This question was originally posted on math.SE 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 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.
Best Answer
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, 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.