3
$\begingroup$

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 $N_1 \cap N_2$?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.