Is every normal subgroup of a finitely generated free group a normal closure of a finite set? 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?
 A: 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 $N=\langle\langle [a^{-n} b a^n,b] : n \in \mathbb{N} \rangle\rangle$.
A: 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,)
A: 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.
