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What are the homology groups of an abelian group? I know there are simple answers in certain cases (e.g. I believe $H_2(A; \mathbb{Z}) = \wedge^2 A$), but it's surprisingly difficult to find any references (at least, one's that are online and free).

I'm mostly interested in integer coefficients, but would also be interested if there are simple answers for other coefficient rings. (e.g. I think $H_n(A; \mathbb{Q}) \cong \wedge^n A \otimes \mathbb{Q}$?)

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  • $\begingroup$ There is a universal coefficient theorem for group homology so other coefficient rings are fairly easy to handle. $\endgroup$ – Dan Rust Jul 26 '14 at 11:35
  • $\begingroup$ Eventhough the result is "known", I don't think there's a complete description in literature. The integral (co)homology is quite complicated when the order of $A$ gets larger. On the other hand, the mod $p$ (co)homology are easy and generally skipped from any standard texts. $\endgroup$ – Quang Hoang Jul 26 '14 at 12:28
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While I do not know any complete description, there are several quite general results in Section V.6 of the book Cohomology of groups of K. S. Brown. Here is a summary. In what follows, $R$ is a commutative ring with unit and $G$ is an abelian group.

  • There is a graded homomorphism $ \psi \colon \Lambda^*(G \otimes R) \to H_*(G;R)$ which is natural in $G$.

  • The map $\psi$ is always an isomorphism in degrees $0$ and $1$. If $R$ is a principal ideal domain of characteristic $0$ (meaning its fraction field has characteristic $0$), then it is also an isomorphism in degree $2$.

  • If $R$ is a principal ideal domain, $\psi$ is injective. Moreover, if $G$ is finitely generated it is a split injection.

  • If $R$ is a principal ideal domain and every prime $p$ such that $G$ has $p$-torsion is invertible in $R$, then $\psi$ is an isomorphism. In particular, it is an isomorphism if:

a) $ R = \mathbb{Q}$

b) $ R = \mathbb{Z}/p$ and $G$ is $p$-torsion-free.

c) $ R = \mathbb{Z}$ and $G$ is torsion-free.

Finally, it is also shown that if $G$ has $p$-torsion, then $$H_*(G;\mathbb{Z}/p) \cong \Lambda^*(G \otimes \mathbb{Z}/p) \otimes \Gamma(\text{Tor}(G,\mathbb{Z}/p))$$ where $\Gamma(S)$ denotes the divided polynomial algebra over $S$.

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  • $\begingroup$ No, actually that part about R is not needed, I'll edit it $\endgroup$ – Goa'uld May 19 '16 at 22:21

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