I am asked to prove the following statement:
Suppose $G$ is a group with a finite presentation $\langle S\mid R\rangle$ and that $\langle S \mid R'\rangle$ is a second presentation of $G$ with the same set of generators. Show that there exists a finite subset $R''$ of $R'$ such that $\langle S \mid R''\rangle$ is still a presentation of $G$.
My attempt: Because the generators are identical in $\langle S \mid R\rangle$ and $\langle S \mid R'\rangle$, every relation in $R$ has to follow from relations in $R'$. Since $R$ is finite, we have then a finite number of relations in $R'$ from which the relations in $R$ follow. This finite set in $R'$ is the subset $R''$ we are looking for.
I did assume here that every relation in $R$ follows from a finite number of relations in $R'$, but I'm not sure if this is correct.
Thanks in advance!
[This is my first post here, I hope it's all up to standard]
Yes your argument is correct but informal.
For formal arguments with presentations, it is better to use relators rather than relations. A relation $w_1=w_2$ corresponds to a relator $w_1w_2^{-1}$ i.e. a word that evaluates to the trivial element of the group.
Saying that every relator $w$ in $R$ follows from the relators of $R'$ means that $w$ can be written (in the free group $F(S)$ on $S$) as a product of conjugates of elements of $R'$. That is, for some $n \ge 0$, there exist elements $g_1,\ldots, g_n \in F(S)$ and $w_1,\ldots,w_n \in R'$ such that $$w =_{F(S)} w_1^{g_1} w_2^{g_2} \cdots w_n^{g_n},$$ where $w_i^{g_i} := g_i^{-1}w_ig_i$. But now it is evident that $w$ follows from the finite set of relators $\{w_1,\ldots,w_n\}$ of $R'$.