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).