On pg 195 of the book "Algebraic Topology", Hatcher says the following:
Every abelian group $H$ has a free resolution of the form $0\to F_1\to F_0\to H\to 0$, with $F_i=0$ for $i>1$.
He justifies this by saying that the kernel of the map $F_0\to H$ is a subgroup of the free group $F_0$, and hence also free. This would imply that $F_1$ is free, and hence the map $F_1\to F_0$ would have a trivial kernel.
Why isn't this true for non-abelian groups? Wouldn't the same argument be applicable even there?
It will also be true for non abelian group since a subgroup of a non abelian free group is also free.