If w* is any element of P*, let N* be the normal subgroup which the word R* = w*aw*-la-l generates in P*. (P* is the fundamental group of the torus of genus g.) Does the factor group P*IN* have an element of finite order (does it have torsion)? The question arose in connection with the Poincare conjecture [5], and is answered here in the negative. The method of proof grew out of the work of Magnus. For inspiration thanks are due and gratefully offered to Gilbert Baumslag. To tackle the problem, the first known fact to observe is that an extension of a torsion-free group by the free cyclic group is always torsion-free. Let H* be the normal subgroup in P*IN* generated by the symbols b, x1, .., X2g2. Because the words G* and R* above, which define P*, respectively P*IN*, contain the symbol a to exponent sum zero, the group P*IN* is extension of the group H* by the free cyclic group generated by a. One needs, then, to show that H* is torsion-free. The next known fact to observe is that a generalized free product of torsion-free groups, with free amalgamated subgroups, is again torsionfree. A series of lemmas pave the way to a presentation of H*-given in the theorem of this paper-as such a free product. To start with, I define (an infinite set of) new generators xij for P* and rewrite the words G* and R* in their terms (as two infinite sets Gj and Rj). Then I replace the new words by equivalent ones (namely G5 and Aj), defined in Lemma 3, that suit my purpose. The salient properties of the new words are established in the last lemmas. Define, for integral k, the following new symbols: