This question was originally posted on [math.SE](https://math.stackexchange.com/questions/136821/can-the-n-string-sphere-braid-group-embed-in-to-the-n1-string-sphere-braid-gr) by myself nearly a year ago. I've been thinking again about the problem after it recently received a little attention, but little progress was made in finding a solution. So, I feel it's sufficiently *difficult* to post on mathoverflow.


It's an easy exercise to show that if $B\_n$ is Artin's classical braid group on $n$ strings, then $B\_n$ can be embedded in $B\_{n+1}$ (and in a canonical way). A similar statement can be proved for the pure braid group $P\_n$. This property is useful for proving various properties of the classical braid groups. One therefore asks if a similar property holds for braid groups on other surfaces such as the sphere.


Let $P\mathcal{S}\_n$ be the pure $n$-string braid group on the sphere $S^2$. Fox's definition of this group is the fundamental group of the configuration space $F\_{n}S^2=\prod\_n S^2\setminus\{(x\_1,\ldots,x\_n)|\exists i\neq j, x\_i=x\_j\}$ and then the full braid group $\mathcal{S}\_n$ is defined to be the fundamental group of the configuration space $B\_nS^2=F\_nS^2/\Sigma\_n$ where $\Sigma\_n$ is the action of the symmetric group by permuting coordinates of the elements of $F\_nS^2$.


It was proven by [Fadell and Van Buskirk](https://projecteuclid.org/journals/duke-mathematical-journal/volume-29/issue-2/The-braid-groups-of-E2-and-S2/10.1215/S0012-7094-62-02925-3.short) that the braid group $\mathcal{S}\_n$ has presentation given by the braid generators $\sigma\_1,\ldots,\sigma\_{n-1}$ and relations
$$\begin{eqnarray}\sigma\_i\sigma\_j\sigma\_i&=&\sigma\_j\sigma\_i\sigma\_j&(\mbox{ for }|i-j|=1)\\
\sigma\_i\sigma\_j&=&\sigma\_j\sigma\_i&(\mbox{ for }|i-j|>1)\\
\gamma&=&1&\end{eqnarray}$$
where $\gamma=(\sigma\_1\sigma\_1\ldots\sigma\_{n-1})(\sigma\_{n-1}\ldots\sigma\_2\sigma\_1)$.


With that framework now built up, my question is, can $\mathcal{S}\_n$ be embedded in to $\mathcal{S}\_{n+1}$ for $n\geq 3$ (and similarly for their pure counter parts)? The naive 'add a string on the end' map will not work because, for instance, the braid $\gamma$ becomes non-trivial when a string is added on the end.


I would think that the answer is no because of the dependence of the relation $\gamma=1$ on $n$, but a proof eludes me.



Can the n-string sphere braid group embed in to the (n+1)-string sphere braid group?
**Revision:** For $n>6$, there is no embedding of $\mathcal{S}\_n \hookrightarrow \mathcal{S}\_{n+1}$. 


First, recall that there is an extension $\mathbb{Z}/2\mathbb{Z} \to \mathcal{S}\_n \to Mod(S\_{0,n})$, where $Mod(S\_{0,n})$ is the (orientation preserving) mapping class group of the $n$-punctured sphere.


Inside $Mod(S\_{0,n})$, there is a subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$, which is a rotation of order $n-2$ of $S^2$, and fixes the north and south poles. The $n$ punctures include the north and south poles and one orbit of size $n-2$. The preimage of this group in $\mathcal{S}\_n$ is isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ or to $\mathbb{Z}/2(n-2)\mathbb{Z}$ (of course if $n$ is odd, these are isomorphic). Then we get a corresponding subgroup of $\mathcal{S}\_{n+1}$ from the injection. Project this group to $Mod(S\_{0,n+1})$ via the map $\mathcal{S}\_{n+1}\to Mod(S\_{0,n+1})$. The kernel of this projection will have size at most $2$. By the [Nielsen Realization Theorem](http://en.wikipedia.org/wiki/Nielsen_realization_problem), any finite subgroup of $Mod(S\_{0,n+1})$ must preserve a complete hyperbolic metric of finite area on $S\_{0,n+1}$, and in particular by uniformization extends to a finite group of conformal automorphisms of $S^2$ which permutes $n+1$ marked points. The finite subgroups of $PSL\_2(\mathbb{C})$ lie inside a conjugate of $SO(3)$, so the image is an abelian subgroup of $SO(3)$. The only abelian subgroups of $SO(3)$ are cyclic (or $\mathbb{Z}/2\mathbb{Z}^2$), so the image is either $\mathbb{Z}/(n-2)\mathbb{Z}$ or $\mathbb{Z}/2(n-2)\mathbb{Z}$ (in which case we may take an index 2 subgroup isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$; here we need $n>6$ to conclude that the image is not $\mathbb{Z}/2^2$). However, for $n>5$, there is no subgroup of $SO(3)$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ which permutes $n+1$ points, and therefore there is no such subgroup of $Mod(S\_{0,n+1})$, a contradiction. To see this, note that a cyclic group of rotations of $S^2$ isomorphic to $\mathbb{Z}/(n-2)\mathbb{Z}$ has two fixed points, and every other orbit of size $n-2$. Thus, there must be some $k$ and $e$ such that there are $k$ orbits of size $n-2$, and $e$ orbits of size $1$, where $e\leq 2$. If $k\leq 1$, then we get $n+1=k(n-2)+e \leq n$, a contradiction. If $k\geq 2$, then $n+1=k(n-2)+e \geq 2(n-2)$, so $n\leq 5$, a contradiction. 


