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?



On the group homology
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](http://www.math.nus.edu.sg/~matberic/ch1.ps). 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.


