This may be a silly question as I do not know much about presentations of groups.
Let $\langle S\mid R\rangle $ be a finite presentation of a group. Is it always possible to get a finite presentation $\langle S’\mid R’\rangle$ of an isomorphic group with $|S’|=|R’|$, at least if $|R|>0$?
I know that if $|S|>|R|$ I can simply double some relations until I get the equality, but it is not clear to me if I can do something analogous when $|S|<|R|$, since when adding a new generator I must also add a new relation.
Best Answer
The answer is no. The smallest example is the presentation of the Klein $4$-group $$\langle x,y \mid x^2 = y^2 = 1, xy=yx\rangle.$$ In any presentation $\langle S \mid R \rangle$ of this group we have $|R| \ge |S| + 1$.
I don't know whether there is an elementary proof of that. It follows from a more general result that, for a finite presentation $\langle S \mid R \rangle$ of a finite group $G$, we have $|R| \ge |S| + |d(M(G))|$, where $d(M(G))$ is the smallest number of generators of the Schur Multiplier $M(G)$ of $G$.
$M(G)$ is the (unique up to isomorphism) largest group $M$ for which there is a group $E$ with $M \le Z(E) \cap [E,E]$ and $E/M \cong G$. For the Klein 4-group, we have $|M(G)| = 2$, where, for the covering group $E$ we can take either $D_8$ or $Q_8$.
Interestingly, $Q_8$ does have a balanced presentation (i.e. $|S|=|R|$), namely $\langle x,y \mid x^2=y^2, y^{-1}xy=x^3\rangle$.
There is a lot of literature on this topic. Search for balanced presentations, and the deficiency of a presentation.