Artin's presentation of **braid group on three strands** is:
$$ B\_3 = \langle l,r : lrl = rlr \rangle $$
where you should think of "$l$" as the positive crossing between the left and middle strands and "$r$" as the positive crossing of the right and middle strands:



```
     |   |   |         |   |   |
      \ /    |         |    \ /
 l =   \     |     r = |     \
      / \    |         |    / \
     |   |   |         |   |   |

```

Then there is a surjection $B\_3 \to S\_3$ given by $l \mapsto (12)$ and $r \mapsto (23)$. ($S\_3$ is the symmetric group on three letters: it is generated by $(12)^2 = 1 = (23)^2$ and the braid relation above.) The **pure braid group** $PB\_3$ is the kernel of this surjection.


The **six-crossing braid** $b = lr^{-1}lr^{-1}lr^{-1}$ is an element of the pure braid group. Let $N$ be the minimal normal subgroup of $B\_3$ that contains $b$. Certainly $N \subseteq PB\_3$.



> 
> **Question:** Do we have $N = PB\_3$?
> 
> 
> 


### Motivation


The motivation for my question comes from a neat trick that Conway showed us years ago. It leads to a more nuanced question than what I asked that I will pose as its own question if the answer above is "no". My memory is that at the time Conway did not know the answer to the more nuanced question, which suggests that the answer above cannot be "yes".


Take a long and reasonably thin rectangle of paper, and score it with two cuts, so that you have three strips of paper that are connected at both ends, so that you end up with a (framed, oriented, ...) "theta graph":



```
 |--------|
 |        |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |  |  |  |
 |        |
 |--------|

```

Then with some finagling, you can in fact "tie" the six-crossing braid in those three strands, without ripping the paper further, by passing the bottom through itself a few times. (The trick is that it's easier to unbraid than to braid, so make your paper long enough that you can put $bb^{-1}$ into it, and then unbraid the $b^{-1}$.) Put another way: you can first put a hair-tie at the end of your ponytail, and then braid your hair.


The harder question is to characterize all braids that you can put on the above "theta graph". The following facts are essentially obvious:


* Any "braiding" of the theta graph is pure.
* The set of braidings of the theta graph is a subgroup of $B\_3$.
* The set of braidings of the theta graph is closed under conjugating by arbitrary braids.


Therefore, the set $T$ of braidings of the theta graph is a normal subgroup of $B\_3$, with $N \subseteq T \subseteq PB\_3$. In particular, a positive answer to the question above characterizes $T$.



Is the pure braid group on three strands generated as a normal subgroup of the braid group by the six-crossing braid?
No.


Let $P\_n$ be the pure braid group on $n$ strands. Forgetting the first strand gives a projection $P\_3\to P\_2$. The kernel is the free group $F\_2=\pi\_1(\mathbb{C}\setminus
\{2,3\},1)$ given by moving the first strand around the other two while they are held in place. Since $P\_2$ is just $\mathbb{Z}$, we have a short exact sequence


$1\to F\_2\to P\_3\to \mathbb{Z}\to 1$


Your element $b$ is contained in the kernel of this sequence (see the picture below) so its normal closure is contained in $F\_2\lt P\_3$. For an appropriate basis $\langle x,y\rangle=F\_2$, the element $b$ represents $xyx^{-1}y^{-1}$.



```
\ |  |
 \   |
  \  |
   \ |
  |  |
  |  |\
  |  | |
  |   /
  |  /
  | / 
   / |
  /  |
 /   |
| |  |
 \|  |
  |  |
  | \
  |  \
  |   \
  |  | |
  |  |/
  |  |
  |/ |
  |  |
  |  |
/ |  |

```

In fact, removing *any* of the three strands results in a trivial 2-strand braid (as can be seen from the picture). Closing up the ends of the braid thus results in a 3-component link so that if any one component is removed, the other two are unlinked. This is the well-known [Borromean rings](http://en.wikipedia.org/wiki/Borromean_rings). There are three different surjections $P\_3\to \mathbb{Z}$, given by forgetting each strand separately, and the group of [Brunnian braids](http://en.wikipedia.org/wiki/Brunnian_link#Brunnian_braids) is exactly the intersection of their kernels. Since $xyx^{-1}y^{-1}$ normally generates the commutator subgroup of $F\_2$, the normal closure $N$ of $b$ will be exactly the group of Brunnian braids. In particular, $N$ is an infinite-rank free group.


