod _{i\in I} g_i^{p_i}, g_{k}^{q_{k}}]$ for some $k\in J$. Applying (4) to these elements will produce a product of conjugates of elements of the form

$$\begin{aligned}{}[\sigma , \omega ], [g_l^{p_l}, \omega ], [\sigma , g_k^{q_k}],\text { or }[g_l^{p_l}, g_k^{q_k}] \end{aligned}$$

for some $l\in I$ and $k\in J$. The first 3 of these are clearly in ${{\,\mathrm{Alt}\,}}(X_n)$. The 4th is also in ${{\,\mathrm{Alt}\,}}(X_n)$ from our assumption that $\mathbf{c}\in I_n$ together with Lemma 3.5. Hence, from Remark 3.4, a commutator of any two elements in $F_\mathbf{c}$ is in ${{\,\mathrm{Alt}\,}}(X_n)$. $\square$

Lemma 3.9

Let ${ \mathbf{c}}=(c_2, \ldots , c_n)\in I_n$. Then $d(F_\mathbf{{c}})=d(H_n)+1$.

Proof

Recall that $d(G_\mathbf{c})=n-1$. Including an element from ${{\,\mathrm{FSym}\,}}(X_n)\setminus {{\,\mathrm{Alt}\,}}(X_n)$ to a minimal generating set for $G_\mathbf{c}$ therefore yields a generating set of $F_\mathbf{c}$ of size n. So we need only show that, with the given hypothesis, no generating set of $F_\mathbf{c}$ of size $n-1$ exists. From Lemma , the abelianization of $F_\mathbf{c}$ is $\mathbb {Z}^{n-1}\times C_2$. Thus $F_\mathbf{c}$ cannot be generated by $n-1$ elements. $\square$