Is every normal subgroup of a finitely generated free group a normal closure of a finite set?

Question

Let $G$ be a group, $S\subset G$ a subset, then the smallest normal subgroup of $G$ which contains $S$ is called the normal closure of $S$, and denoted by $S^G$.

My question is, if $G$ is a free group of rank $n$ for $n\in\mathbb N$, then is every normal subgroup of $G$ is a normal closure of a finite sunset? Furthermore, does there exist a bound on the cardinality of such a finite subset?

Answer 1

The question is equivalent to: Is every finitely generated group finitely presented? The answer is No. See SE/547087 for instance. A counterexample is $G=\langle a,b \rangle$ with $N=\langle\langle [a^{-n} b a^n,b] : n \in \mathbb{N} \rangle\rangle$.

Answer 2

As has already been answered, the answer is no in general. But of course finite groups are finitely presented! In particular, if you know that your subgroup is of finite index $k$, then the Nielsen-Schreier index formula states that it is free of rank $$k(n-1)+1,$$

which gives a bound for the cardinality of a normal generating subset.

Answer 3

It is a classic result of B.H.Neumann that there is a continuum (that is, $2^{\aleph_0}$) of pairwise non-isomorphic two-generated groups^†. That is, there is a continuum of groups with presentations of the form $\langle a, b\mid \mathbf{r}\rangle$, where $\mathbf{r}$ is the set of relators (so the set $\mathbf{r}$ normally generates the kernel of $F(a, b) \twoheadrightarrow G\cong\langle a, b;\mathbf{r}\rangle$).

It is relatively easy to convince yourself that if the set $\mathbf{r}$ is finite then there are only countably many such presentations (and so countable many two-generated, finitely presentable groups). What is less obvious is that there are only countably many recursively enumerable sets $\mathbf{r}$. Hence, of the continuum of two-generated groups, most are not even recursively presentable never mind finitely presentable!

I should say that the idea of "recursively presentable groups" seems rather odd, but there is a rather spectacular result of Graham Higman which says that a finitely generated group is recursively presentable if and only if it can be embedded into a finitely presentable group.

(I should say that this all implies basically the same results for free groups of rank greater than two.)

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