where a(r) are positive integers and where Fr and Frij are elements in k such that F1, . . , F, are pairwise distinct. The purpose of this note is to prove this characterization in the case k = C. It is well known that a singularity has one characteristic pair if and only if the link of this singularity is a torus knot. On the other hand, according to Burde and Ziechang ([2]), torus knots are the only knots whose groups have a non-trivial center. We show that the existence of a factorization (1) implies non-triviality of the center of the group of the link of singularity F. First, observe that the fundamental group of the link of the singularity F(T, U) at the right hand side of (1) has as the center an infinite cyclic group. Indeed this singularity is equivalent to Hs = (U FrT)a(r). The complement to its link is fibred over 2[s points] with fibre 5' which represents the generator of the center. (Here and in what follows, Sn denotes a sphere of dimension n). The map is just given by (T, U) -, (T: U). In fact it is easy to show that the fundamental group of this link has a presentation