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 $B_n$.
Thank you for your help
A. Popoff
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.
[Note: this answer is essentially the same as Steve D's comment.]