If $G$ is finitely generated and $G/N$ is finitely presented , then $N$ is normal closure of finite set.

Question

Suppose $G$ is a finitely generated group and $N$ is a normal subgroup of $G$ such that $G/N$ is a finitely presented .Then show that $N$ is a normal closure of some finite set of $G$.

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[My attempt]

Let $\pi : G \to G/N$ be a natural quotient map.

Since $G/N$ is a finitely presented, there exist a exact sequence of group $$ 1 \; \to \; \langle R\rangle \; \to \; F(S) \; \stackrel{f}{\to} \; G/N \; \to \; 1 $$ where, $F(S)$ is free group on $S$ and $R$ is a relation set. ($F$, $S$ are finite) ($\langle \cdot\rangle$: normal closure notation)

Since $F(S)$ is free, we construct map $\phi: F(S) \to G $ such that $f=\pi \; \circ \; \phi$.

Therefore, we obtain $\phi(R) \subset \ker\pi = N$ so $\langle \phi(R)\rangle \;\subseteq N$.

I'm stuck here.

How can I prove that $ N \subseteq \; \langle \phi(R)\rangle $ ?

Thank you for your attention.

Answer 1

I believe there's no escaping the result, which I learned from the paper

B. H. Neumann, Some remarks on infinite groups, J. London Math. Soc. 12 (1937), 120–127,

but which I understand is a basic fact in universal algebra, that

if a group is finitely presented with respect to a particular finite set of generators, then it is so with respect to any finite set of generators.

Suppose $G$ is generated by the finite set $S$. Let $F$ be the free group on $S$. Then we have natural surjective morphisms $$ F \overset{f}{\to} G \overset{\pi}{\to} G/N. $$ Let $K$ the kernel of the composite map. By the above result, there is a finite subset $T$ of $F$ such that $K$ is the normal closure in $F$ of $T$. Now the image of $K$ under $f$ is $N$, so $N$ is the normal closure in $G$ of the finite set $f(T)$ .

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Addendum Contrary to my expectations, you can indeed avoid the result I quoted, see Derek Holt's comment below.