Let $G$ be a group satisfying $H_1(G;\mathbb{Z})$ is free abelian group and $H_i(G;\mathbb{Z})=0$ for $i\geq 2$.
Is it true that $G$ is free group?
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2 Answers
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Perfect, locally free groups exist. Such a thing has vanishing $H_1(G,\mathbb Z)$, has $H_p(G,M)=0$ for all $p\geq2$ and all $M$, and is not free.
A. J. Berrick constructs an explicit example here. If you prefer an example where $H_1(G,\mathbb Z)$ is free and non-zero, just consider the free product of a perfect, locally free group with a non-trivial free group.
answered Apr 22, 2011 at 4:31
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Groups with $H_i(G;\mathbb{Z})=0$ for all $i>0$ are called acyclic groups. There are lots of them. As Mariano Suarez-Alvarez says, any perfect locally-free group is acyclic. There are also finitely-presented examples. The Higman group $G=\langle a,b,c,d\,\,:\,\,a^b=a^2,\,b^c=b^2,\, c^d=c^2,\, d^a=d^2\rangle$ is an example. These groups play a vital role in the proof of the Kan-Thurston theorem: see for example the article `The topology of discrete groups' by Baumslag, Dyer and Heller (J Pure Appl Alg 16 (1980) 1-47) which discusses the Higman group.
