Braid Group, B_4->> S_4 onto, do I know kernel is P_4, pure braid group? 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? 
 A: 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 $P_n$.
