MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G_0$ be a finitely generated group, and suppose there are groups $G_i$ and $K_i$ as in the following short exact sequences

$1\to K_i\to G_{i+1}\to G_i\to 1$

with $K_i$ free and nonabelian (you may assume finitely generated), and $G_i$ commutative transitive. (If $a$ is nontrivial and $b$ and $c$ both commute with $a$, then $b$ and $c$ commute.) Does it follow that $\mathrm{rank}(G_i)\to\infty$ as $i\to\infty$? Are there examples of extensions of this sort where the rank doesn't increase?

share|cite|improve this question
What's your definition of rank? – HJRW Oct 3 '10 at 18:17
Minimal number of elements in a generating set. – nolte Oct 3 '10 at 18:30
up vote 11 down vote accepted

I assume that you consider the infinite cyclic group to be free. Then take the free nilpotent group $G_n$ of class $c\gg 1$ with 2 generators. It has an infinite cyclic central subgroup $K_n$, the factor-group $G_{n-1}=G_n/K_n$ is nilpotent and torsion-free, it has an infinite cyclic subgroup $K_{n-1}$, and so on. Every group $G_i$ is 2-generated, the chain can be arbitrary long ($n$ depends on $c$).

Edit 1: Since you do not want to consider $\mathbb Z$ free enough, here is another example. Take the Baumslag-Solitar group $BS(2,3)=\langle a,t | ta^2t^{-1}=a^3\rangle$. It is non-Hopfian, and has a free normal subgroup $K$ such that $BS(2,3)/K$ is isomorphic to $BS(2,3)$. So all $G_i$ are isomorphic to $BS(2,3)$, all $K_i$ are infinitely generated free groups. Is that what you want?

Edit 2: Since you have another condition now, "commutative transitive", then here is another example. Take a non-elementary torsion-free hyperbolic group $G$. It has a free normal subgroup $N$ such that $G/N$ is still non-elementary torsion-free and hyperbolic (that is proved by Olshanskii, and also by several others, including Delzant). You can continue as long as you wish. All $G_i$ will be hyperbolic and torsion-free (hence commutative-transitive), all $K_i$ will be infinitely generated free groups.

Just to anticipate a future change in the formulation of the question: if you really insist that $K_i$ are finitely generated, the question becomes harder and I am not sure the answer is still the same.

Edit 3: Since you want to have an infinite sequence, here is what to do. For every hyperbolic group $G$ with 2 generators, there exists another 2-generated hyperbolic group $G'$ and a free normal subgroup $N \le G'$ such that $G'/N=G$. This can be done using Rips' construction. In the original Rips' construction (and in all modifications) $N$ was not free, because he wanted $N$ to be finitely generated. But if you do not want $N$ to be finitely generated, it is easy to modify Rips' construction to make $N$ free. Using this you can construct your sequence $G_0=G, G_1=G', G_2=G_1', ...$.

Edit 4:In fact Rips' construction does not quite work because the number of generators increases. Certainly if $G$ is free of rank 2, $G'$ cannot be of rank 2. But here is another construction. Take a (torsion-free) lacunary hyperbolic group given by an infinite presentation satisfying a small cancelation condition (see Let $r_1,r_2,...$ be the presentation of $G$. Then $G$ is commutative transitive (it is easily deduced from the fact that $G$ is an inductive limit of hyperbolic groups and surjective homomorphisms). Now the group $G'$ given by the same presentation but without $r_1$ is again lacunary hyperbolic, $G$ is a factor-group of $G'$ over the normal subgroup $N$ generated by $r_1$. It is possible to prove that $N$ is free. Indeed, if some product of conjugates of $r_1$ is equal to 1 in $G$, consider the corresponding van Kampen diagram. The boundary of that diagram has parts labeled by $r_1$ and parts labeled by the conjugators. By Greendlinger lemma, if the diagram has cells, it must have a cell with more than, say, $90\%$ of its boundary common with the boundary of the diagram (take the small cancelation condition $C'(1/300)$). Then more than a half of that part of the boundary must be inside a conjugator, the conjugator can be shortened, and a shorter product of conjugates of $r_1$ is equal to 1 in $G'$. Since $G'$ is lacunary hyperbolic again and satisfies the same small cancelation condition as $G$, we can repeat the construction. Since the presentation is infinite, the process will continue indefinitely.

Edit 5: A more clean way to prove that$N$ is free in Edit 4 is the following. suppose that some product of conjugates of $r_1$ is equal to 1 in $G'$. Consider the corresponding van Kampen diagram $\Delta$. Its boundary label is equal to 1 in the group given by 1 relator $r_1$. Consider the diagram $\Psi$ corresponding to that equality. Now identify the boundaries of $\Psi$ and $\Delta$. We get a diagram over the presentation of $G$ on a sphere: $\Delta$ occupies the northern hemisphere, $\Psi$ occupies the southern hemisphere, and the product of conjugates of $r_1$ labels the equator. Reduce that diagram. Since we can assume that $\Psi$ is reduced, and the $r_1$-cells are only in the south, the $r_1$-cells won't cancel. Hence we shall have a reduced non-empty diagram over the presentation of $G$ on a sphere which is impossible because of the Greendlinger lemma (the boundary of the spherical diagram is empty).

share|cite|improve this answer
I should have been more specific. $K_i$ should be nonabelian. I apologize. – nolte Oct 3 '10 at 18:36
Again, I should learn to be more specific. The $G_i$ should be commutative transitive. Editing to reflect. Thanks. – nolte Oct 3 '10 at 19:08
Regarding Edit 2, I am trying to extend $G_0$ to $G_1$, $G_1$ to $G_2$, etc., not take further and further quotients of $G_0$. I think examples where $\mathrm{rank}(G_i)$ is bounded independently of $i$, with the ranks of $K_i$ not necessarily finite, would be very interesting as well. – nolte Oct 3 '10 at 19:49
Example with hyperbolic groups give you the sequence in the opposite order: $G_n, G_{n-1},...$ where $G=G_n$. Here $G_n/free group=G_{n-1}, G_{n-1}/free group=G_{n-2}$, etc. – Mark Sapir Oct 3 '10 at 19:55
You are supposed to go the other way, and build $G_{n+1}$, $G_{n+2}$, $G_{n+3}$, etc. – nolte Oct 3 '10 at 20:01

Your Answer


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.