Consider the Kernel $K_n$ of the natural group homomorphism from the $n$-th braid group to the symmetric group. Then one can delete the $m$-th braid. This is a well defined homomorphism $d_m:K_n\rightarrow K_{n-1}$. So is there for every $n\in \mathbb{N}$ a braid $1\neq b\in K_n$ with $d_m(b)=0$ for all $m$.

This is clearly true for $n=2$, as $K_1$ is trivial and it is also true for $n=2$ (The "standard" braid does the job). What about higher $n$. Is there a nice construction, that works for every $n$ ?