Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

In my research, I work with certain finitely presented quotients of Coxeter groups. These are the automorphism groups of abstract polytopes, which are combinatorial generalizations of "usual" polytopes. (Essentially, an abstract polytope is an incidence complex.) Now, in this context, there is a useful combinatorial operation that has a nice effect on the automorphism groups. In fact, it's easy to generalize the operation on groups, so I'm curious whether any work has been done with this.

Let $G = \langle X \mid R \rangle$ and $H = \langle X \mid S \rangle$ be finitely presented groups. (I'm not sure that the finiteness of the presentation is essential, but let's assume it for now.) In other words, we have that $G = F(X) / \overline{R}$ and $H = F(X) / \overline{S}$, where $F(X)$ is the free group on $X$ and $\overline{R}$ is the normal closure of $R$ in $F(X)$. Then if $K$ naturally covers $G$ and $H$ (that is, if the identity map on $X$ extends to (surjective) homomorphisms from $K$ to $G$ and $K$ to $H$), we have that $K$ covers the group $F(X) / (\overline{R} \cap \overline{S})$. Similarly, if $G$ and $H$ naturally cover $K$, then the group $F(X) / (\overline{R} \overline{S})$ with presentation $\langle X \mid R \cup S \rangle$ naturally covers $K$ as well.

Therefore, the group $F(X) / (\overline{R} \cap \overline{S})$ is the minimal natural cover of $G$ and $H$, and $F(X) / (\overline{R} \overline{S})$ is the maximal natural quotient of $G$ and $H$. The first group is the fibre product of $G$ and $H$ over the second group.

These seem like such natural operations that I would guess they have been studied before, but I am having trouble finding anything. Any references would be greatly appreciated.

EDIT: Let me expand a little bit. As Mark Sapir points out, I'm essentially looking at the lattice of normal subgroups of F(X). I've looked a little at the general theory of subgroup lattices, but I'm really only interested in normal subgroups of F(X) or other finitely presented groups. Also, I find it difficult to work with the normal subgroups of F(X) directly, whereas it's not too hard to work with the quotients by these normal subgroups via presentations. So I'm hoping to find something that's somewhat more specific than just subgroup lattices; ideally, something that works with presentations.

Here are some examples of the type of questions I'm interested in:

  1. Is there a simple way to write down a presentation for $F(X) / (\overline{R} \cap \overline{S})$ without changing the generators?
  2. Given G and H, what are some conditions on the relations under which $\overline{R} \overline{S} = F(X)$? (This obviously won't be all-inclusive, but some instructive examples would be nice.)
share|improve this question
    
I have updated my answer. –  Mark Sapir Nov 5 '10 at 11:38
add comment

1 Answer

up vote 5 down vote accepted

What you are studying is the lattice of normal subgroups of the free group $F(X)$. The normal subgroup lattices of groups have been studied a lot. For example, this lattice is complete and modular. See the references here.

Update. About your newer, more concrete, questions. Even if $R$ and $S$ are finite, the normal subgroup $\bar R\cap \bar S$ may not be finitely generated. If the membership problem for $\bar R$ and $\bar S$ is decidable, then in principle one can decide the membership problem for the intersection, so you will find a generating set of the intersection - the whole intersection. But I am sure that the problem whether the intersection is finitely generated is undecidable (although I did not think about a proof).

The question whether $\bar R\bar S=F_k$ is the triviality problem which is undecidable (by a theorem of Adian-Rabin) even when $R$ is empty. For example, take any presentation of the trivial group, call a subset of it $R$ and the complement $S$. You get the equality $\bar R\bar S=F_k$. Another way: take a presentation of a finite group of exponent $n$, say, $S_3$ has exponent $6$ and presentation $\langle a,b \mid a^2=b^3=1, aba=b^{-1}\rangle$, and add relations that "kill" the generators, say $a^5=1, b^7=1$. The first set of relations is $R$, the second is $S$.

share|improve this answer
    
Thanks for your updated answer. Regarding your last paragraph: I know there's no general algorithm to determine whether $\overline{R} \overline{S} = F_k$. What I'm interested in is finding simple conditions on R and S that are sufficient to guarantee that $\overline{R} \overline{S} = F_k$. But maybe that question is too broad to have a good answer. Mostly, I was just hoping to find a treatment of the normal subgroup lattice of F(X) from the point of view of presentations of the quotient groups. –  Gabe Cunningham Nov 5 '10 at 14:55
    
@Gabe: Indeed, the question is too broad. I do not think one can say anything about the case $\bar R\bar S=F_k$ except something which would be clearly equivalent to the definition. If you have any specific information about $R$ or $S$, the answer could be more specific as well. For example, $R$ and $S$ may have some geometric meaning, then you can search for a geometric answer. –  Mark Sapir Nov 5 '10 at 15:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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