# Embedding groups into groups with some vanishing homology groups

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?

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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 ? –  Ralph Apr 22 '11 at 8:04
@Ralph: Your free group embeds in a free group on two generators. –  Tom Goodwillie Apr 22 '11 at 12:35
@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. –  Andy Putman Apr 22 '11 at 15:53
@Tom, @Andy: Thanks for the info. –  Ralph Apr 22 '11 at 18:33

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.

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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.

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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 :)

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Whoops, I forgot a condition (namely, that the target group is finitely generated). Still, +1! –  Andy Putman Apr 22 '11 at 4:50
Ah! Well, I'll leave this up just because the concept of algebraically closed groups is fun :) –  Mariano Suárez-Alvarez Apr 22 '11 at 5:15
I agree! (and have to write some more to get over the character limit) –  Andy Putman Apr 22 '11 at 5:40