A finitely presented group $G$ is given by a presentation $\langle S,R\rangle$, where $R$ is finite. Show that $S$ is finite.

Question

We were asked to prove this theorem in an exercise. This is what I have thus far:

Suppose $S$ were infinite. Denote the set of symbols of $S$ that do not occur in any relation by $S'$. Then the free group $F_{S'}$ is isomorphic to a subgroup of $G$. Because $R$ is finite and $S$ is infinite, $S'$ must be infinite so $F_{S'}$ has infinite rank.

This is where I'm stuck. I initially thought that this would be enough to conclude that $G$ has no finite presentation, but finitely presented groups can have free groups with infinite rank as subgroups, right?

Any help is appreciated.

Answer 1

A finitely presented group must have a presentation with a finite set of generators (be finitely generated) and a finite set of relations (finitely related). You have shown that if $S$ is infinite and $R$ is fintite (for a given $S$ and $R$), then $G$ can not be finitely presented as it would not be finitely generated. It follows that if $G$ is finitely presented, $S$ must be finite.

Answer 2

Just kill the generators $S^{\prime}$.

More formally: Let $G=\langle S; R\rangle$ be finitely presentable and assume that $R$ is finite. Suppose $S$ is infinite. Then kill every generator which appears (or its inverse appears) in any relator from $R$. You are left with an infinitely generated free group, hence your group $G$ cannot be finitely generated, and hence $G$ is not finitely presentable. This is a contradiction, and so $S$ must be finite.

(If you are comfortable with free products: If $S$ is infinite then define $S_1$ as follows. $$S_1:=\{g\in S; s\text{ or }s^{-1}\text{ occurs in some relator from }R\}$$ Define $S_2$ to be such that $S=S_1\coprod S_2$. Then $G=\langle S_1, S_2; R\rangle$ is the free product $\langle S_1; R\rangle\ast F(S_2)$. Because $S$ is infinite but $S_1$ is finite, $S_2$ must be infinite, hence $F(S_2)$ is infinitely generated free. What the above paragraph does is kill the generators $S_1$ and you are left with $F(S_2)$.)