# Braid groups and representations

I was wondering if $\mathbb{Z} \wr S_n$, where $\mathbb{Z}$ is the usual group of integers, $S_n$ the symmetric group on n elements and $\wr$ the wreath product of two groups, contains the braid group $B_n$.

I was also wondering if $n+1$-dimensional matrices of the form :

$$\begin{bmatrix} 1&0&0 \\\\ 1&0&1 \\\\ 1&1&0 \end{bmatrix}$$ for B2

$$\begin{bmatrix} 1&0&0&0 \\\\ 1&0&1&0 \\\\ 1&1&0&0 \\\\ 0&0&0&1 \end{bmatrix}$$ and $$\begin{bmatrix} 1&0&0&0 \\\\ 0&1&0&0 \\\\ 1&0&0&1 \\\\ 1&0&1&0 \end{bmatrix}$$ for B3

and so on... form a representation of the braid group $B_n$.

A. Popoff

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Checking whether something is a braid group representation is straightforward; you just have to check the braid relations. Have you done that yet? – Qiaochu Yuan Mar 12 '11 at 21:33
Hi... yes I checked and it works fine, but then why bothering with complicated representations like the Burau one, when this one is simpler.... – user8167 Mar 12 '11 at 21:36
Perhaps it is not faithful? (In fact, it shouldn't be faithful. The linearity of the braid groups was until recently an open problem so a faithful linear representation shouldn't be easy to write down.) I guess the Burau representations aren't faithful either... – Qiaochu Yuan Mar 12 '11 at 21:37
I have to admit, I didn't check for faithfulness, it just arose while doing some other stuff with braid groups... – user8167 Mar 12 '11 at 21:39
Your hypothesis is not correct: for n=3, your wreath product is solvable, but $B_3$ contains a non-abelian free subgroup. – user641 Mar 12 '11 at 22:13

It is certainly not true that $\mathbb{Z}\wr S_n$ contains $B_n$, for any $n>2$. Indeed, by definition, $\mathbb{Z}\wr S_n$ has a free abelian subgroup of finite (indeed, $n!$) index.
On the other hand, the kernel of the natural map of pure braid groups $PB_n\to PB_{n-1}$ obtained by forgetting a strand is the free group of rank $n-1$ (this follows from the Birman Exact Sequence, I suppose), so if $n>2$, $PB_n$ (and hence $B_n$) contains a non-abelian free group. If it were contained in $\mathbb{Z}\wr S_n$ then this non-abelian free group would have an abelian subgroup of finite index, which is absurd.
Hi... I have trouble following your proof; wouldn't it be possible that $\mathbb{Z}\wr S_n$ contains both ? Here's what I was trying on GAP: a:=FreeGroup("a"); b:=SymmetricGroup(3); g:=WreathProduct(a,b); u:=g.1*g.3*g.1; v:=g.3*g.1*g.2*g.1*g.2; uvu=vuv; (returns true) – user8167 Mar 13 '11 at 21:02
There is no question that your wreath product contains a quotient of the the braid group ($S_n$ for example!!). – user641 Mar 13 '11 at 21:27
Alexandre: suppose that $|G:K|=n$ and $H$ is a subgroup of $G$. It is an easy exercise to check that $|H:H\cap K|\leq n$. So if $G$ has a subgroup of finite index that's abelian, then $H$ must also have a subgroup of finite index that's abelian. But non-abelian free groups certainly don't have abelian finite-index subgroups, for all sorts of reasons (eg growth). – HJRW Mar 13 '11 at 22:36