11
$\begingroup$

Which finite subsets $S \subset \mathbb{N}$ have the following property : every countable group $G$ embeds into a finitely generated group $\Gamma$ such that $H_i(\Gamma;\mathbb{Z})=0$ for all $i \in S$.

The only positive answer I know here is that $S=\{1\}$ works since every countable group can be embedded into a simple group. I don't know any negative answers.

I'm especially interested in singleton sets $S$ (in particular, $S=\{2\}$ and $S=\{3\}$).

Also, is the question easier if I restrict myself to finitely generated or finitely presentable groups?

$\endgroup$
3
  • $\begingroup$ Does every countable group embed into a finitely generated group (I think for example of the free group with infinitely countable generators) ? If not, what should hold for those groups that doesn't embed accordingly ? $\endgroup$
    – Ralph
    Apr 22, 2011 at 8:04
  • $\begingroup$ @Ralph: Your free group embeds in a free group on two generators. $\endgroup$ Apr 22, 2011 at 12:35
  • $\begingroup$ @Ralph : You can use a sequence of HNN extensions to embed any countable group into a 2 generator group. I'm pretty sure that this is proven in Rotman's book on group theory (and many other places), but I'm not in my office right now so I don't have a reference handy. $\endgroup$ Apr 22, 2011 at 15:53

3 Answers 3

5
$\begingroup$

See Baumslag, G.; Dyer, E.; Miller, C. F. On the integral homology of finitely presented groups. Bull. Amer. Math. Soc. (N.S.) 4 (1981), no. 3, 321–324, and the full version Baumslag, G.; Dyer, E.; Miller, C. F., III On the integral homology of finitely presented groups. Topology 22 (1983), no. 1, 27–46. Lemma 4 in particular.

$\endgroup$
10
$\begingroup$

To add to Mark Sapir's post, the answer is precisely given as Corollary 5.6 of $\textit{The Topology of Discrete Groups}$ by Baumslag, Dyer, Heller (JPAA 16, 1980):

"Every countable group can be embedded in a 7-generator acyclic group."

Thus all possible $S$ work.

$\endgroup$
6
$\begingroup$

Every countable group can be embedded in a countable algebraically closed group, and the latter is acyclic.

It follows that all subsets of $\mathbb N$ have the property you want :)

$\endgroup$
3
  • $\begingroup$ Whoops, I forgot a condition (namely, that the target group is finitely generated). Still, +1! $\endgroup$ Apr 22, 2011 at 4:50
  • $\begingroup$ Ah! Well, I'll leave this up just because the concept of algebraically closed groups is fun :) $\endgroup$ Apr 22, 2011 at 5:15
  • $\begingroup$ I agree! (and have to write some more to get over the character limit) $\endgroup$ Apr 22, 2011 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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