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 $N_1 \cap N_2$?


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.


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