Let $G = \left< a\_1,b\_1, ... , a\_g, b\_g | [a\_1,b\_1] \cdots [a\_g,b\_g] \right>$ be the fundamental group of a surface of genus $g$. Let $N\_1$ and $N\_2$ be two normal subgroups of $G$ that are described explicitly as the normal closures of two finite sets of words say $H\_1 = \left< x\_1,...,x\_n \right> ^N$ and $H\_2 = \left< y\_1,...,y\_m \right>^N$. 


Is it always true that $N\_1 \cap N\_2$ will be finitely normally generated? Is there a procedure for finding a normal generating set for $N\_1 \cap N\_2$? 



Normal generating set for the intersection of two normal subgroups of a surface group
I believe that the intersection is not always finitely generated as a normal subgroup. The factor-group $F\_g$ of $G$ over the normal subgroup $M=\langle a\_1,...,a\_g\rangle^N$ is freely generated by the images of $b\_1,...,b\_g$ which we shall denote by the same letters: $b\_1,...,b\_g$. Since $M$ is finitely normally generated, it is enough to consider your question for the free group $F\_g$. 
Suppose that $g>2$. Let $N\_1=\langle b\_1\rangle^N, N\_2=\langle b\_2\rangle^N$. Then $N\_i$ consists of all words in $F$ which become trivial after removing $b\_i$, $i=1,2$. The intersection of $N\_1$ and $N\_2$ consists of all words which become trivial both after removing $b\_1$ entrees and after removing $b\_2$ entrees. I do not have time to check it now, but I think that $N\_1 \cap N\_2$ is not finitely generated as a normal subgroup. It is generated as a normal subgroup by all commutators $[b\_1, b\_2^{\pm u}]$ where $u$ is an arbitrary word in the generators of $F\_g$. The factor group $F\_g/(N\_1\cap N\_2)$ is not finitely presented. I will elaborate on that when I have time. Or perhaps somebody can do that for me. 


**Update.** Here is the proof that $N=N\_1\cap N\_2$ is not finitely generated as a normal subgroup. As I wrote above $N$ is normally generated by all commutators $[b\_1, b\_2^u]$ where $u$ is an arbitrary word in $b\_1,,,,,b\_g$. For simplicity assume that $g=3$ (the general case is not significantly different) and denote $b\_1, b\_2, b\_3$ by $a,b,c$. I claim that $N$ is normally generated by commutators $[a, b^{\pm c^n}]$, $n\in \mathbb{Z}$. Clearly each of these commutators is in $N$. We need to show that all relations $[a,b^u]=1$ follow from relations $[a, b^{\pm c^n}]=1$. So suppose that all relations $[a, b^{\pm c^n}]=1$ hold in some group $G=\langle a,b,c\rangle$. In particular, $ab=ba$. Consider a relation $[a, b^u]=1$ for some word $u$. Suppose first that $u$ does not contain the letter $a^{\pm 1}$. Since $b^u=b^{b^ku}$ for every $k$, we can assume that the total degree of $b$ in $u$ is 0. Therefore $u^{-1}$ is a product of conjugates of the form $b^{\pm c^n}$. Note that the relation $[a, b^{\pm u}]=1$ is equivalent to the relation $[a^{u^{-1}},b^{\pm 1}]=1$. Since by our assumption $b^{\pm c^n}$ commutes with $a$, the relation $[a^u,b^{\pm 1}]=1$ follows. 


Now suppose that $u$ contains occurrences of $a^{\pm 1}$. Since $b$ and $a$ commute, the relation $[a, b^u]=1$ is equivalent to $[a, b^{a^ku}]=1$ for every $k$, hence we can assume that the total degree of $a$ in $u$ is 0. Hence $u$ is a product of conjugates of the form $a^{\pm v}$ where $v$ is a word in $b,c$. By what we prove in the previous paragraph, $a^{\pm v}$ commutes with $b$ (since $v$ does not have occurrences of $a^{\pm 1}$. Hence $b^u=b$ modulo the relations $[a, b^{c^n}]=1$ and we are done. 


We need to show that the factor group $G=F\_3/N$ is not finitely presented. By our claim, $G$ has the presentation $\langle a,b,c \mid [a, b^{c^n}]=1, n\in \mathbb{Z}\rangle$. But this is the HNN double (extension with centralizer) of the free group $F\_3$ along the infinitely generated subgroup $H=\langle b^{c^n}, n\in \mathbb{Z}\rangle$, so $G$ is indeed not finitely presented.


