Braid Groups Mapped to Symmetric Groups How can I construct five elements in terms of the Braid Generators $\sigma_1 \sigma_2$ that are in the kernel of the homomorphism from the braid group on three strands to the symmetric group on three letters? I tried, but my braids keep turning out to be the identity braid.
 A: Hint: Every element $\delta$ of the kernel of the map from the braid group $B_n$ to the symmetric group $S_n$ is such that every strand of $\delta$ begins and ends in the same position (why?).
A: By definition, the kernel of the canonical homomorphism $B_n\to S_n$ is the pure braid group on $n$ letters $P_n$, which is the group of all braids whose strings start and end on the same point of the disk (they don't permute the end points of the strings). Given this, you should be able to see that elements like $\sigma_i^2$ or $(\sigma_1\sigma_2\sigma_1\sigma_2)^3$ are mapped to the trivial permutation because they do not permute the ends of the strings.
The best way to create elements in the pure braid group is to pick an arbitrary non-trivial element $g\in B_n$, see what element $g$ gets mapped to under the canonical homomorphism to $S_n$ and then, because $S_n$ is finite, the image of $g$ must have finite order, so taking some power $g^k$ will give an element of the pure braid group. Because $B_n$ has no torsion elements, $g^k$ will never be trivial when $g$ is non-trivial.
