A question about the free resolution of an abelian group.

Question

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?

In my very partial and perhaps totally outdated understanding, resolutions typically consist of modules. So $H$ is being treated no so much as an abelian group here, as a $\Bbb Z$-module. — Lubin, Jul 24 '17 at 17:08

score 0 · Answer 1 · answered Jul 24 '17 at 16:58

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It will also be true for non abelian group since a subgroup of a non abelian free group is also free.

answered Jul 24 '17 at 16:58

Tsemo Aristide

89,587

So you're saying that the free resolution of all groups contains only two non-zero free groups, $F_0$ and $F_1$? – Jul 24 '17 at 17:03
the argument is true for non commutative free groups. – Tsemo Aristide Jul 24 '17 at 17:04

A question about the free resolution of an abelian group.

1 Answers1