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?
Braid Group, B_4->> S_4 onto, do I know kernel is P_4, pure braid group?
-
0The 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
1 Answers
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 $P_n$.
-
0Sorry, 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