Second homology group of perfect group knowing its cohomology

Question

Is it true that, if $G$ is a finitely generated (infinite) perfect group with $H^2(G)\simeq \mathbb{Z} \times \mathbb{Z}$, then $H_2(G) \simeq \mathbb{Z} \times \mathbb{Z}$?

My thoughts:

By the Universal Coefficients Theorem, we know that $H_2(G)$ has rank 2, so it is isomorphic to $ \mathbb{Z}^2 \oplus T$, where $T$ is the torsion part. Moreover, since $G$ is perfect, we have that $H_1(G) \simeq 0$, and so for every $\mathbb{Z}$-module $A$ thanks to the Universal Coefficients Theorem we have an isomorphism $H^2(G; A) \simeq {\rm Hom}_{\mathbb{Z}}(H_2(G), A)$. If $T \neq 0$, there exists a prime $p$ such that $H^2(G, \mathbb{Z}_p) \simeq {\rm Hom}_{\mathbb{Z}}(H_2(G), \mathbb{Z}_p)\neq 0$.

I don't know how to procede from this point. Maybe I need some additional hypothesis?

Thanks in advance.

(Where not specified, I work with integer coefficients).

Answer 1

Since $\mathbb Z/p$ is finitely generated, you actually have another "universal coefficients" formula, for $H^2(G,\mathbb Z/p)$ :

following what is discussed here, we get a short exact sequence $0\to H^2(G)\otimes \mathbb Z/p\to H^2(G,\mathbb Z/p)\to \mathrm{Tor}^\mathbb Z_1(H^1(G),\mathbb Z/p)\to 0$

(use $A= $ the usual chain complex that is used to define group (co)homology, so either the standard complex, or the singular chain complex of $BG$, ..., $M=\mathbb Z$, $B= \mathbb Z$ in degree $0$ and $N= \mathbb Z/p$)

$H^1(G)= 0$ because $G$ is perfect, so the short exact sequence actually exhibits an isomorphism $H^2(G,\mathbb Z/p) \cong (\mathbb Z/p)^2$ (in your example)

But now, $\hom(H_2(G),\mathbb Z/p) \cong (\mathbb Z/p)^2\oplus \hom(T,\mathbb Z/p)$ (because $\hom(\mathbb{Z,Z}/p) \cong \mathbb Z/p$ !)

so if the two are isomorphic, it means $\hom(T,\mathbb Z/p) = 0$. This holds for all $p$, so $T=0$

Note that in general, it is not true that a finitely generated group has finitely generated homology (see e.g. this paper), so you need a better argument to say that $H_2(G)$ is of the form $\mathbb Z^2\oplus T$. Your argument (and my help) works under this assumption

If your group is finitely presented, then its $H_1$ and $H_2$ are finitely generated (this follows by taking a specific CW-model for $BG$ which has only finitely many $1$- and $2$- cells) - so this whole argument works in this case.

In fact, the result does not hold without some additional assumption. Indeed, theorem H in On the integral homology of finitely presented groups (Baumslag, Dyer, Miller, see here) implies in particular that given any finitely generated abelian group $A$ and any countable abelian group $B$, there is a finitely generated group $G$ with $H_1(G) = A,H_2(G) = B$.

Taking $A=0, B= \mathbb Z^2\oplus \mathbb Q$ gives a finitely generated group $G$ which is perfect, and such that $H^2(G) = \hom(H_2(G),\mathbb Z)= \mathbb Z^2 \oplus \hom(\mathbb{Q,Z}) = \mathbb Z^2$, so a counterexample to your claim.

In conclusion :

If you assume $G$ to be finitely presented, or at least $H_2(G)$ finitely generated [so if you have some additional assumption], then your claim holds, by the above. However, with no additional assumption, the claim does not hold; and there is a counterexample for each countable abelian group $B$ with $\hom(B,\mathbb Z)\cong \mathbb Z^2$