Let $\Gamma\_{g,n}$ denote the mapping class group of an oriented surface of genus $g$ and with $n$ marked points. We assume that elements of $\Gamma\_{g,n}$ are not allowed to permute the marked points. I am interested in the case $g=0$.


In [Farb & Margalit](https://doi.org/10.1515/9781400839049), on page 114, it is claimed that $\Gamma\_{g,n}$ can be generated by $2g+n$ Dehn twists along the curves drawn on Figure 4.10. I was wondering if this statement is also true in the case $g=0$.


In [math/9912248](https://arxiv.org/pdf/math/9912248.pdf), Wajnryb exhibits a family of generators for $\Gamma\_{0,n}$ in Lemma 23. The curves $\alpha\_{i,j}$ illustrated on Figure 12 are a family of $n(n-1)/2$ generators.


My questions are the following:


1. Is the minimal number of Dehn twists generators of $\Gamma\_{0,n}$ known ?
2. What would be the $n$ generators of $\Gamma\_{0,n}$ if the construction of Farb-Margalit cited above applies in the case $g=0$ ?



Minimal number of (Dehn twists) generators of the mapping class group of a marked sphere
The minimum number of Dehn twist generators (and in fact the minimum number of generators of any kind) for $\Gamma\_{0,n}$ is ${n-1 \choose 2} - 1$. Here's why.


A presentation for $\Gamma\_{0,n}$ is known, and can be found in Lemma 4.1 of [this paper](http://nyjm.albany.edu/j/2017/23-9v.pdf) by Rebecca R. Winarski and myself. In the paper, $\operatorname{PMod}(\Sigma\_0,\mathcal B(n))$ is the group $\Gamma\_{0,n}$.


Number the $n$ marked points. The generators $A\_{i,j}$ are Dehn twists about curves that surround only the $i$th and $j$th marked points (see Figure 3 from the paper). These are essentially the same curves in the paper by Wajnryb that is linked in the question.


The generating set used in our paper is the set $\{A\_{i,j} \mid 1 \leq i < j \leq n-1\}$, and one of the relations (relation (5)) is $$(A\_{1,2}A\_{1,3} \cdots A\_{1,n-1})\cdots(A\_{n-3,n-2}A\_{n-3,n-1})(A\_{n-2,n-1}) = 1. $$ In this relation, each $A\_{i,j}$ appears exactly once, so you can use a Tietze transformation to eliminate one of the generators. We are now left with a generating set consisting of ${{n-1}\choose{2}} - 1$ Dehn twists.


The other 4 relations are all commutation relations (that is, of the form $[W,X] = 1$), so we can conclude that the abelianization of $\Gamma\_{0,n}$ is a free abelian group of rank ${n-1 \choose 2} - 1$. Therefore $\Gamma\_{0,n}$ cannot be generated by less than ${n-1 \choose 2} - 1$ elements.


I guess this also answers your question 2, in the sense that the result stated in Farb & Margalit does not hold for genus 0.


