It has been asked here, whether a retract of a finitely presented group is again finitely presented, i.e. if $G$ is a finitely presented group and $H$ is a group which fits into a split exact sequence $1\rightarrow K\rightarrow G\rightarrow H\rightarrow 1$, is then $H$ again finitely presented? The answer is yes and has several solutions (see loc. cit. and in particular Wall: Finiteness conditions for CW-complexes Lemma 1.3).

A group $G$ is (per def) of type $F_n$ iff there is a model of $K(G,1)$ with a compact $n$-skeleton. $G$ is finitely generated resp. presented iff $G$ is of type $F_1$ resp. $F_2$. So the following question is a generalization of the above: If $H$ is a retract of $G$ and $G$ is of type $F_n$, is then also $H$ of type $F_n$?

  • $\begingroup$ Yes, this is true. Check Geoghegan's book "Topological methods in group theory". If it is not there, I can write a proof at MO. Once you set up the correct (pro) homotopy groups for this problem, the proof becomes essentially the same as for the statement that a retract of an $n$-connected space is also $n$-connected. $\endgroup$ – Misha May 17 '13 at 19:02
  • $\begingroup$ Tanks for your comment, Misha. I checked the book before I posted the question, but I couldn't find it. Maybe I'm just blind. $\endgroup$ – Werner Thumann May 17 '13 at 19:10
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    $\begingroup$ A brief proof is that a group is of type $F_n$ if and only if it is coarsely $n-1$-connected. (I am almost sure Ross has this in his book, if not, we have it in our lectures on geometric group theory.) A Lipschitz retraction applied to the coarse $k$-th homotopy group implies vanishing for the subgroup, by the same argument as for the usual homotopy groups. $\endgroup$ – Misha May 17 '13 at 19:48
  • $\begingroup$ Thank you, Misha. I will have a closer look at the material and argument you proposed. $\endgroup$ – Werner Thumann May 21 '13 at 8:36

I close this question by giving the answer of Misha in the comments: The statement is true because a group is of type $F_n$ if and only if it is coarsely $n-1$-connected. Thanks to Misha for pointing this out.


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