I recently had the following problem as a homework assignment.
Let $S$ be a finite generating set of a group $G$. Suppose that G is finitely presenting over $S$, and let $R$ be any set of relations over $S$ which also presents $G$; that is, $G = \langle S ∣ R \rangle$. Then a finite subset of $R$ presents $G$ over $S$.
After clarifying with my teacher, I determined that the question should be phrased in the following way.
Let $S$ be finite, and suppose $G = F_S/\langle \langle T \rangle \rangle$ for some finite set $T$ of relators. Suppose $\langle \langle R \rangle \rangle = \langle \langle T \rangle \rangle$. Then there exists a finite subset $R' \subseteq R$ such that $\langle \langle R' \rangle \rangle = \langle \langle T \rangle \rangle$.
This is not too hard to prove. However, I had originally interpreted the question differently, and spent many hours scratching my head trying to solve a different problem. My original interpretation was the following.
Let $\langle S ∣ T \rangle$ be a finite presentation of $G$; that is, $G \cong F_S / \langle \langle T \rangle \rangle$ and $S,T$ are finite. Suppose $\langle S ∣ R \rangle$ is another presentation of $G$; that is, $G \cong F_S/ \langle \langle R \rangle \rangle$. Then there exists a finite subset $R' \subseteq R$ such that $G \cong F_S/ \langle \langle R' \rangle \rangle$.
My question is: If we interpret the problem my way, is it a true statement? That is, is the above statement in bold true? I've tried to solve it for a while to no avail, so I think it's probably false. I don't have an idea about how to construct a counterexample, though.