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00footnotetext: 2000 Mathematics Subject Classification: 57M07, 20F05, 20F3800footnotetext: Keywords: Mapping class groups, punctured surfaces, involutions, generating sets

On the Involution Generators of the Mapping Class Group of a Punctured Surface

Tüli̇n Altunöz, Mehmetci̇k Pamuk, and Oğuz Yıldız Department of Mathematics, Middle East Technical University and Faculty of Engineering, Başkent University, Ankara, Turkey [email protected] Department of Mathematics, Middle East Technical University, Ankara, Turkey [email protected] Department of Mathematics, Middle East Technical University, Ankara, Turkey [email protected]
Abstract.

Let Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) denote the mapping class group of a connected orientable surface of genus gg with pp punctures. For every even integer p10p\geq 10 and g14g\geq 14, we prove that Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) can be generated by three involutions. If the number of punctures pp is odd and 9\geq 9, we show that Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) for g13g\geq 13 can be generated by four involutions. Moreover, we show that for an even integer p4p\geq 4 and 3g63\leq g\leq 6, Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) can be generated by four involutions.

1. Introduction

Let Σg,p\Sigma_{g,p} denote a connected orientable surface of genus-gg with pp-punctures, when p=0p=0 we write Σg\Sigma_{g}. The mapping class group of Σg,p\Sigma_{g,p} is the group of isotopy classes of orientation-preserving homeomorphisms of Σg,p\Sigma_{g,p} preserving the set of punctures.

Here is a brief history of generating sets for Mod(Σg,p){\rm Mod}(\Sigma_{g,p}): Dehn [4] showed that Mod(Σg){\rm Mod}(\Sigma_{g}) can be generated by 2g(g1)2g(g-1) Dehn twists. About a quarter century later, Lickorish [15] gave a generating set consisting of 3g13g-1 Dehn twists. Later, Humphries [8] reduced the number of Dehn twist generators to 2g+12g+1. He also proved that the number 2g+12g+1 is minimal for g2g\geq 2. Johnson [10] proved that the same set of Dehn twists also generates Mod(Σg,1){\rm Mod}(\Sigma_{g,1}). In the presence of multiple punctures, Gervais [7] proved that Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) can be generated by 2g+p2g+p Dehn twists for p1p\geq 1.

If it is not required that the generators are Dehn twists, then it is possible to obtain smaller generating sets for Mod(Σg,p){\rm Mod}(\Sigma_{g,p}): For g3g\geq 3 and p=0p=0, Lu [17, Theorem 1.31.3] proved that Mod(Σg){\rm Mod}(\Sigma_{g}) can be generated by three elements.For g1g\geq 1 and p=0p=0 or 11, a minimal (since the group is not abelian) generating set of two elements, a product of two Dehn twists and a product of 2g2g Dehn twists, was first given by Wajnryb [25]. Korkmaz [12, Theorem 55] improved this result by showing that one of these two generators can be taken as a Dehn twist. He also showed that this group is generated by two elements of finite order  [12, Theorem 1414]. For g3g\geq 3, Kassabov obtained a generating set of involution elements where the number of generators depends on gg and the parity of pp (see  [11, Theorem 1]). Later, Monden [20] removed the parity condition on pp for g=7g=7 and g=5g=5. For g1g\geq 1 and p2p\geq 2, Monden [21] also gave a generating set for Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) consisting of three elements. Recently, he [22] gave a minimal generating set for Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) containing two elements for g3g\geq 3.

Note that any infinite group generated by two involutions must be isomorphic to the infinite dihedral group whose subgroups are either cyclic or dihedral. Since Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) contains nonabelian free groups, it cannot be generated by two involutions. In this paper, we obtain the following result (cf.  [11, Remark 5]):

Theorem A.

For every even integer p8p\geq 8 and g14g\geq 14, Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) can be generated by three involutions. Moreover, for every even integer p4p\geq 4 and for g=3,4,5g=3,4,5 or 66, Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) can be generated by four involutions.

At the end of the paper, we also show that Theorem A also holds for the cases p=2p=2 or p=3p=3. For surfaces with odd number of punctures, we have the following result:

Theorem B.

For every odd integer p9p\geq 9 and g13g\geq 13, Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) is generated by four involutions. Moreover, for every odd integer p5p\geq 5 and for g=3,4,5g=3,4,5 or 66, Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) can be generated by five involutions.

The paper is organized as follows: In Section 2, we quickly provide the necessary background on mapping class groups. The proofs of Theorem A and Theorem B are given in Section 3.

Acknowledgements. This work was supported by the Scientific and Technological Research Council of Turkey (TÜBİTAK)[grant number 120F118].

2. Background and Results on Mapping Class Groups

Let Σg,p\Sigma_{g,p} denote a connected orientable surface of genus gg with pp punctures specified by the set P={z1,z2,,zp}P=\{z_{1},z_{2},\ldots,z_{p}\} of pp distinguished points. If pp is zero then we omit it from the notation and write Σg\Sigma_{g}. The mapping class group Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) of the surface Σg,p\Sigma_{g,p} is defined to be the group of the isotopy classes of orientation preserving self-diffeomorphisms of Σg,p\Sigma_{g,p} which fix the set PP. The mapping class group Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) of the surface Σg,p\Sigma_{g,p} is defined to be the group of isotopy classes of all orientation preserving self-diffeomorphisms of Σg,p\Sigma_{g,p} which fix the set PP. Let Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) denote the subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) consisting of elements which fix the set PP pointwise. It is obviuos that we have the following exact sequence:

1Mod0(Σg,p)Mod(Σg,p)Symp1,1\longrightarrow{\rm Mod}_{0}(\Sigma_{g,p})\longrightarrow{\rm Mod}(\Sigma_{g,p})\longrightarrow Sym_{p}\longrightarrow 1,

where SympSym_{p} denotes the symmetric group on the set {1,2,,p}\{1,2,\ldots,p\} and the last projection is given by the restriction of the isotopy class of a diffeomorphism to its action on the punctures.

Let βi,j\beta_{i,j} be an embedded arc that joins two punctures ziz_{i} and zjz_{j} and does not intersect δ\delta on Σg,p\Sigma_{g,p}. Let Di,jD_{i,j} denote a closed regular neighbourhood of βi,j\beta_{i,j}, which is a disk with two punctures. There exists a diffeomorphism Hi,j:Di,jDi,jH_{i,j}:D_{i,j}\to D_{i,j}, which interchanges the punctures such that Hi,j2H_{i,j}^{2} is equal to the right handed Dehn twist about Di,j\partial D_{i,j} and is the identity on the complement of the interior of Di,jD_{i,j}. Such a diffeomorphism is said to be the (right handed) half twist about βi,j\beta_{i,j}. It can be extended to a diffeomorphism of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}). Throughout the paper we do not distinguish a diffeomorphism from its isotopy class. For the composition of two diffeomorphisms, we use the functional notation; if ff and gg are two diffeomorphisms, then the composition fgfg means that gg is applied first and then ff.
For a simple closed curve aa on Σg,p\Sigma_{g,p}, following  [1, 13], we denote the right-handed Dehn twist tat_{a} about aa by the corresponding capital letter AA. Let us also remind the following basic facts of Dehn twists that we use frequently throughout the paper: Let aa and bb be simple closed curves on Σg,p\Sigma_{g,p} and fMod(Σg,p)f\in{\rm Mod}(\Sigma_{g,p}).

  • If aa and bb are disjoint, then AB=BAAB=BA (Commutativity).

  • If f(a)=bf(a)=b, then fAf1=BfAf^{-1}=B (Conjugation).

Let us finish this section by noting that we denote the conjugation relation fgf1fgf^{-1} by fgf^{g} for any f,gMod(Σg,p)f,g\in{\rm Mod}(\Sigma_{g,p}).

3. Involution generators for Mod(Σg,p){\rm Mod}(\Sigma_{g,p})

Let us start this section by recalling the following set of generators given by Korkmaz [13, Theorem 55].

Theorem 3.1.

If g3g\geq 3, then the mapping class group Mod(Σg){\rm Mod}(\Sigma_{g}) can be generated by the four elements RR, A1A21A_{1}A_{2}^{-1}, B1B21B_{1}B_{2}^{-1}, C1C21C_{1}C_{2}^{-1}.

Let us also recall the following well known result from algebra.

Lemma 3.2.

Let GG and KK be groups. Suppose that the following short exact sequence holds,

1N𝑖G𝜋K1.1\longrightarrow N\overset{i}{\longrightarrow}G\overset{\pi}{\longrightarrow}K\longrightarrow 1.

Then the subgroup Γ\Gamma contains i(N)i(N) and has a surjection to KK if and only if Γ=G\Gamma=G.

In our case where G=Mod(Σg,p)G={\rm Mod}(\Sigma_{g,p}) and N=Mod0(Σg,p)N={\rm Mod}_{0}(\Sigma_{g,p}), we have the following short exact sequence:

1Mod0(Σg,p)Mod(Σg,p)Symp1.1\longrightarrow{\rm Mod}_{0}(\Sigma_{g,p})\longrightarrow{\rm Mod}(\Sigma_{g,p})\longrightarrow Sym_{p}\longrightarrow 1.

Therefore, we obtain the following useful result which follows immediately from Lemma 3.2. Let Γ\Gamma be a subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}). If the subgroup Γ\Gamma contains Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) and has a surjection to SympSym_{p} then Γ=Mod(Σg,p)\Gamma={\rm Mod}(\Sigma_{g,p}).

Refer to caption
Figure 1. The involutions ρ1\rho_{1} and ρ2\rho_{2} if g=2kg=2k and p=2bp=2b.

Throughout the paper, we consider the embeddings of Σg,p\Sigma_{g,p} into 3\mathbb{R}^{3} in such a way that it is invariant under the rotations ρ1\rho_{1} and ρ2\rho_{2}. Here, ρ1\rho_{1} and ρ2\rho_{2} are the rotations by π\pi about the zz-axis (see Figures 1 and 2). Note that Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) contains the element R=ρ1ρ2R=\rho_{1}\rho_{2} which satisfies the following:

  • (i)

    R(ai)=ai+1R(a_{i})=a_{i+1}, R(bi)=bi+1R(b_{i})=b_{i+1} for i=1,,g1i=1,\ldots,g-1 and R(bg)=b1R(b_{g})=b_{1},

  • (ii)

    R(ci)=ci+1R(c_{i})=c_{i+1} for i=1,,g2i=1,\ldots,g-2,

  • (iii)

    R(z1)=zpR(z_{1})=z_{p} and R(zi)=zi1R(z_{i})=z_{i-1} for i=2,,pi=2,\ldots,p.

We want to note here that, in the following lemmata, where we present generating sets for surfaces with even number of punctures, we mainly follow the proof of [13, Theorem 55]. We use them in the proof of Theorem A and then for surfaces with odd number of punctures we explain how our arguments are modified.

Lemma 3.3.

For every even integer g=2k14g=2k\geq 14 and every even integer p=2b10p=2b\geq 10, the subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements

ρ1,ρ2 and ρ1Hb1,bHb+1,b1Ck3Bk1AkAk+21Bk+31Ck+41\rho_{1},\rho_{2}\textrm{ and }\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1}

contains the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} for i=1,,gi=1,\ldots,g.

Proof.

Consider the models of Σg,p\Sigma_{g,p} depicted in Figure 1. Let F1:=Hb1,bHb+1,b1Ck3Bk1AkAk+21Bk+31Ck+41F_{1}:=H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1} and let Γ\Gamma be the subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements ρ1\rho_{1}, ρ2\rho_{2} and ρ1F1\rho_{1}F_{1}. One can see that the elements R=ρ1ρ2R=\rho_{1}\rho_{2} and F1=ρ1ρ1F1F_{1}=\rho_{1}\rho_{1}F_{1} are contained in the subgroup Γ\Gamma. Let F2F_{2} be the element obtained by the conjugation of F1F_{1} by R3R^{-3}. Since

R3(ck3,bk1,ak,ak+2,bk+3,ck+4)=(ck6,bk4,ak3,ak1,bk,ck+1)R^{-3}(c_{k-3},b_{k-1},a_{k},a_{k+2},b_{k+3},c_{k+4})=(c_{k-6},b_{k-4},a_{k-3},a_{k-1},b_{k},c_{k+1})

and

R3(zb1,zb,zb+1)=(zb+2,zb+3,zb+4),R^{-3}(z_{b-1},z_{b},z_{b+1})=(z_{b+2},z_{b+3},z_{b+4}),
F2=F1R3=Hb+2,b+3Hb+4,b+31Ck6Bk4Ak3Ak11Bk1Ck+11Γ.F_{2}=F_{1}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}C_{k-6}B_{k-4}A_{k-3}A_{k-1}^{-1}B_{k}^{-1}C_{k+1}^{-1}\in\Gamma.

Let F3F_{3} be the element F1F1F21F_{1}^{F_{1}F_{2}^{-1}}, that is F3=Hb1,bHb+1,b1Ck3Ak1BkAk+21Bk+31Ck+41Γ.F_{3}=H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}A_{k-1}B_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1}\in\Gamma.

Since we use repeatedly similar calculations in the remaining parts of the paper, let us provide some details here. It can be shown that the diffeomorphism F1F21F_{1}F_{2}^{-1} maps the curves {ck3,bk1,ak,ak+2,bk+3,ck+4}\{c_{k-3},b_{k-1},a_{k},a_{k+2},b_{k+3},c_{k+4}\} to the curves {ck3,bk1,ak,bk+2,ak+3,ck+4}\{c_{k-3},b_{k-1},a_{k},b_{k+2},a_{k+3},c_{k+4}\}, respectively. Also it follows from the factorizations of half twists Hb1,bHb+1,b1H_{b-1,b}H_{b+1,b}^{-1} and Hb4,b3Hb2,b31H_{b-4,b-3}H_{b-2,b-3}^{-1} commute and we get

F3\displaystyle F_{3} =\displaystyle= F1F1F21\displaystyle F_{1}^{F_{1}F_{2}^{-1}}
=\displaystyle= (F1F21)(Hb1,bHb+1,b1Ck3Bk1AkAk+21Bk+31Ck+41)(F1F21)1\displaystyle(F_{1}F_{2}^{-1})(H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1})(F_{1}F_{2}^{-1})^{-1}
=\displaystyle= Hb1,bHb+1,b1Ck3Ak1BkAk+21Bk+31Ck+41.\displaystyle H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}A_{k-1}B_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1}.

The subgroup Γ\Gamma contains the following elements:

F4\displaystyle F_{4} =\displaystyle= F3R3=Hb+2,b+3Hb+4,b+31Ck6Ak4Bk3Ak11Bk1Ck+11,\displaystyle F_{3}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}C_{k-6}A_{k-4}B_{k-3}A_{k-1}^{-1}B_{k}^{-1}C_{k+1}^{-1},
F5\displaystyle F_{5} =\displaystyle= F3F3F4=Hb1,bHb+1,b1Bk3Ak1BkAk+21Bk+31Ck+41.\displaystyle F_{3}^{F_{3}F_{4}}=H_{b-1,b}H_{b+1,b}^{-1}B_{k-3}A_{k-1}B_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1}.

From these, we obtain the element F5F31=Bk3Ck31F_{5}F_{3}^{-1}=B_{k-3}C_{k-3}^{-1}, which is contained in Γ\Gamma. By conjugating this element with powers of RR, we conclude that

BiCi1Γfori=1,,g1.B_{i}C_{i}^{-1}\in\Gamma\ \textrm{for}\ i=1,\ldots,g-1.

The subgroup Γ\Gamma also contains the element F1F31=Bk1AkBk1Ak11F_{1}F_{3}^{-1}=B_{k-1}A_{k}B_{k}^{-1}A_{k-1}^{-1}. After conjugating with R3R^{3} and considering the inverse, we have Ak+2Bk+3Ak+31Bk+11ΓA_{k+2}B_{k+3}A_{k+3}^{-1}B_{k+1}^{-1}\in\Gamma. This in turn implies that for i=1,,g1i=1,\ldots,g-1, the elements

AiBi+1Ai+11Bi1Γ.A_{i}B_{i+1}A_{i+1}^{-1}B_{i}^{-1}\in\Gamma.

We also have the following elements in Γ\Gamma:

F6\displaystyle F_{6} =\displaystyle= F1(Ak+2Bk+3Ak+31Bk+21)=Hb1,bHb+1,b1Ck3Bk1AkBk+21Ak+31Ck+41,\displaystyle F_{1}(A_{k+2}B_{k+3}A_{k+3}^{-1}B_{k+2}^{-1})=H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}B_{k+2}^{-1}A_{k+3}^{-1}C_{k+4}^{-1},
F7\displaystyle F_{7} =\displaystyle= F6R3=Hb+2,b+3Hb+4,b+31Ck6Bk4Ak3Bk11Ak1Ck+11 and\displaystyle F_{6}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}C_{k-6}B_{k-4}A_{k-3}B_{k-1}^{-1}A_{k}^{-1}C_{k+1}^{-1}\textrm{ and }
F8\displaystyle F_{8} =\displaystyle= F6F6F7=Hb1,bHb+1,b1Ck3Bk1AkCk+11Ak+31Ck+41,\displaystyle F_{6}^{F_{6}F_{7}}=H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}C_{k+1}^{-1}A_{k+3}^{-1}C_{k+4}^{-1},

Hence, we can conclude that F81F6=Ck+1Bk+21ΓF_{8}^{-1}F_{6}=C_{k+1}B_{k+2}^{-1}\in\Gamma. Again conjugating with RR implies that

CiBi+11Γ,for alli=1,,g1.C_{i}B_{i+1}^{-1}\in\Gamma,\ \textrm{for all}\ i=1,\ldots,g-1.

Furthermore, we can see that Γ\Gamma contains the following elements:

F9\displaystyle F_{9} =\displaystyle= (Bk3Ck31)F3(Ck+4Bk+51)=Hb1,bHb+1,b1Bk3Ak1BkAk+21Bk+31Bk+51,\displaystyle(B_{k-3}C_{k-3}^{-1})F_{3}(C_{k+4}B_{k+5}^{-1})=H_{b-1,b}H_{b+1,b}^{-1}B_{k-3}A_{k-1}B_{k}A_{k+2}^{-1}B_{k+3}^{-1}B_{k+5}^{-1},
F10\displaystyle F_{10} =\displaystyle= F9R3=Hb+2,b+3Hb+4,b+31Bk6Ak4Bk3Ak11Bk1Bk+21,\displaystyle F_{9}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}B_{k-6}A_{k-4}B_{k-3}A_{k-1}^{-1}B_{k}^{-1}B_{k+2}^{-1},
F11\displaystyle F_{11} =\displaystyle= F9F9F10=Hb1,bHb+1,b1Bk3Ak1BkBk+21Bk+31Bk+51.\displaystyle F_{9}^{F_{9}F_{10}}=H_{b-1,b}H_{b+1,b}^{-1}B_{k-3}A_{k-1}B_{k}B_{k+2}^{-1}B_{k+3}^{-1}B_{k+5}^{-1}.

From these, we obtain F11F91=Bk+21Ak+2ΓF_{11}F_{9}^{-1}=B_{k+2}^{-1}A_{k+2}\in\Gamma. By the action of RR, we can conclude that

AiBi1Γfor alli=1,,g.A_{i}B_{i}^{-1}\in\Gamma\ \textrm{for all}\ i=1,\ldots,g.

This completes the proof by Theorem 3.1 since the subgroup Γ\Gamma contains the elements

A1A21\displaystyle A_{1}A_{2}^{-1} =\displaystyle= (A1B11)(B1C11)(C1B21)(B2A21),\displaystyle(A_{1}B_{1}^{-1})(B_{1}C_{1}^{-1})(C_{1}B_{2}^{-1})(B_{2}A_{2}^{-1}),
B1B21\displaystyle B_{1}B_{2}^{-1} =\displaystyle= (B1C11)(C1B21) and\displaystyle(B_{1}C_{1}^{-1})(C_{1}B_{2}^{-1})\textrm{ and }
C1C21\displaystyle C_{1}C_{2}^{-1} =\displaystyle= (C1B21)(B2C21).\displaystyle(C_{1}B_{2}^{-1})(B_{2}C_{2}^{-1}).

    

Refer to caption
Figure 2. The involutions ρ1\rho_{1} and ρ2\rho_{2} for g=2k+1g=2k+1 and p=2bp=2b.

If gg is odd and pp is even, we have the following result:

Lemma 3.4.

For every odd integer g=2k+115g=2k+1\geq 15 and even integer p=2b10p=2b\geq 10, the subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements

ρ1,ρ2 and ρ1Hb1,bHb+1,b1Ck3BkAk+1Ak+21Bk+31Ck+51\rho_{1},\rho_{2}\textrm{ and }\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+5}^{-1}

contains the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} for i=1,,gi=1,\ldots,g.

Proof.

Consider the models for Σg,p\Sigma_{g,p} as shown in Figure 2. Let Γ\Gamma denote the subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements ρ1\rho_{1}, ρ2\rho_{2} and ρ1G1\rho_{1}G_{1}, where G1=Hb1,bHb+1,b1Ck3BkAk+1Ak+21Bk+31Ck+51G_{1}=H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+5}^{-1}. The elements R=ρ1ρ2R=\rho_{1}\rho_{2} and G1=ρ1ρ1G1G_{1}=\rho_{1}\rho_{1}G_{1} belong to the subgroup Γ\Gamma. Let G2G_{2} denote the conjugation of G1G_{1} by R3R^{-3},

G2=G1R3=Hb+2,b+3Hb+4,b+31Ck6Bk3Ak2Ak11Bk1Ck+21.G_{2}=G_{1}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}C_{k-6}B_{k-3}A_{k-2}A_{k-1}^{-1}B_{k}^{-1}C_{k+2}^{-1}.

It is easy to verify that the element

G3\displaystyle G_{3} =\displaystyle= G1G1G2\displaystyle G_{1}^{G_{1}G_{2}}
=\displaystyle= Hb1,bHb+1,b1Bk3BkAk+1Ak+21Ck+21Ck+51\displaystyle H_{b-1,b}H_{b+1,b}^{-1}B_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}C_{k+2}^{-1}C_{k+5}^{-1}

is contained in Γ\Gamma. Let

G4=G3R3=Hb+2,b+3Hb+4,b+31Bk6Bk3Ak2Ak11Ck11Ck+21.G_{4}=G_{3}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}B_{k-6}B_{k-3}A_{k-2}A_{k-1}^{-1}C_{k-1}^{-1}C_{k+2}^{-1}.

Thus we get the element

G5=G3G3G41=Hb1,bHb+1,b1Bk3Ck1Ak+1Ak+21Ck+21Ck+51,\displaystyle G_{5}=G_{3}^{G_{3}G_{4}^{-1}}=H_{b-1,b}H_{b+1,b}^{-1}B_{k-3}C_{k-1}A_{k+1}A_{k+2}^{-1}C_{k+2}^{-1}C_{k+5}^{-1},

which is contained in GG. This implies that G3G51=BkCk11ΓG_{3}G_{5}^{-1}=B_{k}C_{k-1}^{-1}\in\Gamma. By conjugating BkCk11B_{k}C_{k-1}^{-1} with powers of RR, we see that

Bi+1Ci1Γ,B_{i+1}C_{i}^{-1}\in\Gamma,

for all i=1,,g1i=1,\ldots,g-1. In particular, the element Ck+5Bk+61ΓC_{k+5}B_{k+6}^{-1}\in\Gamma. Hence, the subgroup Γ\Gamma contains the following element:

G6=G1(Ck+5Bk+61)=Hb1,bHb+1,b1Ck3BkAk+1Ak+21Bk+31Bk+61.G_{6}=G_{1}(C_{k+5}B_{k+6}^{-1})=H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}B_{k+6}^{-1}.

Then, we see that the elements

G7\displaystyle G_{7} =\displaystyle= G6R3=Hb+2,b+3Hb+4,b+31Ck6Bk3Ak2Ak11Bk1Bk+31 and\displaystyle G_{6}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}C_{k-6}B_{k-3}A_{k-2}A_{k-1}^{-1}B_{k}^{-1}B_{k+3}^{-1}\textrm{ and}
G8\displaystyle G_{8} =\displaystyle= G6G6G7=Hb1,bHb+1,b1Bk3BkAk+1Ak+21Bk+31Bk+61\displaystyle G_{6}^{G_{6}G_{7}}=H_{b-1,b}H_{b+1,b}^{-1}B_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}B_{k+6}^{-1}

are contained in Γ\Gamma, which implies that the subgroup Γ\Gamma contains the element G6G81=Ck3Bk31G_{6}G_{8}^{-1}=C_{k-3}B_{k-3}^{-1}. By the action of RR we see that

CiBi1ΓC_{i}B_{i}^{-1}\in\Gamma

for all i=1,,g1i=1,\ldots,g-1. Moreover, we get

G9\displaystyle G_{9} =\displaystyle= (Bk2Ck31)G6=Hb1,bHb+1,b1Bk2BkAk+1Ak+21Bk+31Bk+61Γ,\displaystyle(B_{k-2}C_{k-3}^{-1})G_{6}=H_{b-1,b}H_{b+1,b}^{-1}B_{k-2}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}B_{k+6}^{-1}\in\Gamma,
G10\displaystyle G_{10} =\displaystyle= G9R3=Hb+2,b+3Hb+4,b+31Bk5Bk3Ak2Ak11Bk1Bk+31Γ and\displaystyle G_{9}^{R^{-3}}=H_{b+2,b+3}H_{b+4,b+3}^{-1}B_{k-5}B_{k-3}A_{k-2}A_{k-1}^{-1}B_{k}^{-1}B_{k+3}^{-1}\in\Gamma\textrm{ and}
G11\displaystyle G_{11} =\displaystyle= G9G9G10=Hb1,bHb+1,b1Ak2BkAk+1Ak+21Bk+31Bk+61Γ.\displaystyle G_{9}^{G_{9}G_{10}}=H_{b-1,b}H_{b+1,b}^{-1}A_{k-2}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}B_{k+6}^{-1}\in\Gamma.

From these, we have G9G111=Bk2Ak21ΓG_{9}G_{11}^{-1}=B_{k-2}A_{k-2}^{-1}\in\Gamma so that

BiAi1Γ,B_{i}A_{i}^{-1}\in\Gamma,

for i=1,,gi=1,\ldots,g, by the action of RR. The remaining part of the proof can be completed as in the proof of Lemma 3.3.       

In the following four lemmata, we give generating sets for smaller genera.

Lemma 3.5.

For g=6g=6 and every even integer p4p\geq 4, the group generated by the elements

ρ1,ρ2,ρ2B2A3A41B51 and ρ1Hb1,bHb+1,b1C3C41\rho_{1},\rho_{2},\rho_{2}B_{2}A_{3}A_{4}^{-1}B_{5}^{-1}\textrm{ and }\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{3}C_{4}^{-1}

contains the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} for i=1,,gi=1,\ldots,g.

Proof.

Consider the models for Σ6,p\Sigma_{6,p} as shown in Figure 1. Let Γ\Gamma be the subgroup of Mod(Σ6,p){\rm Mod}(\Sigma_{6,p}) generated by the elements ρ1\rho_{1}, ρ2\rho_{2}, ρ2F1\rho_{2}F_{1} and ρ1E1\rho_{1}E_{1} where F1=B2A3A41B51F_{1}=B_{2}A_{3}A_{4}^{-1}B_{5}^{-1} and E1=Hb1,bHb+1,b1C3C41E_{1}=H_{b-1,b}H_{b+1,b}^{-1}C_{3}C_{4}^{-1}. Hence the elements R=ρ1ρ2R=\rho_{1}\rho_{2}, F1=ρ2ρ2F1F_{1}=\rho_{2}\rho_{2}F_{1} and E1=ρ1ρ1E1E_{1}=\rho_{1}\rho_{1}E_{1} are contained in the subgroup Γ\Gamma.

The subgroup Γ\Gamma contains the following elements:

F2\displaystyle F_{2} =\displaystyle= F1R=B3A4A51B61Γ,\displaystyle F_{1}^{R}=B_{3}A_{4}A_{5}^{-1}B_{6}^{-1}\in\Gamma,
F3\displaystyle F_{3} =\displaystyle= F1F1F2=B2B3A41A51Γ\displaystyle F_{1}^{F_{1}F_{2}}=B_{2}B_{3}A_{4}^{-1}A_{5}^{-1}\in\Gamma
F4\displaystyle F_{4} =\displaystyle= F3R=B3B4A51A61Γ and\displaystyle F_{3}^{R}=B_{3}B_{4}A_{5}^{-1}A_{6}^{-1}\in\Gamma\textrm{ and}
F5\displaystyle F_{5} =\displaystyle= F3F3F41=B2B3B41A51Γ.\displaystyle F_{3}^{F_{3}F_{4}^{-1}}=B_{2}B_{3}B_{4}^{-1}A_{5}^{-1}\in\Gamma.

Hence we get the element F51F3=B4A41ΓF_{5}^{-1}F_{3}=B_{4}A_{4}^{-1}\in\Gamma. By the action of RR, for all i=1,,6i=1,\ldots,6,

AiBi1Γ.A_{i}B_{i}^{-1}\in\Gamma.

Moreover, we have

F6\displaystyle F_{6} =\displaystyle= E1E1F3=Hb1,bHb+1,b1B3C41Γ and\displaystyle E_{1}^{E_{1}F_{3}}=H_{b-1,b}H_{b+1,b}^{-1}B_{3}C_{4}^{-1}\in\Gamma\textrm{ and}
F7\displaystyle F_{7} =\displaystyle= E1E1F1=Hb1,bHb+1,b1C3B51Γ.\displaystyle E_{1}^{E_{1}F_{1}}=H_{b-1,b}H_{b+1,b}^{-1}C_{3}B_{5}^{-1}\in\Gamma.

This implies that F6E11=B3C31ΓF_{6}E_{1}^{-1}=B_{3}C_{3}^{-1}\in\Gamma and F71E1=B5C41ΓF_{7}^{-1}E_{1}=B_{5}C_{4}^{-1}\in\Gamma and so

BiCi1Γ and Bi+1Ci1Γ,B_{i}C_{i}^{-1}\in\Gamma\textrm{ and }B_{i+1}C_{i}^{-1}\in\Gamma,

for all i=1,,5i=1,\ldots,5, by conjugating these elements with powers of RR. The proof can be completed as in the proof of Lemma 3.3.       

Lemma 3.6.

For g=5g=5 and every even integer p4p\geq 4, the group generated by the elements

ρ1,ρ2,ρ1Hb1,bHb+1,b1A3A41 and ρ2A2B2C2C31B41A41\rho_{1},\rho_{2},\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}A_{3}A_{4}^{-1}\textrm{ and }\rho_{2}A_{2}B_{2}C_{2}C_{3}^{-1}B_{4}^{-1}A_{4}^{-1}

contains the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} for i=1,,gi=1,\ldots,g.

Proof.

Consider the models for Σ5,p\Sigma_{5,p} as shown in Figure 2. Let Γ\Gamma denote the subgroup of Mod(Σ5,p){\rm Mod}(\Sigma_{5,p}) generated by the elements ρ1\rho_{1}, ρ2\rho_{2}, ρ1F1\rho_{1}F_{1} and ρ2E1\rho_{2}E_{1} where F1=Hb1,bHb+1,b1A3A41F_{1}=H_{b-1,b}H_{b+1,b}^{-1}A_{3}A_{4}^{-1} and E1=A2B2C2C31B41A41E_{1}=A_{2}B_{2}C_{2}C_{3}^{-1}B_{4}^{-1}A_{4}^{-1}. Thus the elements R=ρ1ρ2R=\rho_{1}\rho_{2}, F1=ρ1ρ1F1F_{1}=\rho_{1}\rho_{1}F_{1} and E1=ρ2ρ2E1E_{1}=\rho_{2}\rho_{2}E_{1} are in the subgroup Γ\Gamma.

One can obtain the following elements:

F2\displaystyle F_{2} =\displaystyle= F1R1=Hb,b+1Hb+2,b+11A2A31\displaystyle F_{1}^{R^{-1}}=H_{b,b+1}H_{b+2,b+1}^{-1}A_{2}A_{3}^{-1}
F3\displaystyle F_{3} =\displaystyle= F2E1=Hb,b+1Hb+2,b+11B2A31 and\displaystyle F_{2}^{E_{1}}=H_{b,b+1}H_{b+2,b+1}^{-1}B_{2}A_{3}^{-1}\textrm{ and}
F4\displaystyle F_{4} =\displaystyle= F3E1=Hb,b+1Hb+2,b+11C2A31,\displaystyle F_{3}^{E_{1}}=H_{b,b+1}H_{b+2,b+1}^{-1}C_{2}A_{3}^{-1},

which are contained in Γ\Gamma. Thus we get that F2F31=A2B21ΓF_{2}F_{3}^{-1}=A_{2}B_{2}^{-1}\in\Gamma and F3F41=B2C21ΓF_{3}F_{4}^{-1}=B_{2}C_{2}^{-1}\in\Gamma. By conjugating these elements with powers of RR, we see that

AiBi1Γ and BjCj1Γ,A_{i}B_{i}^{-1}\in\Gamma\textrm{ and }B_{j}C_{j}^{-1}\in\Gamma,

which also implies that AiCi1ΓA_{i}C_{i}^{-1}\in\Gamma for all i=1,,5i=1,\ldots,5 and j=1,,4j=1,\ldots,4. Finally, it can be verified that

E1(a3,c3)=(a3,b4)E_{1}(a_{3},c_{3})=(a_{3},b_{4})

so that the group Γ\Gamma contains the element

(A3C31)E1=A3B41.(A_{3}C_{3}^{-1})^{E_{1}}=A_{3}B_{4}^{-1}.

Hence AiBi+11ΓA_{i}B_{i+1}^{-1}\in\Gamma for all i=1,,5i=1,\ldots,5 by the action of RR. The rest of the proof is similar to that of Lemma 3.3.       

Lemma 3.7.

For g=4g=4 and every even integer p4p\geq 4, the group generated by the elements

ρ1,ρ2,ρ2B1A2A31B41 and ρ1Hb1,bHb+1,b1C2C31\rho_{1},\rho_{2},\rho_{2}B_{1}A_{2}A_{3}^{-1}B_{4}^{-1}\textrm{ and }\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{2}C_{3}^{-1}

contains the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} for i=1,,gi=1,\ldots,g.

Proof.

Let us consider the models for Σ4,p\Sigma_{4,p} as shown in Figure 1 and let Γ\Gamma be the subgroup of Mod(Σ4,p){\rm Mod}(\Sigma_{4,p}) generated by the elements ρ1\rho_{1}, ρ2\rho_{2}, ρ2F1\rho_{2}F_{1} and ρ1E1\rho_{1}E_{1} where F1=B1A2A31B41F_{1}=B_{1}A_{2}A_{3}^{-1}B_{4}^{-1} and E1=Hb1,bHb+1,b1C2C31E_{1}=H_{b-1,b}H_{b+1,b}^{-1}C_{2}C_{3}^{-1}. Thus it is clear that the elements R=ρ1ρ2R=\rho_{1}\rho_{2}, F1=ρ2ρ2F1F_{1}=\rho_{2}\rho_{2}F_{1} and E1=ρ1ρ1E1E_{1}=\rho_{1}\rho_{1}E_{1} belong to the subgroup Γ\Gamma. We have the element

F2=E1E1F1=Hb1,bHb+1,b1C2B41Γ.F_{2}=E_{1}^{E_{1}F_{1}}=H_{b-1,b}H_{b+1,b}^{-1}C_{2}B_{4}^{-1}\in\Gamma.

Thus the subgroup Γ\Gamma contains the elements FE121=B4C31F{{}_{2}}^{-1}E_{1}=B_{4}C_{3}^{-1} and ρ1(B4C31)ρ1=B2C21\rho_{1}(B_{4}C_{3}^{-1})\rho_{1}=B_{2}C_{2}^{-1}. By conjugating these elements with powers of RR, we get

Bi+1Ci1Γ and BiCi1ΓB_{i+1}C_{i}^{-1}\in\Gamma\textrm{ and }B_{i}C_{i}^{-1}\in\Gamma

for all i=1,2,3i=1,2,3. One can also obtain that the subgroup Γ\Gamma contains the following elements:

F3\displaystyle F_{3} =\displaystyle= (C1B11)F1=C1A2A31B41,\displaystyle(C_{1}B_{1}^{-1})F_{1}=C_{1}A_{2}A_{3}^{-1}B_{4}^{-1},
F4\displaystyle F_{4} =\displaystyle= F3R(B1C11)=C2A3A41B11(B1C11)=C2A3A41C11 and\displaystyle F_{3}^{R}(B_{1}C_{1}^{-1})=C_{2}A_{3}A_{4}^{-1}B_{1}^{-1}(B_{1}C_{1}^{-1})=C_{2}A_{3}A_{4}^{-1}C_{1}^{-1}\textrm{ and}
F5\displaystyle F_{5} =\displaystyle= F3F3F4=C1A2A31A41.\displaystyle F_{3}^{F_{3}F_{4}}=C_{1}A_{2}A_{3}^{-1}A_{4}^{-1}.

Thus we obtain that F5F31=A4B41ΓF_{5}F_{3}^{-1}=A_{4}B_{4}^{-1}\in\Gamma. By the action of RR, AiBi1ΓA_{i}B_{i}^{-1}\in\Gamma for all i=1,2,3,4i=1,2,3,4. The remaining part of the proof is very similar to that of Lemma 3.3.       

Lemma 3.8.

For g=3g=3 and every even p4p\geq 4, the group generated by the elements

ρ1,ρ2,ρ1Hb1,bHb+1,b1A2A31 and ρ2A1B1C1C21B31A31\rho_{1},\rho_{2},\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}A_{2}A_{3}^{-1}\textrm{ and }\rho_{2}A_{1}B_{1}C_{1}C_{2}^{-1}B_{3}^{-1}A_{3}^{-1}

contains the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} for i=1,2,3i=1,2,3.

Proof.

Consider the models for Σ3,p\Sigma_{3,p} as shown in Figure 2. Let Γ\Gamma be the subgroup of Mod(Σ3,p){\rm Mod}(\Sigma_{3,p}) generated by the elements ρ1\rho_{1}, ρ2\rho_{2}, ρ1F1\rho_{1}F_{1} and ρ2E1\rho_{2}E_{1} where F1=Hb1,bHb+1,b1A2A31F_{1}=H_{b-1,b}H_{b+1,b}^{-1}A_{2}A_{3}^{-1} and E1=A1B1C1C21B31A31E_{1}=A_{1}B_{1}C_{1}C_{2}^{-1}B_{3}^{-1}A_{3}^{-1}. Thus the elements R=ρ1ρ2R=\rho_{1}\rho_{2}, F1=ρ1ρ1F1F_{1}=\rho_{1}\rho_{1}F_{1} and E1=ρ2ρ2E1E_{1}=\rho_{2}\rho_{2}E_{1} are contained in the subgroup Γ\Gamma. We get the elements

F2\displaystyle F_{2} =\displaystyle= F1R1=Hb,b+1Hb+2,b+11A1A21Γ,\displaystyle F_{1}^{R^{-1}}=H_{b,b+1}H_{b+2,b+1}^{-1}A_{1}A_{2}^{-1}\in\Gamma,
F3\displaystyle F_{3} =\displaystyle= F2E1=Hb,b+1Hb+2,b+11B1A21Γ and\displaystyle F_{2}^{E_{1}}=H_{b,b+1}H_{b+2,b+1}^{-1}B_{1}A_{2}^{-1}\in\Gamma\textrm{ and}
F4\displaystyle F_{4} =\displaystyle= F3E1=Hb,b+1Hb+2,b+11C1A21Γ.\displaystyle F_{3}^{E_{1}}=H_{b,b+1}H_{b+2,b+1}^{-1}C_{1}A_{2}^{-1}\in\Gamma.

From these, the subgroup Γ\Gamma contains the elements F2F31=A1B11F_{2}F_{3}^{-1}=A_{1}B_{1}^{-1} and F3F41=B1C11F_{3}F_{4}^{-1}=B_{1}C_{1}^{-1}, which implies that A1C11ΓA_{1}C_{1}^{-1}\in\Gamma. Hence

AiBi1Γ,BjCj1Γ and AjCj1ΓA_{i}B_{i}^{-1}\in\Gamma,B_{j}C_{j}^{-1}\in\Gamma\textrm{ and }A_{j}C_{j}^{-1}\in\Gamma

for all i=1,2,3i=1,2,3 and j=1,2j=1,2, by the action of RR. We also have the following element

(A2C21)E1=A2B31,(A_{2}C_{2}^{-1})^{E_{1}}=A_{2}B_{3}^{-1},

which is contained in Γ\Gamma. This implies that

AiBi+11ΓA_{i}B_{i+1}^{-1}\in\Gamma

for i=1,2i=1,2 by the action of RR. One can complete the proof as in the proof of Lemma 3.3.       

Remark 3.9.

Our results in lemmata 3.33.8 are also valid for surfaces with odd number of punctures. To see that our proofs also work for such surfaces we refer the reader to Figures 33 and 55 in [2].

Refer to caption
Figure 3. The curves ei,je_{i,j} and γi\gamma_{i} on the surface Σg,p\Sigma_{g,p}.

Now, in the remainder of the paper let Γ\Gamma be the subgroup of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements given explicitly in lemmata 3.33.8 with the conditions mentioned in these lemmata. The proof of the following lemma is similar to that of   [2, Lemma 4.64.6], nevertheless we give a proof for the sake of completeness of the paper.

Lemma 3.10.

The group Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) is contained in the group Γ\Gamma.

Proof.

It follows from the subgroup Γ\Gamma contains the elements AiA_{i}, BiB_{i} and CjC_{j} for all i=1,,gi=1,\ldots,g and j=1,,g1j=1,\ldots,g-1 by lemmata 3.33.8 that it is sufficient to prove that Γ\Gamma contains the Dehn twists Ei.jE_{i.j} for some fixed ii (j=1,2,,p1j=1,2,\ldots,p-1). Let us first note that Γ\Gamma contains AgA_{g} and R=ρ1ρ2R=\rho_{1}\rho_{2}. Consider the models for Σg,p\Sigma_{g,p} as shown in Figures 1 and 2. By the fact that the diffeomorphism RR maps aga_{g} to e1,p1e_{1,p-1}, we get

RAgR1=E1,p1Γ.RA_{g}R^{-1}=E_{1,p-1}\in\Gamma.

The diffeomorphism ϕi=Bi+1Γi1CiBi\phi_{i}=B_{i+1}\Gamma_{i}^{-1}C_{i}B_{i} which maps each ei,je_{i,j} to ei+1,je_{i+1,j} for j=1,2,,p1j=1,2,\ldots,p-1 (see Figure 3). By the proof of  [2, Lemma 4.54.5], the group Γ\Gamma contains the element ϕg\phi_{g}. Thus we have

ϕg1ϕ2ϕ1E1,p1(ϕg1ϕ2ϕ1)1=Eg,p1H.\phi_{g-1}\cdots\phi_{2}\phi_{1}E_{1,p-1}(\phi_{g-1}\cdots\phi_{2}\phi_{1})^{-1}=E_{g,p-1}\in H.

Likewise, the diffeomorphism RR sends eg,p1e_{g,p-1} to e1,p2e_{1,p-2}. Then we obtain

REg,p1R1=E1,p2Γ.RE_{g,p-1}R^{-1}=E_{1,p-2}\in\Gamma.

It follows from

ϕg1ϕ2ϕ1E1,p2(ϕg1ϕ2ϕ1)1=Eg,p2Γ\phi_{g-1}\cdots\phi_{2}\phi_{1}E_{1,p-2}(\phi_{g-1}\cdots\phi_{2}\phi_{1})^{-1}=E_{g,p-2}\in\Gamma

that

R(Eg,p2)R1=E1,p3ΓR(E_{g,p-2})R^{-1}=E_{1,p-3}\in\Gamma

Continuing in this way, we conclude that the elements E1,1,E1,2,E_{1,1},E_{1,2}, ,E1,p1\ldots,E_{1,p-1} belong to Γ\Gamma, which completes the proof.       

Proof of Theorem A.

Consider the surface Σg,p\Sigma_{g,p} as in Figures 1 and 2.

If g=2k14g=2k\geq 14 and p=2b10p=2b\geq 10: In this case, consider the surface Σg,p\Sigma_{g,p} as in Figure 1. Since

ρ1(ck3)=ck+4,ρ1(bk1)=bk+3 and ρ1(ak)=ak+2,\rho_{1}(c_{k-3})=c_{k+4},\rho_{1}(b_{k-1})=b_{k+3}\textrm{ and }\rho_{1}(a_{k})=a_{k+2},

we get

ρ1Ck3ρ1=Ck+4,ρ1Bk1ρ1=Bk+3 and ρ1Akρ1=Ak+2.\rho_{1}C_{k-3}\rho_{1}=C_{k+4},\rho_{1}B_{k-1}\rho_{1}=B_{k+3}\textrm{ and }\rho_{1}A_{k}\rho_{1}=A_{k+2}.

Also, since ρ1Hb1,bρ1=Hb+1,b\rho_{1}H_{b-1,b}\rho_{1}=H_{b+1,b}, it is easy to see that ρ1Hb1,bHb+1,b1Ck3Bk1AkAk+21Bk+31Ck+41\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1} is an involution. Therefore, the generators of the subgroup Γ\Gamma given in Lemma 3.3 are involutions.

If g=2k+113g=2k+1\geq 13 and p=2b10p=2b\geq 10: In this case, consider the surface Σg,p\Sigma_{g,p} as in Figure 2. It follows from

ρ1(ck3)=ck+5,ρ1(bk)=bk+3 and ρ1(ak+1)=ak+2,\rho_{1}(c_{k-3})=c_{k+5},\rho_{1}(b_{k})=b_{k+3}\textrm{ and }\rho_{1}(a_{k+1})=a_{k+2},

that we have

ρ1Ck3ρ1=Ck+5,ρ1Bkρ1=Bk+3 and ρ1Ak+1ρ1=Ak+2.\rho_{1}C_{k-3}\rho_{1}=C_{k+5},\rho_{1}B_{k}\rho_{1}=B_{k+3}\textrm{ and }\rho_{1}A_{k+1}\rho_{1}=A_{k+2}.

Also, by the fact that ρ1Hb1,bρ1=Hb+1,b\rho_{1}H_{b-1,b}\rho_{1}=H_{b+1,b}, it is easy to see that the element ρ1Hb1,bHb+1,b1Ck3BkAk+1Ak+21Bk+31Ck+51\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+5}^{-1} is an involution. We conclude that the generators of the subgroup Γ\Gamma given in Lemma 3.4 are involutions.

If g=3,4,5g=3,4,5 or g=6g=6 and p=2b4p=2b\geq 4: It follows from

  • ρ2(b2)=b5\rho_{2}(b_{2})=b_{5}, ρ2(a3)=a4\rho_{2}(a_{3})=a_{4} and ρ1(c3)=c4\rho_{1}(c_{3})=c_{4} if g=6g=6,

  • ρ1(a3)=a4\rho_{1}(a_{3})=a_{4}, ρ2(a2)=a4,ρ2(b2)=b4\rho_{2}(a_{2})=a_{4},\rho_{2}(b_{2})=b_{4} and ρ2(c2)=c3\rho_{2}(c_{2})=c_{3} if g=5g=5,

  • ρ2(b1)=b4\rho_{2}(b_{1})=b_{4}, ρ2(a2)=a3\rho_{2}(a_{2})=a_{3} and ρ1(c2)=c3\rho_{1}(c_{2})=c_{3} if g=4g=4

  • ρ1(a2)=a3\rho_{1}(a_{2})=a_{3}, ρ2(a1)=a3,ρ2(b1)=b3\rho_{2}(a_{1})=a_{3},\rho_{2}(b_{1})=b_{3} and ρ2(c1)=c2\rho_{2}(c_{1})=c_{2} if g=3g=3 and

  • ρ1Hb1,bρ1=Hb+1,b\rho_{1}H_{b-1,b}\rho_{1}=H_{b+1,b} if g=3,4,5g=3,4,5 or g=6g=6

that the following elements:

  • ρ2B2A3A41B51\rho_{2}B_{2}A_{3}A_{4}^{-1}B_{5}^{-1} and ρ1Hb1,bHb+1,b1C3C41\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{3}C_{4}^{-1} if g=6g=6,

  • ρ1Hb1,bHb+1,b1A3A41\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}A_{3}A_{4}^{-1} and ρ2A2B2C2C31B41A41\rho_{2}A_{2}B_{2}C_{2}C_{3}^{-1}B_{4}^{-1}A_{4}^{-1} if g=5g=5,

  • ρ2B1A2A31B41\rho_{2}B_{1}A_{2}A_{3}^{-1}B_{4}^{-1} and ρ1Hb1,bHb+1,b1C2C31\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{2}C_{3}^{-1} if g=4g=4 and

  • ρ1Hb1,bHb+1,b1A2A31\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}A_{2}A_{3}^{-1} and ρ2A1B1C1C21B31A31\rho_{2}A_{1}B_{1}C_{1}C_{2}^{-1}B_{3}^{-1}A_{3}^{-1} if g=3g=3

given in lemmata 3.53.8 are involutions.

Next, we show that the subgroup Γ\Gamma is equal to the mapping class group Mod(Σg,p){\rm Mod}(\Sigma_{g,p}). By Lemma 3.10, the group Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) is contained in the group Γ\Gamma. Hence, by Lemma 3.2, we need to prove that Γ\Gamma is mapped surjectively onto SympSym_{p}. The element ρ1ρ2Γ\rho_{1}\rho_{2}\in\Gamma has the image (1,2,,p)Symp(1,2,\ldots,p)\in Sym_{p}.

As proven above, the Dehn twists AiA_{i}, BiB_{i} and CiC_{i} belong to the subgroup Γ\Gamma. Thus, it can be easily observed that the factorization of half twists Hb1,bHb+1,b1H_{b-1,b}H_{b+1,b}^{-1} are contained in subgroup Γ\Gamma. Therefore, the group Γ\Gamma also contains the following element:

Rb2(Hb1,bHb+1,b1)R2b=H1,2H3,21,R^{b-2}(H_{b-1,b}H_{b+1,b}^{-1})R^{2-b}=H_{1,2}H_{3,2}^{-1},

which has the image (1,2,3)Symp(1,2,3)\in Sym_{p}. This completes the proof since the elements (1,2,,p)(1,2,\ldots,p) and (1,2,3)(1,2,3) of SympSym_{p} generate the whole group SympSym_{p} if pp is even [9, Theorem B].

When the number of punctures is odd, we introduce an additional involution ρ3\rho_{3} (depicted in Figure 4) to our generating set. The main reason behind adding an extra involution is for generating the symmetric group SympSym_{p}. We want to point out that aside from generating SympSym_{p}, all of our proofs in the case of even number of punctures work for odd number of punctures. For ρ1\rho_{1} and ρ2\rho_{2}, we distribute punctures as in Figures 33 and 55 in [2] (see also Remark 3.9).

Refer to caption
Figure 4. The involution ρ3\rho_{3} on the surface Σg,p\Sigma_{g,p} for p=2b+1p=2b+1.
Proof of Theorem B.

For the first part of the proof we show that

  • (i)

    For every even integer g=2k14g=2k\geq 14 and every odd integer p=2b+19p=2b+1\geq 9, the subgroup Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements

    ρ1,ρ2 and ρ1Hb1,bHb+1,b1Ck3Bk1AkAk+21Bk+31Ck+41,ρ3.\rho_{1},\rho_{2}\textrm{ and }\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k-1}A_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1},\rho_{3}.
  • (ii)

    For every odd integer g=2k+115g=2k+1\geq 15 and odd integer p=2b+19p=2b+1\geq 9, the subgroup Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) of Mod(Σg,p){\rm Mod}(\Sigma_{g,p}) generated by the elements

    ρ1,ρ2 and ρ1Hb1,bHb+1,b1Ck3BkAk+1Ak+21Bk+31Ck+51,ρ3.\rho_{1},\rho_{2}\textrm{ and }\rho_{1}H_{b-1,b}H_{b+1,b}^{-1}C_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+5}^{-1},\rho_{3}.

Note that, it is enough to prove that the subgroup generated by the elements above mapped surjectively onto SympSym_{p}. For this, consider the images of the elements ρ1\rho_{1}, ρ2\rho_{2} and ρ3\rho_{3}

(1,p1)(2,p2)(b,b+1),\displaystyle(1,p-1)(2,p-2)\ldots(b,b+1),
(1,p)(2,p1)(b,b+2),\displaystyle(1,p)(2,p-1)\ldots(b,b+2),
(2,p1)(3,p2)(b,b+2).\displaystyle(2,p-1)(3,p-2)\ldots(b,b+2).

This finishes the proof for the first part,since these elements generate SympSym_{p}, see [20, Lemma 6]. For the second part of the theorem, note that adding ρ3\rho_{3} to the corresponding generating set given in Theorem A, finishes the proof.

As a last observation, one can prove that Theorem A also holds for the cases p=2p=2 or p=3p=3. In theses cases, the generating set of Γ\Gamma can be chosen as

{ρ1,ρ2,ρ1Ck3Bk1AkAk+21Bk+31Ck+41}ifg=2k14,{ρ1,ρ2,ρ1Ck3BkAk+1Ak+21Bk+31Ck+51}ifg=2k+113.{ρ1,ρ2,ρ2B2A3A41B51,ρ1C3C41}ifg=6.{ρ1,ρ2,ρ1A3A41,ρ2A2B2C2C31B41A41}ifg=5.{ρ1,ρ2,ρ2B1A2A31B41,ρ1C2C31}ifg=4.{ρ1,ρ2,ρ1A2A31,ρ2A1B1C1C21B31A31}ifg=3..\begin{array}[]{lll}\{\rho_{1},\rho_{2},\rho_{1}C_{k-3}B_{k-1}A_{k}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+4}^{-1}\}&\textrm{if}&g=2k\geq 14,\\ \{\rho_{1},\rho_{2},\rho_{1}C_{k-3}B_{k}A_{k+1}A_{k+2}^{-1}B_{k+3}^{-1}C_{k+5}^{-1}\}&\textrm{if}&g=2k+1\geq 13.\\ \{\rho_{1},\rho_{2},\rho_{2}B_{2}A_{3}A_{4}^{-1}B_{5}^{-1},\rho_{1}C_{3}C_{4}^{-1}\}&\textrm{if}&g=6.\\ \{\rho_{1},\rho_{2},\rho_{1}A_{3}A_{4}^{-1},\rho_{2}A_{2}B_{2}C_{2}C_{3}^{-1}B_{4}^{-1}A_{4}^{-1}\}&\textrm{if}&g=5.\\ \{\rho_{1},\rho_{2},\rho_{2}B_{1}A_{2}A_{3}^{-1}B_{4}^{-1},\rho_{1}C_{2}C_{3}^{-1}\}&\textrm{if}&g=4.\\ \{\rho_{1},\rho_{2},\rho_{1}A_{2}A_{3}^{-1},\rho_{2}A_{1}B_{1}C_{1}C_{2}^{-1}B_{3}^{-1}A_{3}^{-1}\}&\textrm{if}&g=3.\\ \end{array}.

It can be easily proven that the group Γ\Gamma contains Mod0(Σg,p){\rm Mod}_{0}(\Sigma_{g,p}) by the similar arguments in the proofs of lemmata 3.33.8. The element ρ1ρ2Γ\rho_{1}\rho_{2}\in\Gamma has the image (1,2,,p)Symp(1,2,\ldots,p)\in Sym_{p}. Hence, this element generates SympSym_{p} for p=2p=2. If p=3p=3, we distribute the punctures as in   [11, Figure 11]. Then the element ρ1\rho_{1} has the image (1,3)(1,3), which generate SympSym_{p} together with the element (1,2,3)(1,2,3). Therefore, the group Γ\Gamma is mapped surjectively onto SympSym_{p} for p=2,3p=2,3. One can conclude that the group Γ\Gamma is equal to Mod(Σg,p){\rm Mod}(\Sigma_{g,p}).

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