The Torsion Generating Set Of The Extended Mapping Class Groups In Low Genus Cases
Abstract.
We prove that for genus , the extended mapping class group can be generated by two elements of finite orders. But for , cannot be generated by two elements of finite orders.
Key words and phrases:
mapping class group, generator, torsion2010 Mathematics Subject Classification:
57N05, 57M20, 20F381. Introduction
Korkmaz has proved that the mapping class group can be generated by two elements of finite orders in [3]. Using the notation that (m, n are integers) to mean a group can be generated by two elements whose orders are and respectively, Korkmaz’s result says:
torsion generating set  

consisting of two elements  
It is an open problem listed in [4] that whether the extended mapping class group can be generated by two torsion elements. In [1], the author partially solved such a problem when the genus . In this paper, we deal with .
When , the method and idea in the process of calculation in this paper are mostly the same as those in [1] and [3]. The reason for and should be treated separately is as the follow. When the genus is high, there will be plenty of space to find a simple closed curve satisfying two conditions: (1) it lies in the periodic orbit; (2) it does not intersect with some given curves. When the genus is less than 5, we cannot do this. So we use other treatment carefully. When , we use the presentation of to prove it cannot be generated by two elements of finite orders. So we can summarize the result as follow:
torsion generating set  
consisting of two elements  
impossible  
still unknown  
2. Preliminary
Notations.
(a) We use the convention of functional notation, namely, elements of the mapping class group are applied right to left, i.e. the composition means that is applied first.
(b) A Dehn twist means a righthand Dehn twist.
(c) We denote the curves by lower case letters , , , (possibly with subscripts) and the Dehn twists about them by the corresponding capital letters , , , . Notationally we do not distinguish a diffeomorphism/curve and its isotopy class.
Humphries generators and the gon.
Humphries have proved the following theorem ([2]).
Theorem 2.1.
Let be the curves as on the lefthand side of figure 1. Then the mapping class group is generated by ’s and .
The genus surface can be looked as a gon, whose opposite edges are glued together in pairs. vertices of the gon are glued to be two vertices.
We can also draw the curves on the gon as the righthand side of figure 1. There is a natural rotation of the gon that sends to . In this paper, we will use the curve as figure 1 shows. Denote . They are also used in this paper.
We need the intersection numbers between the curves . Consider the index in modulo classes. When viewing these curves in the gon, we need to be careful. Some times though two such curves meet at the vertex of the gon, They do not really intersect. We can perturb them a little to cancel the intersection point. The intersection numbers between are listed as follow:

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .

if and only if .
Some torsion elements
Obviously we have . Take the reflection of the regular gon satisfying . We can check . See figure 2.
In [1] we know for . We will see it is also true for .
3. The main result and the proof
Theorem 3.1.
Let as before. For , .
Proof.
Denote the subgroup generated by and as . We will prove that includes all the elements in . Similar to [1], The proof of the lemma has 4 steps.
Step 1. For every , we prove is in .
Step 2. For every , we prove is in .
Step 3. Using lantern relation, we prove that for every , is in .
Step 4. .
The motivation of step 2 and step 3 is as follow. There is a lantern on the surface where the curves in the lantern relation appear as , , , , , , showed on the upper side of figure 3. The lantern relation can be also written as . So one Dehn twist can be decomposed into the product of pairs of Dehn twists. Draw the lantern in the gon as on the lower side of figure 3. We will find some of the pairs of Dehn twists we use can be expressed as the form . When the , we cannot find a lantern on the surface.
The proof of Step 1:
The reason for is . After conjugating with , we have for every , is in .
The proof of step 2:
Suppose the genus .
We already know does not intersect with or . So maps the pair of curves to the pair of curves . Since is in , is in . We also have for every , is in . See figure 4.
We know does not intersect with . We can check . So maps the pair of curves to the pair of curves , maps the pair of curves to the pair of curves . This means maps the pair of curves to the pair of curves . See figure 5.
Hence is in . After conjugating some power of and multiplying some , we have for every , is in .
Suppose the genus .
We know that does not intersect with or . So maps the pair of curves to the pair of curves . We can also check when the genus is 3, . So is in . See figure 6.
After conjugating with some power of and multiplying some , we have for every , and are in . We also have for every , is in .
We know does not intersect with or . So maps the pair of curves to the pair of curves . Then is in . For every , is also in . See figure 7.
We know does not intersect with or . So maps the pair of curves to the pair of curves . Then is in . See figure 8.
We know does not intersect with or . So maps the pair of curves to the pair of curves . Then is in . See figure 9.
We can check that . So is in . Conjugating with some power of and multiplying , we have for every , is in . Multiplying it by , we have for every , is in .
The proof of step 3:
We want to show for every , is in .
Recall lantern relation, we have , or , where and are the curves showed in figure 3. By the result of step 2, and are in . What we need is to prove is also in . Notice . We only need to show there exist some such that and are in .
Suppose .
We can check that . We also know does not intersect with . So maps to . Hence is in . See figure 10.
We can check . Since does not intersect with , maps to . Hence is in . See figure 11.
Suppose .
The fact still holds. When we cannot find some that does not intersect with simultaneously. We use some curves instead.
At first we find does not intersect with . So maps to , is in . See figure 12.
is also in . Then we find does not intersect with or . So maps to . Hence is in . See figure 13.
Similarly, The fact still holds. When , we can find does not intersect with . So maps to . Hence is in . And then multiply , we have in . See figure 14.
The proof of step 4:
Since both and are in , by Humphries’s result, contains the mapping class group . Now is an orientation reversing element. is an index 2 subgroup of . So for . ∎
Theorem 3.2.
For , is . It cannot be generated by two elements of finite orders.
Proof.
We only need to prove that cannot be generated by two elements of finite orders.
It is well known that , where and are the dihedral group of order 6 and order 4 respectively (see, for example, [5] or [6]). It has a presentation as
Every element in can be written as a reduced form in one of the following 3 types:

,

, or

.
Here each and each .
For an element in type (2), can be conjugated to
.
So its conjugacy class is the same as an element in type (3) with a shorter word length.
For an element in type (3), can be conjugated to . So its conjugacy class is the same as an element in type (1) or (2) with a shorter word length.
For an element in type (1), since the word length its power will be larger, it must not be of finite order. So an element of finite order in must be conjugated to one of the following 8 elements: .
By adding in a new relation , we get a quotient group homomorphic to the Cartesian product , which is a finite group with the presentation
For the convenience of calculation, we can think of as a permutation subgroup of the symmetric group :
element in  permutation  element in  permutation 

1  ()  (45)  
(123)  (123)(45)  
(132)  (132)(45)  
(12)  (12)(45)  
(13)  (13)(45)  
(23)  (23)(45) 
We can check all the possible images in of the conjugacy classes in as follow:
Conjugacy classes in  elements in 

,  , 
, ,  , , 
,  , 
We have the following fact: if two elements in can generate , the only possible cases are and .
Each one of has order 2. If an element in is mapped to , , or , it must also have order 2. But two elements of order 2 can only generate a dihedral group, not . ∎
Remark 3.3.
Though cannot be generated by two torsion elements, it can be generated by two elements. In fact, since
we have
So the extended mapping class group for case: (1) can be generated by two elements; (2) cannot be generated by two elements of finite orders. This is different from the case .
References
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