Let $B_n$ be the braid group on $n$ strands. As is well known, if $\sigma_i$ is the operation of crossing the string in position $i$ over the string in position $i+1$, then the elements $\sigma_1,\dots,\sigma_{n-1}$ generate $B_n$ and the relations $\sigma_i\sigma_j=\sigma_j\sigma_i$ ($|i-j|\geq 2$) and $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}$ give a presentation of the group.
I am interested in a group that can be obtained from $B_n$ by adding another set of relations. For each $k$, define $T_k$ to be the "twist" of the first $k$ strands. Geometrically, you take hold of the bottom of the first $k$ strands and rotate your hand through 360 degrees in such a way that strands to the left go over strands to the right. In terms of the generators, $T_k=(\sigma_1\dots\sigma_{k-1})^k$, since $\sigma_1\dots\sigma_{k-1}$ takes the $k$th most strand and lays it across the next $k-1$ strands. Let us also define $S_k$ to be the right-over-left twist of the strands from $k+1$ to $n$. That is, $S_k=(\sigma_{k+1}\dots\sigma_{n-1})^{-(n-k)}$. (Also, let us take $T_1$ and $S_{n-1}$ to be the identity.)
I am interested in the group you get if you start with $B_n$ and add in the relation $T_kS_k=1$ for every $k$. Let me make some utterly trivial observations.
If $n=2$ then we get the cyclic group $C_2$. That's because $\sigma_1$ is the only generator and $T_2S_2=\sigma_1^2$. If $n=3$ then we get $S_3$. That's because $T_1S_1=\sigma_2^{-2}$ and $T_2S_2=\sigma_1^{-2}$, so the relations we are adding are $\sigma_1^2$ and $\sigma_2^2$, and it is well known that those, together with the braid relations, give a presentation of the symmetric group.
Beyond that I don't know what to say, though I've convinced myself (without a proof) that when $n=4$ the group is infinite: in general, it seems that the extra relations can be used to do only a limited amount of untwisting. (I do have a proof that there are pure braids that cannot be reduced to the identity once we have four strands. It's a fairly easy exercise and I won't give it here.)
What exactly is my question? Well, I'd be interested to know whether the word problem in this group is soluble in reasonable time. In the service of that, I'd like to know whether this group is one that people have already looked at, or whether it at least belongs to a class of groups that people have already looked at. (E.g., perhaps the solubility of the word problem follows from some general theory.) And is there some nice way of characterizing the subgroup of $B_n$ that we are quotienting by? That is, which braids belong to the normal closure of the set of braids $T_kS_k$? (One way of answering this would be to characterize their normal forms.)
The motivation for the question comes from part of an answer that Thurston gave to a question I asked about unknots. It seems to me that this question ought to be relevant to the untying of unknots, but easier.
One final remark: the word problem in $B_n$ can be solved in polynomial time. (If my understanding is correct, this is a result of Thurston that built on work of Garside.) Since adding more relations makes more braids equal to the identity but also gives more ways of converting a word into another, it is not clear whether the problem I am asking should be easier or harder than the word problem for braid groups. However, my hunch is that it is harder (for large $n$, that is).
Best Answer
If I understand your group correctly, your quotient $G_n = B_n / \langle T_kS_k : k = 1,2,\cdots, n\rangle$ can be thought of as a fairly geometric object -- it's the mapping class group of a sphere with $n$ marked points.
Let $\pi_0 Diff(S^2,n+1)$ be the mapping class group of $S^2$ that preserves $n+1$ points. There is an epi-morphism $\pi_0 Diff(S^2,n+1) \to \Sigma_{n+1}$, so we can talk about the subgroup of $\pi_0 Diff(S^2,n+1)$ that fixes one of the $n+1$ permuted points, call it $A_n$. I believe
$$ A_n \simeq B_n / \langle T_n \rangle$$
and I believe Ian Agol mentioned this. I'm using angle brackets to indicate "normal subgroup generated by".
But we still have to account for the other relators you want to mod out by, $T_1S_1, T_2S_2, \cdots, T_{n-1}S_{n-1}$.
Let $X_n$ be the subgroup of $Diff(S^2,n+1)$ that stabilizes one of the $n+1$ points. Then there is a map $X_n \to Diff(S^2,n)$ given by forgetting that point. This map is the fibre of a locally-trivial fibre bundle with base-space a $n$-times punctured sphere. Moreover, this expresses $\pi_0 Diff(S^2,n)$ as a quotient of $A_n$, and the relators are precisely the remaining ones you list (this requires a bit of a computation).
So I believe the above is an argument your group $G_n$ is $\pi_0 Diff(S^2,n)$. This explains why you get $\mathbb Z_2$ in the $n=2$ case, and $\Sigma_3$ in the $n=3$ case. This group is linear for all $n$. There's two published proofs of this, one by Bigelow and myself and another by Korkmaz. The representation is obtained as a tweaking of the Lawrence-Krammer representation.