Could the number of defining relators of a finitely presented group increase

Question

Do there exist finitely generated groups $G$ and $H$ with following properies:

1. $G=\langle g_1,\dotsc,g_n\rangle$ is not finitely presented, Let $R:=\{r_1,r_2,\dots\}$ be the set of its defining realtors ($r_k=w_k(g_1,\dotsc,g_n)$, $k\in\mathbb N$).
2. $H$ is finitely presented,
3. For each finite subset $S$ of $R$, there is a generating set of $H$ with defining relators as $S$. That's mean if $ S=\{r_{j_1},\dotsc,r_{j_m}\}$, then there is a generating set $\{h_1,\dotsc,h_n\}$ of $H$ with relators $w_{j_i}(h_1,\dotsc,h_n)$, $i=1,\dotsc,m$?

Answer 1

You are in particular asking for a finitely presented group $H$ with infinitely many different surjections onto a finitely generated group $G$. The obvious thing to do is to take $H$ to be non-Hopfian -- the most famous example is $BS(2,3)=\langle a,b\mid b^{-1}a^2ba^{-3}\rangle$ -- which means that there is an epimorphism $\eta:H\to H$ with non-trivial kernel. Now take $K_n=\ker\eta^n$, take $K_\infty=\underrightarrow{\lim} \ker \eta^n$ and take $G=H/K_\infty$.

Note that $G$ must be infinitely presented. Indeed, if not then $K$ is the normal closure of finitely many elements $\{k_i\}$; but then there is some $N$ such that $\{k_i\}\subseteq\ker\eta^N$, whence $\ker{\eta^{N+1}}=\ker{\eta^N}$, which is absurd.

If I've understood your question correctly (I think you probably only want to consider some increasing sequence of finite subsets $S\subseteq R$ that exhaust $R$, not all possible finite subsets $S$ of $R$) then you want to take $A$ to be a generating set for $H$ and, for each $n$, take $A_n=\eta^n(A)$, a generating set for $H/K_n\cong H$. Since $H$ is finitely presented, for each such $n$ there is a finite presentation $\langle A\mid R_n\rangle$ for $H$, where $R_n\in F(A)$ normally generates $K_n$. In particular, the union $R_\infty=\bigcup_nR_n$ normally generates $K_\infty$.