TITLE: Can any presentation of a finitely presented group be reduced to a finite one?
QUESTION [12 upvotes]: Suppose $G = \langle x_1, \ldots, x_n \mid p_1, \ldots, p_m \rangle$ is a finitely presented group, and let $\langle A \mid R \rangle$ be another presentation of $G$, with $A$ and $R$ possibly infinite.  Do there always exist finite subsets $A' \subset A$, $R' \subset R$ such that $G = \langle A' \mid R' \rangle$?
I feel like the answer should be "yes."  Here's my idea: we can write each $x_i$ as a product of finitely many $a \in A$ (and their inverses); denote by $A_i$ this finite set of $a$'s.  Then the finite set $A_1 \cup \cdots \cup A_n$ generates $G$.
Similarly, each relator $p_i$ can be derived from a finite set of relators $R_i \subset R$.  Here's my problem, though: how do I know that any relator $w$ in the letters $A_1 \cup \cdots \cup A_n$ can be reduced using these $R_i$?  Not every such $w$ blocks off into $x_i$-chunks.

REPLY [5 votes]: Not every presentation of a finitely presented group can be reduced to a finite one.
In his highly advisable notes on Geometric Group theory, Charles F. Miller III provides the presentation:
\begin{equation}
\langle a,b,c_o,c_1,c_2,\ldots  \mid a^4=1, b^3 =1, c_0^{-1} b c_0 = a^2, c_1^{-1} c_0 c_1 =b, c_2^{-1} c_1 c_2 = c_0 , c_3^{-1}c_2 c_3 = c_1,\ldots\rangle
\end{equation}
which defines a cyclic group of order $2$, but any of its finite subpresentations define a group of order $3$, a group of order $4$, or an infinite group.