Showing that a group ISN'T a subgroup of another

[Talith]

I've encountered a problem in my work where I'd like to show that a group G ISN'T isomorphic to a subgroup of another group H which is similar to G in terms of generators and relations (although I haven't shown this yet so I can't use this fact). I was wondering if anyone could suggest some techniques for solving this kind of problem. The motivation for wanting to show this is that I want to show that if G has a torsion element, we can't automatically assume that H has a torsion element because G isn't isomorphic to a subgroup of H - in a previous example, this actually was the case and so showing H had torsion was easy.

At the moment, G and H are defined in a very geometric way and the most we know about the groups at the moment is that G can be generated by k elements and H can be generated by k+1 elements. There also exist surjective homomorphisms f:G->S_k and g:H->S_k+1 where S_n is the symmetric group on n elements. If it helps, we also know that G has an element of order 2. I know a few other things in terms of relations amongst the generators, but I made this post so that people could suggest techniques, as opposed to providing an answer - so I'll withhold that information for now.

[Giallo]

Do you know the order of both groups? For example if |G| and |H| are relatively prime, then G is not a subgroup of H and H is not one of G (if I recall correctly)

[++$_]

Do you know the order of both groups? For example if |G| and |H| are relatively prime, then G is not a subgroup of H and H is not one of G (if I recall correctly)

That is correct (in fact, in that case G and H have no isomorphic nontrivial subgroups), but since he is worried about the existence of torsion elements I assume that both groups are at least potentially infinite.

[Talith]

Yeh the groups I'm working with are infinite.

[Lul Thyme]

In general, subgroups are not too well-behaved.
For example, the minimal number of generators can increase, etc...

There are some properties that are preserved for subgroups though : nilpotency (and class cannot increase), solvability (same with length), etc...
You could do things like check if the derived subgroup of one is isomorphic to a subgroup of the derived subgroup of the other. You can also generalise this to a variety of other characteristic subgroups and some of the usual series (derived series, lower central series, etc...)

I'm more used to finite groups, but I'm sure there are other similar invariant and characteristic subgroups which are more applicable in the infinite case.

[jestingrabbit]

If you know that H doesn't have an order two element then you're done, but I'm assuming you don't have that.

Also, do you mean that the concrete group H does not have G as a subset, or do you mean that there is not a subset of H that is isomorphic to G?

The second is harder, so I'll assume that. The basic idea that I would try for is to demonstrate a property of G that a superset of G would share. So, if you have some elements of G that commute, and the commutator of H is empty, you're done. Or you have elements that are such that abc = e, and a^2 * b^2 * c^2 = e, but there are no such elements in H. That's what I would go with anyway.

[Talith]

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 S² 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 S² 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 R² 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 a₁, a₂, ..., 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 = a₁ a₂ ... a_n-2 a_n-1 a_n-1 a_n-2 ... a₂ a₁ = 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.