  We describe an algebraic proof of the well-known topological fact that
$\pi_1(SO(n)) \cong Z/2Z$. The fundamental group of $SO(n)$ appears in our
approach as the center of a certain finite group defined by generators and
relations. The latter is a factor group of the braid group $B_n$, obtained by
imposing one additional relation and turns out to be a nontrivial central
extension by $Z/2Z$ of the corresponding group of rotational symmetries of the
hyperoctahedron in dimension $n$.
