Hopf's formula

Remarks

Idea of the proof: the short exact sequence $1 \to R \to F \to G \to 1$ gives the exact sequence of low-degree terms $$H_2(F, \Z) \to H_2(G, \Z) \to R / [F, R] \to F / [F, F] \to G / [G, G] \to 0.$$ Since $F$ is free, $H_2(F, \Z) = 0$, so $H_2(G, \Z)$ is the kernel of $R / [F, R] \to F / [F, F]$, which is $(R \cap [F, F]) / [F, R]$.

In particular, the right side does not depend on the chosen presentation of $G$.

Example: $G = \Z^2 = \langle a, b \mid aba^{-1}b^{-1} \rangle$, with $F$ free on $a, b$. Here $R = [F, F]$, because the normal closure of $aba^{-1}b^{-1}$ in $F$ is the commutator subgroup $[F, F]$. So $H_2(\Z^2, \Z) \cong [F, F] / [F, [F, F]]$, which is infinite cyclic, generated by the image of $aba^{-1}b^{-1}$. This matches the homology of the torus.

Example: $G = \Z / m\Z$, with $F = \Z$ and $R = m\Z$. Then $[F, F]$ is trivial, so $H_2(\Z/m\Z, \Z) = 0$.

Going deeper

For a student

For a finite group $G$, the group $H_2(G, \Z)$ is the Schur multiplier, which controls the central extensions of $G$ and its projective representations. Hopf's formula gives a purely algebraic way to compute it. The exact sequence in the proof is the five-term sequence of the Lyndon–Hochschild–Serre spectral sequence.