This should probably be an easy question, but I don't know how to answer it: Suppose G is a finitely generated presentable group. Suppose a is the absolute minimum of the sizes of all generating sets for G and b is the absolute minimum of the number of relations over all presentations of G. Question: Is it necessary that G has a presentation that simultaneously has a generators and b relations?
The case b = 0 is just the fact that a free group cannot be generated by fewer elements than its free rank.
The problem could probably be interpreted in terms of CW-complexes (where the generators give rise to 1-cells and the relators give rise to 2-cells) but, because of my lack of familiarity with CW-complexes, I don't immediately see how to use these to solve the problem.
It also seems to be related to the notion of "deficiency" of a group, which is the (maximum possible over all presentations) difference #generators – #relations (under the opposite sign convention, the minimum possible difference #relations – #generators).
Best Answer
A stronger question, is the deficiency of $G$ realized for a presentation with the minimal number of generators (rank($G$))? This question is asked in a paper of Rapaport, and proved to be true for nilpotent and 1-relator groups.
Addendum:
The question appears as Question 2, p. 2, of a book by Gruenberg. Lubotzky has answered the analogous question affirmatively in the category of profinite groups (Corollary 2.5). (Note though that this is in the category of profinite presentations, so it does not imply an affirmative answer even for finite groups).