I have this question about the third homology group $H_{3}(G,\mathbb{Z} )$ when $G$ is an abelian group.
I know that there is an injective map of algebras (Kenneth Brown)
$\phi:\bigwedge^{3}(G)\rightarrow H_{3}(G,\mathbb{Z} )$
given by $u\wedge v\wedge w\mapsto [u|v|w]-[u|w|v]+[w|u|v]-[w|v|u]+[v|w|u]-[v|u|w]$
where $[u|v|w]$ is a $3-$cocycle.
My question is: Is there any way to write $H_{3}(G,\mathbb{Z} )$ in term of the algebra $\bigwedge^{3}(G)$?
I was thinking about using the isomorphism of the universal coefficient theorem:
$H_{n}(G,M )\cong (H_{n}(G,\mathbb{Z} )\otimes M)\oplus Tor((H_{n-1}(G,\mathbb{Z} ),M)$
When $M=\mathbb{Z}$ and $n=3$, I do not get anything since the tor part is zero.
Thank you!
-------Update------
I found in this paper
"K3 of a field and the Bloch group" Andrei Suslin
The next lemma. For any abelian group $A$, we have the exact sequence
$0\rightarrow\wedge^{3}(A)\rightarrow H_{3}(A,\mathbb{Z})\rightarrow Tor_{1}^{\mathbb{Z}}(A,A)^{-\sigma}\rightarrow 0$
where $\sigma$ is the involution induced by transposition of arguments, the first homomorphism is induced by the homology multiplication and the second is obtained from the composition
$H_{3}(A,\mathbb{Z})\rightarrow^{\Delta_{A}}H_{3}(A\times A,\mathbb{Z})\rightarrow Tor_{1}^{\mathbb{Z}}(A,A), $
$\Delta_{A}$ being the diagonal map $A\rightarrow A\times A$, $a\mapsto (a,a)$.