Normal subgroups of braid groups There is a lot of normal subgroups in braid groups (for example there is an action of $B_n$ on unitriangular bilinear forms on $R^n$ over arbitrary commutative ring $R$: $b_i\colon e_j\mapsto e_j$, $j\ne i, i+1$, $b_i\colon e_{i+1}\mapsto e_i$, $b_i\colon e_i\mapsto e_{i+1}-(e_i, e_{i+1})e_i$ and set $R=\mathbb Z_m$). 
Is there any classification (with no conditions on terms of classification) of normal subgroups of $B_n$?
update: the interesting case of this classification for me is the case of finite index normal subgroups. For example I don't even know what is the kernel of the action described above. The answer may be useful in algebraic geometry (see A.L. Gorodentsev, TRANSFORMATIONS OF EXCEPTIONAL BUNDLES ON $\mathbb P^n$)
 A: It depends what you mean by classification.  Let's start with the example of the free group of rank $n$, $F_n$.  The normal subgroups of $F_n$ correspond to $n$-generator groups with marked generating sets, which of course is a hopelessly complicated set.  So in some sense, the normal subgroups of $F_n$ are not classifiable.  On the other hand, we have:
Greenberg's Theorem: If $N\lhd F_n$ then precisely one of the following holds.

*

*$N$ is trivial.

*$N$ is of finite index in $F_n$.

*$N$ is infinitely generated.

Of course, the second option is pretty complicated, and the third even more so.  But you might call this a sort of 'classification'.
Now, the pure braid group $PB_n$ admits a natural map onto $PB_{n-1}$, by forgetting a strand.  As observed above, $B_3$ surjects $\mathbb{Z}/2*\mathbb{Z}/3$, which has a free subgroup of finite index.  So every $B_n$ virtually surjects a free group.  This indicates that the normal subgroups of $B_n$ are almost as complicated as the normal subgroups of $F_m$.

The OP has now indicated that he is interested in the normal subgroups of finite index.  The answer above shows that we at least have the following:

Let $n\geq 3$.  For every finite group $Q$ there is a subgroup $N\lhd B_n$ of finite index with $Q\hookrightarrow B_n/N$.

That said, there are probably many more normal subgroups of finite index.  Here's a question in this direction to which I don't know the answer.

Are braid groups residually finite simple?


Further edit:
I think this is what you're looking for.  Here's an article by McReynolds, giving a proof that pure braid groups have the congruence subgroup property.  This can be thought of as a classification of the subgroups of finite index.
A: If you're willing to consider "any classification" you might consider the covering space theory of configuration spaces.
Equivalence classes of regular (finite) connected coverings of the configuration space $C_{n}(\mathbb{R}^{2})$ of n unordered points in $\mathbb{R}^{2}$ completely classify (finite index) normal subgroups of $B_n=B_n(\mathbb{R}^{2})=\pi_{1}(C_{n}(\mathbb{R}^{2}))$ via the usual Galois correspondence.
This may not be very enlightening but may offer a practical approach to studying the lattice of normal subgroups of $B_n$. Hansen's polynomial covering theory might also help to identify interesting subgroups of $B_n$.
A: This does not seem like a realistic hope. When you say a lot of normal subgroups we can be more precise. A group is conjugacy separable if given two elements which are not conjugate then there is a finite quotient group such that the images of the two elements are not conjugate in this finite group. This is stronger than residually finite which is the special case where one of the group elements is the identity.
The point is I think the braid groups are conjugacy seperable, For $B_3$ we have the short exact sequence $$0 \rightarrow \mathbb{Z} \rightarrow B_3 \rightarrow \mathbb{Z}_2 * \mathbb{Z}_3 \rightarrow 0 $$
and the free product of two cyclic groups is conjugacy seperable.
I don't know the argument for $B_n$.
Edit (in response to comments). The result that braid groups are residually finite is stated in
Magnus (Property III) and
Rolfsen (Theorem 2.5)
. The argument is that free groups are residually finite and the automorphism group of a residually finite group is residually finite.
