2·1. Maximality of $\mathrm{LMod}_{p_k}(S_g)$

In this subsection, we discuss the maximality of $\mathrm{LMod}_{p_k}(S_g)$ in $\mathrm{Mod}(S_g)$ when k is prime. In the process, we also give a set of coset representatives of $\mathrm{LMod}_{p_k}(S_g)$ in $\mathrm{Mod}(S_g)$ . Let G be a group and $H\subset G$ be a finite-index subgroup of G. Suppose $[G:H]=n$ , and $\{x_i : 1\leq i\leq n\}$ be the left coset representatives. Let $H_i$ denote the left coset of $G/H$ given by $x_iH$ . Without loss of generality, we assume that $x_1=1$ and $H_1=H$ . We will require the following elementary group-theoretic lemma.

We fix the following notation. Let $I_{2g}$ denote the identity matrix of order 2g, and E(i,j) denote the $2g \times 2g$ matrix whose (i,j)th entry is 1 and all other entries are 0. Let $P_{2i}=\Psi_k(T_{b_i}^{-1})=I_{2g}+E(2i,2i-1), \text{ and }$ , $P_{2i+1}=\Psi_k(T_{c_i}^{-1}T_{a_i}T_{a_{i+1}})=I_{2g}+E(2i+1,2i)+E(2i-1,2i+2),$ and let $\mathcal{L}_g= \{T_{a_1},T_{b_1}, \ldots,T_{a_g},T_{b_g},T_{c_1},\ldots,T_{c_{g-1}}\},$ be the Lickorish generating set [ Reference Lickorish18 ] for $\mathrm{Mod}(S_g)$ (where the curves are as indicated in Figure 1).

Proof. Since $\mathrm{Mod}(S_g)[k] \unicode{x003C} \mathrm{LMod}_{p_k}(S_g)$ , it suffices to show that $\Psi_k(\mathrm{LMod}_{p_k}(S_g))$ is maximal in $\mathrm{Sp}(2g;\mathbb{Z}_k)$ . From Theorem 2·1, the distinct cosets of $\Psi_k(\mathrm{LMod}_{p_k}(S_g))$ in $\mathrm{Sp}(2g,\mathbb{Z}_k)$ are in bijective correspondence with the set

\begin{equation*} \{\{\ell v : \ell \in \mathbb{Z}_k^{\times} \}: v \in H_1(S_g,\mathbb{Z}_k) \text{ is primitive}\}.\end{equation*}

Moreover, each coset is represented by a matrix that maps the vector $e_1 \in H_1(S_g,\mathbb{Z}_k)$ to some vector $\ell v$ , where $\ell \in \mathbb{Z}_k^{\times}$ . The distinct coset representatives are listed in Table 1 below.

Table 1.

The coset representatives of $\Psi_k(\mathrm{LMod}_{p_k}(S_g))$ in $\mathrm{Sp}(2g,\mathbb{Z}_k)$

For $ 1\leq i \leq 2g$ , let $g_{i,j}$ (for $1 \leq j \leq k^{2g-i}$ ) denote the distinct coset representatives of type i with $g_{1,1}$ being the identity matrix. Let $H:=\Psi_k(\mathrm{LMod}_{p_k}(S_g))$ and let $H_{i,j} = \langle H, g_{i,j} \rangle$ . In the view of Lemma 2·3, it suffices to show that $\langle H, g_{i,j}\rangle = \mathrm{Sp}(2g,\mathbb{Z}_k)$ whenever $(i,j) \neq (1,1)$ . Since $\Psi_k(h) \in H$ , for each $h \in \mathcal{L}_g \setminus \{T_{b_1}\}$ , this amounts to showing that $\Psi_k(T_{b_1})\in H_{i,j}$ for all $(i,j)\neq (1,1)$ .

First, we consider the case when $i = 2$ . It is apparent that $H_{2,j}=\langle H, \Psi_k(T_{b_1}T_{a_1}T_{b_1})^{-1}\rangle,$ for each j. By the braid relation and fact that $\Psi(T_{a_1})\in H$ , we have $\Psi({T_{b_1}})\in \langle H, g_{2,j}\rangle$ for each j.

Now we consider the case $i\geq 3$ . It follows immediately that

\begin{equation*} H_{i,j} = \langle H, \Psi(T_{a_1}T_{b_1}T_{c_1}T_{b_2}T_{c_2}\cdots T_{c_{\ell-1}}T_{b_\ell}T_{a_\ell})^{-2} \rangle\end{equation*}

for each j, where $\ell=\lfloor {i+1}/{2}\rfloor$ . Since $\Psi(T_{a_1}T_{b_1}T_{c_1}T_{b_2}T_{c_2}\cdots T_{c_{\ell-1}}T_{b_\ell}T_{a_\ell})^{2}\in H_{i,j}$ , the commutativity relation between Dehn twists (about disjoint curves), would imply that $\Psi(T_{b_1}T_{c_1}T_{a_1}T_{b_1})\in H_{i,j}$ . Thus, we have

\begin{eqnarray*}\Psi((T_{b_1}T_{c_1}T_{a_1}T_{b_1})^{-1}T_{b_2}(T_{b_1}T_{c_1}T_{a_1}T_{b_1})) & = & \Psi(T_{b_1}^{-1}T_{c_1}^{-1}T_{b_2}T_{c_1}T_{b_1}) \\& = & \Psi(T_{b_1}^{-1}T_{b_2}T_{c_1}T_{b_2}^{-1}T_{b_1}) \\& = & \Psi(T_{b_2}T_{c_1}T_{b_1}T_{c_1}^{-1}T_{b_2}^{-1}),\end{eqnarray*}

where the first equality follows from the commutativity (of Dehn twists about disjoint curves) and the last two follow from the braid relation between Dehn twists about curves (that intersect once). Hence, $\Psi({T_{b_1}})\in H_{i,j}$ for each $i \geq 3$ .

Finally, we consider the case $i = 1$ . In order to show that $\Psi(T_{b_1}) \in H_{1,j}$ (for $j\neq 1$ ), we will show that $H_{1,j}$ will contain an element of $g_{i,j^{\prime}}H$ for some $i\geq 2$ . Suppose that $g_{1,j}$ corresponds to the primitive vector $(1,y_2,y_3,\cdots,y_{2g})$ . Choose a minimum n such that $y_n\neq 0$ . If ${\bar{y}}_n y_n\equiv 1 \pmod k$ , then the matrix $M=I_{2g}+({\bar{y}}_n-1)E(1,1)+(y_n-1)E(2,2) \in H$ . Now setting $Q_2 = \Psi(T_{a_1})$ , consider the following matrices for $1 \le i \le g-1$ :

\begin{eqnarray*} {}Q_{2i+2} & = & \Psi(T_{d_i} T_{a_1}^{-1} T_{a_{i+1}}^{-1})^{-1} \\ & = & I_{2g} + E(2i+1, 2) + E(1, 2i+2) \text{ and } \\ Q_{2i+1} & = & \Psi(T_{a_{i+1}} T_{b_{i+1}} T_{a_{i+1}}) ~ Q_{2i+2} ~ \Psi(T_{a_{i+1}} T_{b_{i+1}} T_{a_{i+1}})^{-1} \\ {} & = & I_{2g} - E(2i+2, 2) + E(1, 2i+1), {}\end{eqnarray*}

where the $d_i$ are as shown in Figure 3. Then $\tilde{g}_{1,j}=Q_n^{-\bar{y}_n}g_{1,j}M \in \langle H, g_{n,j^{\prime}}\rangle$ . Thus, our assertion now follows from Lemma 2·3.

Fig. 3.

The curves $d_i$ in $S_g$ .

2·2. $\mathrm{LMod}_{p_k}(S_g)$ is self-normalising.

A subgroup H of a group G is said to be self-normalising if the normaliser $N_G(H) = H$ . In other words, this means that if $H \trianglelefteq K \le G$ then $H=K$ . In this subsection, we show that $\mathrm{LMod}_{p_k}(S_g)$ satisfies this property.

Proof. Let us assume on the contrary that there exists a proper subgroup $H \unicode{x003C} \mathrm{Mod}(S_g)$ in which $\mathrm{LMod}_{p_k}(S_g) \lhd H$ . Now choose an $h \in H \setminus \mathrm{LMod}_{p_k}(S_g)$ . Then by our assumption $h^{-1} f h \in \mathrm{LMod}_{p_k}(S_g)$ for any $f \in \mathrm{LMod}_{p_k}(S_g)$ . By Theorem 2·1, this is equivalent to requiring that $(h^{-1} f h)_{\#} (e_1) = \ell e_1$ , for some $\ell \in \mathbb{Z}_k^{\times}$ . In other words, $(fh)_{\#}(e_1) = \ell (h_{\#}(e_1))$ . Denoting $h_{\#}(e_1) = v$ , we have $f_{\#}(v) = \ell \, v$ , for some $\ell \in \mathbb{Z}_k^{\times}$ . Thus, it suffices to show that for any primitive vector $v \in H_1(S_g,\mathbb{Z}_k)$ such that $v \notin \{\ell e_1 : \ell \in \mathbb{Z}_k^\times\}$ , there exists an $f \in \mathrm{LMod}_{p_k}(S_g)$ such that $f_{\#}(v) \neq \ell v$ , for any $\ell \in \mathbb{Z}_k^{\times}$ .

Now given a $v \in H_1(S_g,\mathbb{Z}_k)$ , consider a $\bar{v} \in H_1(S_g,\mathbb{Z})$ such that $v = \bar{v} \pmod k$ . Let $\bar{v} = \sum_{i=1}^g \alpha_i a_i + \sum_{i=1}^g \beta_i b_i$ . We break our argument into the following two cases.

3·1. Generating $\Psi(\mathrm{Mod}_{p_k}(S_g,e_1))$

We establish the following theorem in this subsection.

Theorem 3·1. $\Psi(\mathrm{Mod}_{p_k}(S_g,e_1))$ is generated by

\begin{equation*} \mathbb{S}^{(g)} = \eta_g(\Gamma_1(k)) \cup \Psi( \{ T_{a_2}, T_{b_2}, \ldots, T_{a_g}, T_{b_g}, T_{c_1}, \ldots, T_{c_{g-1}} \})\,\end{equation*}

where $\eta_g$ is the canonical embedding $\eta_g : \mathrm{SL}(2,\mathbb{Z}) \to \mathrm{Sp}(2g,\mathbb{Z})$ defined by

where $O_{m,n}$ is the $m \times n$ zero block and $I_{2g-2}$ is the $(2g-2) \times (2g-2)$ identity block.

The proof of Theorem 3·1 will require the following technical number-theoretic lemmas.

Proof. Consider $\delta = \gcd (a_{21}, a_{31}, \ldots, a_{(2g)1})$ . Since $\gcd (a_{11}, \delta) = 1$ , we have integers $t_1, t_2$ such that $t_1 a_{11} + t_2 \delta = 1$ . We define $\gamma = \gcd (a_{31}, \ldots, a_{(2g)1})$ . So we have integers $t_3, \ldots, t_{2g}$ such that $t_3 a_{31} + \ldots + t_{2g} a_{(2g)1} = \gamma$ . As $\delta = \gcd(a_{21}, \gamma)$ , we can choose integers $r_1, r_2$ such that $r_1 a_{21} + r_2 \gamma = \delta$ . This equation can be rewritten as

\begin{equation*} r_1 \frac{a_{21}}{\delta} + r_2 \frac{\gamma}{\delta} = 1.\end{equation*}

By Dirichlet prime number theorem, there exists a natural number N such that $r_1 + N {\gamma}/{\delta}$ is a prime number bigger than $t_1$ . Denoting $\beta_1 = r_1 + N {\gamma}/{\delta}$ and $\beta_2 = r_2 - N {a_{21}}/{\delta}$ , we have:

\begin{equation*} \beta_1 \frac{a_{21}}{\delta} + \beta_2 \frac{\gamma}{\delta} = 1.\end{equation*}

From the equations above, we deduce that $t_1 a_{11} + t_2 (\beta_1 a_{21} + \beta_2 \gamma) = 1$ , which implies that

(3·1·1) \begin{equation} {} {} {} {}t_1 a_{11} + t_2 (\beta_1 a_{21} + \beta_2 (t_3 a_{31} + \cdots + t_{2g} a_{(2g)1})) = 1. {} {}\end{equation}

Denoting $\lambda_1 = t_1, \lambda_2 = t_2 \beta_1, \lambda_i = t_2 \beta_2 t_i$ , for $3 \le i \le 2g$ , we further assume that $a_{11} \equiv 1 \pmod k$ and $a_{j1} \equiv 0 \pmod k$ , for $2 \le j \le 2g$ . Thus, Equation (3·1·1) takes the form $\lambda_1 a_{11} + \lambda_2 a_{21} + \ldots + \lambda_{2g} a_{(2g)1} = 1$ , from which it follows that $\lambda_1 \equiv 1 \pmod k$ .

Proof. Since $\gcd(\lambda_1, \lambda_2) = 1$ , there exist $\alpha_1, \alpha_2 \in \mathbb{Z}$ such that $\alpha_1 \lambda_1 + \alpha_2 \lambda_2 = 1.$ Now let us assume that $\lambda_1 \equiv 1 \pmod k$ . Consider the set $\{ \lambda_2 k i ~ | ~ 0 \le i \le \lambda_1 - 1 \}$ .

We first establish that $\lambda_1$ divides $(\lambda_2 k i + 1)$ , for some i. Suppose we assume on the contrary. Then by pigeonhole principle, there exist $0 \le i \ne j \le \lambda_1 - 1$ such that $\lambda_2 k i + 1 \equiv \lambda_2 k j + 1 \pmod {\lambda_1}$ , and hence $\lambda_2 k i \equiv \lambda_2 k j \pmod {\lambda_1}$ . Since both $\lambda_2$ and k are coprime with $\lambda_1$ , we have $i \equiv j \pmod {\lambda_1}$ , which is a contradiction.

Hence, there exists $i_0$ such that $\lambda_1$ divides $(\lambda_2 k i_0 + 1)$ . Finally, by choosing $\alpha_1 = (\lambda_2 k i_0 + 1) / \lambda_1$ and $\alpha_2 = - k i_0$ , our assertion follows.

Let $\mathrm{Stab}_{\mathrm{Mod}(S_g)}(\overrightarrow{a_1})$ denote the subgroup of $\mathrm{Mod}(S_g)$ which preserves the curve $a_1$ with a chosen orientation. Let $\mathrm{PMod}(S)$ denote the pure mapping class group of S. We need the following lemma, which follows from [ Reference Benson and Margalit10 , chapter 4].

Proof. Let $S_g - a_1$ denote the surface obtained from $\overline{S_g \ \backslash \ a_1}$ after capping its boundary components with marked disks. We have the following short exact sequence

(3·1·2) \begin{equation} {} {} {} {}1 \to \mathbb{Z} \to \mathrm{Stab}_{\mathrm{Mod}(S_g)}(\overrightarrow{a_1}) \xrightarrow \pi \mathrm{PMod}(S_g - a_1) \to 1. {}\end{equation}

Since $\mathrm{PMod}(S_g - a_1) = \langle S \rangle$ , where

\begin{equation*} S = \{ T_{c_1}, \ldots, T_{c_{g-1}}, T_{a_2},T_{b_2}, \ldots, T_{a_g},T_{b_g} \}, \end{equation*}

if follows from the sequence (3·1·2) that $\mathrm{Stab}_{\mathrm{Mod}(S_g)}(\overrightarrow{a_1}) = \langle S, T_{a_1}\rangle$ .

We are now in a position to prove Theorem 3·1.

Proof (of Theorem 3·1). Given an $A = (a_{ij}) \in \Psi(\mathrm{Mod}_{p_k}(S_g,e_1))$ , we wish to express A as a product of elements in the generating set

\begin{equation*} \mathbb{S}^{(g)} = \eta_g(\Gamma_1(k)) \cup \Psi( \{ T_{a_2}, T_{b_2}, \ldots, T_{a_g}, T_{b_g}, T_{c_1}, \ldots, T_{c_{g-1}} \}).\end{equation*}

To this end, we iteratively multiply A with matrices from $\mathbb{S}^{(g)}$ so that the end product is the identity matrix $I_{2g}$ . For this purpose let us consider the curves $d_1, d_2, \ldots, d_{g-1}$ as shown in Figure 3. By Lemma 3·4, $\Psi(T_{d_i}) \in \langle \mathbb{S}^{(g)} \rangle$ , for all i.

To begin with, we consider the first column $A(1) = (a_{11}, a_{21}, \ldots, a_{(2g)1})$ of A. Lemma 3·2 provides us a $B = (\lambda_1, \lambda_2,\ldots,\lambda_{2g}) \in \mathbb{Z}^{2g}$ such that $\gcd(\lambda_1, \lambda_2)=1$ and $A(1)\cdot B = 1$ . By Lemma 3·3, we have $\alpha_1, \alpha_2 \in \mathbb{Z}$ , such that $\alpha_1 \equiv 0 \pmod k$ and $\lambda_1\alpha_2 + \lambda_2\alpha_1 = 1.$

Let $I_{2g}$ denote the identity matrix of order 2g, and E(i,j) denote the $2g \times 2g$ matrix whose (i,j)th entry is 1 and all other entries are 0. Consider the following matrices, for $1 \le i \le g-1$ :

\begin{eqnarray*} {}M_1 & = &\eta_g \left( \begin{bmatrix} {} {}\kern-2pt\lambda_1&\lambda_2 \\ {} {}\kern-2pt -\kern-1.5pt\alpha_1&\alpha_2 {}\end{bmatrix} \right),\\ {}M_{2i+1} & = & \Psi(T_{d_i} T_{a_1}^{-1} T_{a_{i+1}}^{-1})^{-1} \\ & = & I_{2g} + E(2i+1, 2) + E(1, 2i+2), \text{ and } \\ M_{2i+2} & = & \Psi(T_{a_{i+1}} T_{b_{i+1}} T_{a_{i+1}}) ~ M_{2i+1} ~ \Psi(T_{a_{i+1}} T_{b_{i+1}} T_{a_{i+1}})^{-1} \\ {} & = & I_{2g} - E(2i+2, 2) + E(1, 2i+1). {}\end{eqnarray*}

Note that $M_1, \Psi(T_{a_1}), \Psi(T_{b_1})^k \in \eta_g(\Gamma_1(k))$ . We consider the element

\begin{equation*} A_1= \Psi(T_{a_1})^{- \sum_{i=1}^{g-1} \lambda_{2i+1} \lambda_{2i+2}} M_{2g}^{\lambda_{2g-1}} M_{2g-1}^{\lambda_{2g}} \ldots M_4^{\lambda_3} M_3^{\lambda_4} M_1 A.\end{equation*}

Then the $(1,1){th}$ entry in $A_1$ , which we denote as $A_1(1,1)$ , equals 1. Now we consider the following matrices, for $1 \le i \le g-1$ :

\begin{eqnarray*}N_{2i+1} & = & \Psi(T_{b_1}^{-k} T_{a_{i+1}}^{-k}) M_{2i+1} \Psi(T_{b_1})^k M_{2i+1}^{-1} \\ {} & = & I_{2g} - k.E(2i+1, 1) + k.E(2, 2i+2) \text{ and} \\N_{2i+2} & = & \Psi(T_{a_{i+1}} T_{b_{i+1}} T_{a_{i+1}}) ~ N_{2i+1} ~ \Psi(T_{a_{i+1}} T_{b_{i+1}} T_{a_{i+1}})^{-1} \\ {} & = & I_{2g} + k.E(2i+2, 1) + k.E(2, 2i+1).\end{eqnarray*}

Next, we consider

\begin{eqnarray*}A_2 & = & N_{2g}^{(-1)^{2g-1} \frac{A_1(2g,1)}{k}} N_{2g-1}^{(-1)^{2g-2} \frac{A_1(2g-1,1)}{k}} \ldots N_4^{(-1)^3 \frac{A_1(4,1)}{k}} N_3^{(-1)^2 \frac{A_1(3,1)}{k}} A_1 \text{ and}\\A_3 & = &\Psi(T_{b_1})^{A_2(2,1)} A_2.\end{eqnarray*}

We note that the first column of $A_3$ is $(1,0,\ldots, 0)$ , and as $A_3 \in \mathrm{Sp}(2g, \mathbb{Z})$ , the second row of $A_3$ is $(0,1,0,\ldots,0)$ .

Finally, we consider

\begin{eqnarray*}A_4 & = & M_{2g}^{(-1)^{2g} A_3(2g,2)} M_{2g-1}^{(-1)^{2g-1} A_3(2g-1,2)} \ldots M_4^{(-1)^4 A_3(4,2)} M_3^{(-1)^3 A_3(3,2)} A_3 \text{ and} \\A_5 & = & {\Psi(T_{a_1})}^{-A_4(1,2)}A_4.\end{eqnarray*}

We see that $A_5$ has the form $\begin{bmatrix} {}I_2 & O_{2 \times 2g-2}\\ {}O_{2g-2 \times 2} & M,\end{bmatrix},$ where $M \in \mathrm{Sp}(2g-2, \mathbb{Z})$ , and $A_5$ is a product of elements from

\begin{equation*} \Psi(\{ T_{a_2},\ldots,T_{a_g}, T_{b_2}, \ldots, T_{b_g}, T_{c_2}, \ldots, T_{c_{g-1}} \}).\end{equation*}

Hence, the theorem follows.

3·2. Generating the kernel of $\, \Psi\vert_{\mathrm{Mod}_{p_k}(S_g,e_1)}$

We begin this subsection by recalling some relevant definitions and tools from [ Reference Johnson16 ].

The integer n is called the length of the above chain $(\gamma_1, \gamma_2, \ldots, \gamma_n)$ .

Given a chain $C = (\gamma_1, \ldots, \gamma_{2k+1})$ of odd length in $S_g$ , there exists a unique pair of curves that constitutes the boundary of a closed regular neighbourhood of C. We call this pair B as the bounding pair associated with the chain C. Moreover, the unique bounding pair map in $\mathcal{I}(S_g)$ associated with such a B is given by the product of opposite twists about the two boundary curves in B with the positive twist being taken about the curve lying to the left of $\gamma_1, \gamma_2, \ldots, \gamma_{2k+1}$ . This is called the bounding pair map associated with C, and is denoted by $ch(\gamma_1, \ldots, \gamma_{2k+1})$ .

We further extend this definition in the following manner. By an ordered subcollection of C, we mean a subtuple $(\gamma_{i_1}, \gamma_{i_2}, \ldots, \gamma_{i_{\ell}})$ of C such that $\ell$ is even. Given an ordered subcollection $(\gamma_{i_1}, \gamma_{i_2}, \ldots, \gamma_{i_{\ell}})$ of C, let $\theta_j$ denote the free homotopy class of the curve $\tilde{\gamma}_{i_j}*\tilde{\gamma}_{i_j+1}*\ldots *\tilde{\gamma}_{i_{(j+1)}-1}$ , where $\tilde{\gamma}_m$ denotes the representative of $\gamma_m$ in $(\pi_1(S_g),*)$ . We define

\begin{equation*} \widehat{ch}(\gamma_{i_1}, \gamma_{i_2}, \ldots, \gamma_{i_{\ell}}):= ch(\theta_1,\ldots,\theta_{\ell-1})\end{equation*}

to be the bounding pair map associated with the subcollection $(\gamma_{i_1}, \gamma_{i_2}, \ldots, \gamma_{i_\ell})$ and $(\theta_1,\ldots,\theta_{\ell-1})$ will be called the chain induced by $(\gamma_{i_1}, \gamma_{i_2}, \ldots, \gamma_{i_{\ell}})$ . For further details, we refer the reader to [ Reference Johnson16 , section 2].

Now we consider the ordered collection $C_g = (a_1,b_1,c_1,b_2,c_2,b_3,\ldots,c_{g-1},b_g,a_g)$ of length $2g+1$ , where the curves $a_i$ , $b_i$ , and $c_i$ are as shown in Figure 4. For simplicity, we rename these curves and fix the notation $C_g := (\alpha_1,\ldots,\alpha_{2g+1})$ . Also consider the curve $\beta$ as shown in Figure 4.

Fig. 4.

The Johnson chain of curves.

We now state the main result from [ Reference Johnson16 ] that is relevant to our context.

Let the genus of a bounding pair map associated with a chain C of length $2k+1$ is k. Following the notation from Section 1, we now establish the main result in this subsection.

Theorem 3·7. For $k \geq 2$ and $g \geq 3$ ,

\begin{equation*} \ker (\Psi\vert_{\mathrm{Mod}_{p_k}(S_g,e_1)}) \subset \langle \mathbb{S}_{g,k} \cup \{T_{a_1},T_{a_2},T_{b_2}, \ldots,T_{a_g},T_{b_g},T_{c_1},\ldots,T_{c_{g-1}}\} \rangle,\end{equation*}

where $\mathbb{S}_{g,k} = \{F_1,\ldots,F_{g-2}\}$ is a collection of bounding pair maps such that for each i, $F_i$ is a bounding pair map associated with an ordered subcollection of $C_g$ of genus i that intersect $a_1$ nontrivially (as shown in Figure 2 ).

Proof. By construction, a bounding pair map F associated with any ordered subcollection of $(\beta, \alpha_5, \ldots, \alpha_{2g+1})$ or an ordered subcollection of $C_g$ that is disjoint from $a_1$ (as stated in Theorem 3·6), will preserve $a_1$ with orientation. Thus, by Lemma 3·4, it follows that $F \in \langle T_{a_1}, T_{c_1}, \ldots, T_{c_{g-1}}, T_{a_2}, T_{b_2}, \ldots, T_{a_g},T_{b_g}\rangle$ .

For $i = 1,2$ , let $P_i = (\alpha_{i1},\ldots,\alpha_{i\ell})$ be an ordered subcollection of $C_g$ that intersects $a_1$ . Then there exists a $G \in \mathrm{Stab}_{\mathrm{Mod}(S_g)}(\overrightarrow{a_1})$ that maps the chain induced by the ordered subcollection $(\alpha_{11},\ldots,\alpha_{1\ell})$ to the chain induced by the ordered subcollection $(\alpha_{21},\ldots,\alpha_{2\ell})$ . Consequently, $\widehat{ch}(P_1)$ is conjugate to $\widehat{ch}(P_2)$ via G.

Hence, it follows that any two bounding pair maps associated with ordered subcollections of $C_g$ of the same length which intersect $a_1$ are conjugate by an element in

\begin{equation*} \mathrm{Stab}_{\mathrm{Mod}(S_g)}(\overrightarrow{a_1}) = \langle T_{a_1}, T_{c_1}, \ldots, T_{c_{g-1}}, T_{a_2}, T_{b_2}, \ldots, T_{a_g},T_{b_g}\rangle,\end{equation*}

and our assertion follows.

3·3. Generating $\mathrm{LMod}_{p_k}(S_g)$

Following the notation from Section 1, we obtain the next result as a direct consequence of Theorems 3·1 and 3·7.

Corollary 3·8. For $k \geq 2$ and $g \geq 3$ ,

\begin{equation*} \mathrm{Mod}_{p_k}(S_g,e_1) = \langle \ \tilde{\mathbb{S}}_k \cup \mathbb{S}_{g,k} \cup \{T_{a_2},T_{b_2}, \ldots,T_{a_g},T_{b_g},T_{c_1},\ldots,T_{c_{g-1}}\}\rangle.\end{equation*}

Furthermore, from Theorem 2·2 and Corollary 3·8, we obtain our main result.

Theorem 3·9 (Main theorem). For $k \geq 2$ and $g \geq 3$ ,

\begin{equation*} \mathrm{LMod}_{p_k}(S_g) = \langle \ \mathbb{S}_k^{\prime\prime} \cup \tilde{\mathbb{S}}_k \cup \mathbb{S}_{g,k} \cup \{T_{a_2},T_{b_2}, \ldots,T_{a_g},T_{b_g},T_{c_1},\ldots,T_{c_{g-1}}\}\rangle.\end{equation*}

3·4. Generating $\mathrm{SMod}_{p_k}(S_{g_k})$

Recalling the notation from Section 1, let $R_{g,k} \in \mathrm{Mod}(S_{g_k})$ be represented by the free $2\pi/k$ rotation of $S_{g_k}$ that induces the cover $p_k$ . Let $\mathrm{SMod}_{p_k}(S_{g_k})$ be the symmetric mapping class group of $p_k$ , which is the normaliser of $\langle R_{g,k} \rangle$ in $\mathrm{Mod}(S_{g_k})$ . By Birman–Hilden theory, we have the following short exact sequence for $g \geq 2$ :

\begin{equation*} 1 \to \langle R_{g,k} \rangle \cong \mathbb{Z}_k \to \mathrm{SMod}_{p_k}(S_{g_k}) \to \mathrm{LMod}_{p_k}(S_g) \to 1.\end{equation*}

By pulling back the generators of $\mathrm{LMod}_{p_k}(S_g)$ in Theorem 3·9, we obtain an explicit generating set $G_{g,k}$ for the normaliser $N_{\mathrm{Mod}(S_{g_k})}(\langle R_{g,k} \rangle)$ of $\langle R_{g,k} \rangle$ in $\mathrm{Mod}(S_{g_k})$ . Moreover, as the centraliser $C_{\mathrm{Mod}(S_{g_k})}(R_{g,k})$ of $R_{g,k}$ is a finite-index subgroup of $N_{\mathrm{Mod}(S_{g_k})}(\langle R_{g,k} \rangle)$ , one can derive a finite generating set for $C_{\mathrm{Mod}(S_{g_k})}(R_{g,k})$ using $G_{g,k}$ , which is in general cumbersome. In this subsection, we derive an explicit finite generating set for the centraliser $C_{\mathrm{Mod}(S_{g_k})}(\langle R_{g,k} \rangle)$ of $\langle R_{g,k} \rangle$ in $\mathrm{Mod}(S_{g_k})$ when $k=2$ .

Proof. For the case $g \geq 3$ and $k=2$ , Theorem 3·9 would imply that $\mathrm{LMod}_{p_2}(S_g)=\mathrm{Mod}_{p_2}(S_g,e_1)$ . So, we have

\begin{equation*} \mathrm{LMod}_{p_2}(S_g) = \langle \tilde{\mathbb{S}}_2 \cup \mathbb{S}_{g,2} \cup \{T_{a_2},T_{b_2}, \ldots,T_{a_g},T_{b_g},T_{c_1},\ldots,T_{c_{g-1}}\}\rangle.\end{equation*}

We will now follow the notation from Sections 1 and 3. It is a well known fact that $\mathbb{S}_2=\langle \Psi(T_{a_1}), \Psi(T_{b_1}^2) \rangle$ (see [ Reference Stein26 ]). Thus, we have $\tilde{\mathbb{S}}_2=\langle T_{a_1}, T_{b_1}^2 \rangle$ . By applying the chain relations in $\mathrm{Mod}(S_g)$ , we see that $F_i \in \mathbb{S}_{g,2}$ has the form $F_i= T_{d_{i1}}T_{d_{i2}}^{-1}$ , where

\begin{equation*} T_{d_{i1}}T_{d_{i2}} = (T_{b_1}T_{c_1}T_{b_2}T_{c_2}\cdots T_{b_i}T_{c_i}T_{b_{i+1}})^{2i+2}=(T_{b_1}^2T_{c_1}T_{b_2}T_{c_2}\cdots T_{b_i}T_{c_i}T_{b_{i+1}})^{2i+1}.\end{equation*}

Since $T_{d_{i2}}$ is conjugate to $T_{b_1}$ by an element in $\mathrm{Stab}_{\mathrm{Mod}(S_g)}(\overrightarrow{a_1})$ , the assertion in (i) follows.

To show (ii), we first observe that $\mathrm{SMod}_{p_2}(S_{2g-1}) = C_{\mathrm{Mod}(S_{2g-1})}(\langle R_{g,2}\rangle)$ . Moreover, the collection of curves $\{a_1,a_2,b_2, \ldots, a_g,b_g,c_1,\ldots,c_{g-1}\}$ in $S_g$ (shown in Figure 1) pulls back under $p_2$ to the collection $\mathcal{C} \setminus \{b_{1,1}\}$ , where $\mathcal{C}$ is the collection of curves in $S_{2g-1}$ shown in Figure 5. Thus, by pulling back the generators of $\mathrm{LMod}_{p_2}(S_g)$ under the epimorphism $\mathrm{SMod}_{p_2}(S_{2g-1}) \to \mathrm{LMod}_{p_2}(S_g)$ , the assertion follows.