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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

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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?2011-03-12
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    Hi... yes I checked and it works fine, but then why bothering with complicated representations like the Burau one, when this one is simpler....2011-03-12
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    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...2011-03-12
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    I have to admit, I didn't check for faithfulness, it just arose while doing some other stuff with braid groups...2011-03-12
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    @Alexandre: as you surely expect, checking faithfulness is a difficult thing!2011-03-12
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    I'll have to ask an expert then... by the way, I was thinking that if my hypothesis concerning the wreath product is true, then Zm () Sn could be considered as "constrained" braid groups where the generators of the braid group have a limited number m of twists... does that make any sense ? Has this ever been studied ?2011-03-12
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    @Alexandre: regarding that, look at http://mathoverflow.net/questions/48849/transpositions-of-order-three: Gjergji mentions a paper which studies such things.2011-03-12
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    (By the way, I don't really see what matrices you have in mind. Your explanation that they are obtained by «swapping pairs of columns and adding» does not tell me anything... Could you write down explicitly a couple more?)2011-03-12
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    Your hypothesis is not correct: for n=3, your wreath product is solvable, but $B_3$ contains a non-abelian free subgroup.2011-03-12
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    Would be :$$\begin{bmatrix} 1&0&0 \\\\ 1&0&1 \\\\ 1&1&0 \end{bmatrix}$$ 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...2011-03-12
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    @Alexandre: if you add them to the body of the question, it is much easier to find them :)2011-03-12
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    Sorry for that... I corrected it, hope it will make more sense...2011-03-12
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    @Steve ... thanks for the answer. I had found a couple of elements from the wreath product that were satisfying the braid relations, but I wasn't looking at the big picture (this whole thing comes from rather applied stuff...)2011-03-12
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    @user8167: "but then why bothering with complicated representations like the Burau one" -- because different representations highlight different properties of the group being represented. If you were just interested in having a single representation, you could always just take the trivial one.2014-01-29

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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.]

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    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; u*v*u=v*u*v; (returns true)2011-03-13
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    There is no question that your wreath product contains a **quotient** of the the braid group ($S_n$ for example!!).2011-03-13
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    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).2011-03-13
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    As Steve suggests, it is in general easy to check if a given map defines a representation (you just have to check the relations), and very difficult to check whether a given representation is faithful. Indeed, in the following paper, Bridson and I exhibit a sequence of representations with the property that no algorithm determines which are faithful: http://arxiv.org/abs/1003.5117 .2011-03-13