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.


What about a group $G$ with generators $x_i$ for $i \ge 1$, and relations $x_i^6=1$ and $x_{i+1}^{-1}x_i^2x_{i+1}=x_i^3$ for all $i \ge 1$.

The conjugation relation implies that $x_i$ is trivial for each $i$, so $|G|=1$.

But in the group $G_i$ generated by $x_1,\ldots,x_i$ with relations involving only these generators, the generator $x_i$ does not get killed off, so $|G_i|=6$.

Added later: As suitangi has pointed out, this does not work, because we can take $A'$ and $R'$ to be empty sets to get a presentation of the trivial group. So here is asecond try.

The generators are still $x_i$ with $i \ge 1$. The relations are now $x_i^6=1$ for $i > 1$, and $x_{i+1}^{-1}x_i^2x_{i+1}=x_i^{-3}$ for all $i \ge 1$.

Now we still have $x_i=1$ for all $i > 1$, and then the relation $x_2^{-1}x_1^2x_2=x_1^{-3}$ reduces to $x_1^5=1$, so $|G|=5$.

But the group $G_i$ is now infinite cyclic for $i=1$, of order $3990$ for $i=2$, and the free product $C_5*C_6$ for $i>2$. More generally, for any nonempty subset $A'$ of $A$, if $i$ is maximal with $x_i \in A'$, then $x_i$ has order $6$ in the group defined by taking only those relations involving generators in $A'$, so no subset of those relations can define a group on $A'$ that is isomorphic to $G$.

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    $\begingroup$ What about the finite subsets $A' = \varnothing$, and $R'=\varnothing$? $\endgroup$ – suitangi May 12 '16 at 9:08
  • $\begingroup$ Yes good point! I will delete it and rethink. $\endgroup$ – Derek Holt May 12 '16 at 9:16
  • $\begingroup$ I like this as an answer to the question. It looks like this question has generated a surprising amount of interest, so hopefully it will be helpful to others. I think it'd be good to leave your "first attempt" there too, since it illustrates how you overcame a technical snag. $\endgroup$ – Axesilo May 12 '16 at 13:14

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.

  • $\begingroup$ Great reference, thanks! I notice he talks about the other cases when a presentation can be reduced, etc. $\endgroup$ – Axesilo May 12 '16 at 13:11

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