Let $A$ be an abelian group and $p$ be a prime. If $p\ne 2,$ there is a very nice functorial description of the homology algebra $H_*(A,\mathbb Z/p):$ $$H_*(A,\mathbb Z/p)\cong \Lambda^*(A/p)\otimes \Gamma^{*/2}({}_pA),$$ where $\Lambda^*$ denotes the exterior algebra, $\Gamma^*$ denotes the divided power algebra and ${}_pA$ denotes $p$-torsion of $A$ (see Theorem 6.6 in Chapter V of Brown's "Cohomology of groups"). I like this isomorphism (like = it is useful for me) because

1) It is functorial.

2) The right part depends only on $A/p$ and ${}_pA$.

There is also such an isomorphism for $p=2$ but it is not functorial. And I do not like it because of this.

For any prime $p$ (including $p=2$) there is a short exact sequence of functors $$0\longrightarrow \Lambda^2(A/p) \longrightarrow H_2(A,\mathbb Z/p)\longrightarrow {}_pA \longrightarrow 0.$$ I like this short exact sequence because it gives a functorial 'description' of $H_2(A,\mathbb Z/2)$ in terms of $A/2$ and ${}_2A.$ The word 'description' here in a weak sense because it is not an isomorphism. But it is ok for me.

Question: Is there a functorial 'description' of $H_n(A,\mathbb Z/2)$ in terms of $A/2$ and ${}_2A$? Here the word 'description' can be in some weak sense.


I only gather answers for special cases. There is always a natural (functorial) injection $\bigwedge^\ast(A/2) \to H_\ast(A,\mathbb Z/2)$, and furthermore it's a natural isomorphism if ${}_2A=0$. But you must know this already, it's Theorem 6.4 of Brown's book.

Some necessary material is found in the following papers (in French) of Cartan, which Ken Brown references precisely at the end of Theorem 6.6. I'm not completely sure how much it gets you, I am not the best at French translations, but it does at the least mention a description when ${}_2A=A$.
Namely, the two papers which concern $p$ odd and $p=2$ respectively:
Détermination des algèbres $H_*(\pi, n; Z_p)$ et $H^*(\pi, n ;Z_p)$, $p$ premier impair
Détermination des algèbres $H_*(\pi, n; Z_2)$ et $H^*(\pi, n; Z_2)$ ; groupes stables modulo $p$

  • $\begingroup$ I know this material (there is a translation in Russian and I know the Russian variant but I think they are the same). As far as I understand, in the case of $p=2$ he studies $H_*(\pi,n,\mathbb Z/2)$ only for $n>1$ but I need $n=1.$ $\endgroup$ – Sergei Ivanov Jan 18 '17 at 9:23
  • $\begingroup$ True, there is a desired result (in the $p=2$ paper) for $n=1$ when $A$ (Cartan's $\pi$) is all 2-torsion, but I guess you really prefer ${}_2A \ne A$. $\endgroup$ – Chris Gerig Jan 18 '17 at 10:18
  • $\begingroup$ Yes, I need a result for all abelian groups. $\endgroup$ – Sergei Ivanov Jan 18 '17 at 10:28

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