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I have an epimorphism $f:B_4\longrightarrow S_4$, from the braid group on 4 strands onto the symmetric group on 4 elements. Is it possible the kernel is not isomorphic to $P_4$, the pure braid group on 4 strands?

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    What is your definition of the pure braid group? Some people _define_ it as the kernel of $B_n\rightarrow S_n$.2010-12-14
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    @Alex Bartel such a *definition* implies we mean the map which tracks the permutations of the strands.@Alex Bartel The kernel is not generally determined by the second and third groups in the short exact sequence. For example, $D_4\times\mathbb Z_2$, where $D_4$ is the dihedral group on four points, maps onto $\mathbb Z_2$ in two ways. One of these has $D_4$ as kernel, one has $\mathbb Z_2^2$.2010-12-14
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    I see, sorry, I didn't realise you were talking about _some_ epimorphism, rather than the standard one.2010-12-14
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    Yeah, just standard pure braid group, possibly non-standard map $B_4\rightarrow S_4$. Don't know if it's possible...where free groups are lurking, strange things can happen.2010-12-14
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    You can feed the presentation of $B_n$ and $S_n$ into GAP, and ask it to enumerate all the epi-morphisms $B_n \to S_n$ up to an automorphism of $S_n$. That'll answer your question in the $n=4$ case. In general I think this is a pretty do-able question by thinking about cycle decompositions in $S_n$ and presentations of $B_n$.2010-12-14
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    FYI, GAP is okay doing this for any epi-morphisms $G \to H$ provided $H$ is finite and $G$ is finitely presented. As your presentations get large GAP uses increasingly massive piles of memory to accomplish the task, but for $B_4$ and $\Sigma_4$ that's well-within the range of modern laptops.2010-12-14
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    Okay, it's a good tip. I wish I knew of a cleaner way, but this is helpful.2010-12-15
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    The GAP program helped, the result is the group must be $P_4$. To satisfy the braid relations and be surjective the Artin generators for $B_4$ must go to transpositions in $S_4$, the first and third being disjoint, the middle one being *adjacent* to both.2010-12-15

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By Ryan Budney's suggestion, I went ahead and proved the general case. When $B_n$ onto $S_n$ the kernel is isomorphic to $P_n$.

A proof sketch is this: relations for the Artin generators in $B_n$ must be satisfied in the image. Relations $b_ib_{i+1}b_i=b_{i+1}b_ib_{i+1}$ can be rewritten as in terms of conjugation so that every $b_i$ has image of a fixed cycle structure.

The relations which impose commutativity of non-adjacent generators imply that non-adjacent generators get sent to permutations with cycles either coincidental or disjoint.

It can be shown that for $n>4$ the images of non-adjacent generators must actually be disjoint: mildly technical, but not difficult. (The $n=4$ case being easily solved by hand or GAP).

Then counting every other odd generator $b_1,b_3,\ldots$, of which there are $\lceil \frac n 2\rceil$ we have that they must be transpositions, since $3\lceil \frac n 2\rceil>n$. Essentially that's it: up to isomorphism of $S_n$ the images of the generators for $B_n$ are the usual transpositions they induce, so kernel is $P_n$.

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    Very nice. . . . .2010-12-16
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    I which book can I fiind this demonstration?2016-01-05
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    Sorry, no, I don't know if this is done anywhere. Why, @MonsieurGalois if I may ask, do you want this? I might be able to recreate the details...2016-01-11