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.)

  • $\begingroup$ 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. $\endgroup$ – Joshua Ruiter May 6 at 23:25
  • $\begingroup$ 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). $\endgroup$ – Eric Wofsey May 7 at 1:26
  • $\begingroup$ 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 $\operatorname{Hom}(A,G^{ab})=0$. $\endgroup$ – Eric Wofsey May 7 at 1:28

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