# The Unapologetic Mathematician

## Braid groups

Okay, time for a group I really like.

Imagine you’re playing the shell game. You’re mixing up some shells on the surface of a table, and you can’t lift them up. How can you rearrange them? At first, you might think this is just a permutation group all over again, but not quite. Let’s move two shells around each other, taking a picture of the table of the table each moment, and then stack those pictures like a flip-book movie. I’ve drawn just such a picture.

We read this movie from the bottom to the top. The shell on the right moves behind the shell on the left as they switch places. It could also have moved in front of the left shell, though, and that picture would show the paths crossing the other way. We don’t want to consider those two pictures the same.

So why don’t we want to? Because the paths those shells trace out look a lot like the strands of knots! The two dimensions of the table and one of time make a three-dimensional space we can use to embed knots. Since these pictures just have a bunch of strands running up and down, crossing over and under each other as they go we call them “braids”. In fact, these movies form a group. We compose movies by running them one after another. We always bring the $n$ shells back where they started so we can always start the next movie with no jumps. We get the identity by just leaving the shells alone. Finally, we can run a movie backwards to invert it.

There’s one such braid group $B_n$ for each number $n$ of shells. The first one, $B_1$ is trivial since there’s nothing to do — there’s no “braiding” going on with one strand. The second one, $B_2$ is just a copy of the integers again, counting how many twists like the one pictured above we’ve done. Count the opposite twist as $-1$. Notice that this is already different from the symmetric groups, where $S_2$ just has the two moves, “swap the letters” or “leave them alone”.

Beyond here the groups $B_n$ and $S_n$ get more and more different, but they’re also pretty tightly related. If we perform a braiding and then forget which direction we made each crossing we’re just left with a permutation. Clearly every permutation can arise from some braiding, so we have an epimorphism from $B_n$ onto $S_n$. In fact, this shows up when we try to give a presentation of the braid group.

Recall that the symmetric group has presentation:

$
$s_is_{i+1}s_is_{i+1}s_is_{i+1} (1\leq i\leq n-2)>$

The generator $s_i$ swaps the contents of places $i$ and $i+1$. The relations mean that swapping twice undoes a swap, widely spaced swaps can be done in either order, and another seemingly more confusing relation that’s at least easily verified. The braid group looks just like this, except now a twist is not its own inverse. So get rid of that first relation:

$$

The fact that we get from the braid group to the symmetric group by adding relations reflects the fact that $S_n$ is a quotient of $B_n$. It’s interesting to play with this projection and compute its kernel.

March 14, 2007

## Commutator Subgroups

If we have a group $G$, it may or may not be abelian. We can measure how nonabelian it is with the commutator subgroup $\left[G,G\right]$. This is the subgroup generated by all the commutators in $G$. Start with all the elements of $G$ of the form $ghg^{-1}h^{-1}$ with $g$ and $h$ any elements of $G$. If $G$ is abelian, these are all trivial.

If $G$ is not abelian, $\left[G,G\right]$ is a nontrivial normal subgroup of $G$ (verify), so we can form the quotient $G/\left[G,G\right]$. This group is abelian. In fact, there’s a universal property floating around. If we have a group $G$, an abelian group $A$, and a homomorphism $f:G\rightarrow A$, there is a unique homomorphism $s:G/\left[G,G\right]\rightarrow A$ so that $f$ is the composition $G\rightarrow G/\left[G,G\right]\rightarrow^sA$. From this we can see that if $N$ is any normal subgroup of $G$ with $G/N$ abelian, then $\left[G,G\right]\subseteq N$.

We can repeat this construction. Define $G^{(0)}$ to be $G$ itself, and $G^{(n+1)}=\left[G^{(n)},G^{(n)}\right]$. This series of subgroups is extremely important in the classification of groups. If eventually it bottoms out at the trivial subgroup we say the group is “solvable”, which recalls the origins of group theory in finding solutions to polynomial equations. We say a group $P$ where $\left[P,P\right]=P$ is “perfect”. If the series of groups ever hits a perfect group then it stops and will never bottom out, so the original group can’t be solvable.

In 1963, John Thompson and Walter Feit proved that a finite group $G$ with an odd number of elements is solvable. Their paper ran over 250 pages, consuming an entire issue of the Pacific Journal of Mathematics. If you’re up for it, you can read it yourself. This result really got the ball rolling on the project to completely classify all finite simple groups. The classification is considered by many to be complete, spanning tens of thousands of pages in over 500 articles. Since it’s so huge a body, some people think that no one person can check it all over and thus verify the classification. The controversy rages on.

March 14, 2007