Thanks for the help Lul Thyme but I think the groups I'm using a bit too unwieldy to check for those properties. I've convinced myself now that this problem is probably a bit too hard for me with my current skill set, so I'll give the full context in the hope that someone can at least offer some insight.

I'm currently writing a project on the braid groups and analogues on closed surfaces. 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} (the homomorphism is give by 'adding a string' on to the end of any braid in B

_{n} and this can be shown to be a monomorphism). It also happens to be the case that B

_{n} has no torsion elements but this isn't at all easy to show (a fully algebraic proof takes a few pages, and a topological proof requires some fairly strong theorems).

I've now moved on to looking at the braid group on the surface of a sphere, we'll denote the n-string braid group on S

^{2} by S

_{n}. There are several equivalent definitions for this group, but perhaps the easiest to understand is to think of a braid in S

_{n} as being n non-intersecting closed curves in S

^{2} x [0,1] with end points on the boundary spheres (in the same way that the classical braid group is characterised by n closed curves in R

^{2} x [0,1]) and then the group operation is given by concatenation of the braids on their end points. A fairly crude image of braid in S

_{n} can be found

here although I haven't drawn the outer sphere so you'll have to imagine a circle extending from the line at the top down and around the entire image.

Call the braid in the image g. Now, if g were just a braid in the classical braid group, then g would be non-trivial. However, as a braid in S

_{n} it turns out to be trivial. You can see this by pulling the first string down and around, behind the central sphere. Note however, that the map f:S

_{n}->S

_{n+1} which 'adds a string' on to the end of any braid does not map g to a trivial braid - the action of pulling the first string around the sphere can no longer be done. So, f isn't a homomorphism like it was in the case for B

_{n}. This lead me to think that S

_{n} could not be embedded in S

_{n+1}. However, I have no idea how to show this.

to help anyone. The Artin presentation for the classical braid group B

_{n} is given as follows.

B

_{n} has generators a

_{1}, a

_{2}, ..., a

_{n-1}and relations:

a

_{i} a

_{j} = a

_{j} a

_{i} if |i-j|>1

a

_{i} a

_{i+1} a

_{i} = a

_{i+1} a

_{i} a

_{i+1} for i in {1,2,...,n-2}

The sphere braid group S

_{n} has the same presentation, but with the added relation g = a

_{1} a

_{2} ... a

_{n-2} a

_{n-1} a

_{n-1} a

_{n-2} ... a

_{2} a

_{1} = 1. Geometrically this relation says that the braid g, in the image linked above, is a trivial braid.

It's this relation g=1 which isn't respected under the map f(a

_{j})=b

_{j} (where b

_{j} is a generator of S

_{n+1}) and causes f to fail to be a homomorphism.

The fact that S

_{n} and S

_{n+1} are so similar seems to be what is making this problem so difficult.