# First group homology for trivial module

Let $$G$$ be a group and $$A$$ be a $$G$$-module. I use $$H^i(G,A)$$ for group cohomology, and $$H_i(G,A)$$ for group homology. It is well known that if $$A$$ is a trivial module, then $$H^1(G,A) \cong \operatorname{Hom}_{\mathbb{Z}}(G^{\operatorname{ab}}, A)$$ Viewing $$\mathbb{Z}$$ as a trivial $$G$$-module, it is also well known that $$H_1(G,\mathbb{Z}) \cong G^{\operatorname{ab}} \cong \operatorname{Hom}_{\mathbb{Z}}(\mathbb{Z}, G^{\operatorname{ab}})$$ A natural generalization of this second isomorphism is to guess that if $$A$$ is any trivial $$G$$-module, then $$H_1(G,A) \cong \operatorname{Hom}_{\mathbb{Z}}(A,G^{\operatorname{ab}})$$ The conjecture is further "supported" by the parallel with the $$H^1$$ isomorphism.

By mimicking the proof of the $$H_1$$ isomorphism, I was able to prove my conjecture in the case where $$A$$ is finitely generated. (I can add this if there is interest.)

My main question is, is this well known? I cannot find any references to this generalization in the sources I know for group cohomology and homology. Secondly, is the general case ($$A$$ not necessarily finitely generated) true?

The correct statement which is well-known is that $$H_1(G,A)$$ is naturally isomorphic to $$G^{ab}\otimes_\mathbb{Z} A$$, for any trivial $$G$$-module $$A$$. Indeed, this follows from the universal coefficient theorem: the complex whose homology is $$H_*(G,A)$$ is just obtained from the complex whose homology is $$H_*(G,\mathbb{Z})$$ by tensoring with $$A$$, and so we have natural short exact sequences $$0\to H_n(G,\mathbb{Z})\otimes_\mathbb{Z} A \to H_n(G,A)\to \operatorname{Tor}(H_{n-1}(G,\mathbb{Z}),A)\to 0.$$ When $$n=1$$, $$H_{n-1}(G,\mathbb{Z})\cong\mathbb{Z}$$ is torsion-free so we just get an isomorphism $$H_1(G,\mathbb{Z})\otimes_\mathbb{Z} A \to H_1(G,A)$$.
(Note that your proposed generalization doesn't look very natural at all: $$H_1(G,A)$$ is covariant in $$A$$ while $$\operatorname{Hom}_\mathbb{Z}(A,G^{ab})$$ is contravariant in $$A$$. It turns out they actually are isomorphic if $$G$$ is finite and $$A$$ is finitely generated, but the isomorphism is not natural and it is not true more generally.)
• Apologies for "natural," I did not mean it in any categorical sense. How does finite generation of $G$ come into play? I don't think I need it in my proof of $H_1(G,A) \cong \operatorname{Hom}_{\mathbb{Z}}(A,G^{\operatorname{ab}})$ as long as $A$ is finitely generated. – Joshua Ruiter May 6 at 23:25
• Er, actually, $G$ needs to be finite, not just finitely generated. This is just the fact that if $A$ and $B=G^{ab}$ are two abelian groups with $A$ finitely generated and $B$ finite, then $\operatorname{Hom}(A,B)$ and $B\otimes A$ happen to be isomorphic (proof: use the classification of finitely generated abelian groups). – Eric Wofsey May 7 at 1:26
• For a simple counterexample, consider $G=\mathbb{Z}$ and $A=\mathbb{Z}/2\mathbb{Z}$. Then $H_1(G,A)\cong G^{ab}\otimes A\cong \mathbb{Z}/2\mathbb{Z}$ but $\operatorname{Hom}(A,G^{ab})=0$. – Eric Wofsey May 7 at 1:28