Let $G$ be a finite group and $H$ a normal subgroup of $G$. I recently stumbled upon the following identity for the Schur multiplier of $G/H$:


$$\operatorname{H}\_2(G/H,\mathbb{Z}) \cong \frac{\overline{H} \cap [\overline{G},\overline{G}]}{[\overline{H},\overline{G}]}$$


Here $\overline{G}$ is a [Schur covering group](https://groupprops.subwiki.org/wiki/Schur_covering_group) of $G$ and if 


$$1 \to K \to \overline{G} \xrightarrow[]{\lambda} G \to 1$$


is a [universal cover](https://en.wikipedia.org/wiki/Schur_multiplier#Relation_to_central_extensions) of $G$, then $\overline{H} = \lambda^{-1}(H)$. Note in particular that for $H=1$ one recovers the fact that $H\_2(G,\mathbb{Z}) \cong K$.


I've found this formula as a special case of an algebro-geometric identity (namely, via an explicit formula for the unramified Brauer group of a smooth compactification of a certain family of tori). My question is the following:



> 
> Is there purely group-theoretic proof of the above identity?
> 
> 
> 


The identity quite resembles [Hopf's formula](https://groupprops.subwiki.org/wiki/Hopf%27s_formula_for_Schur_multiplier) for the Schur multiplier, but I was unable to derive it from that result. I also could not find it in some of the standard references for this topic, e.g. Karpilovski's book on the Schur multiplier.



Reference for Schur multiplier identity
Let $G = F/R$ with $F$ free and $H = S/R$.


Since everything is happening modulo $[F,R]$, I am just going to work modulo $[F,R]$.


Then, by the Hopf formula, $M(G)$ (the Schur Multiplier) is isomorphic to $[F,F] \cap R$, which has free abelian complements $C$ in $R$ with $R = ([F,F] \cap R) \times C$, and $\bar{G} = F/C$. (Note that different complements can give non-isomorphic covering groups $\bar{G}$.)


Now, since $[F,F] \cap C = 1$, we have


$$M(G/H) \cong \frac{[F,F] \cap S}{[F,S]} \cong
\frac{([F,F] \cap S)C/C}{[F,S]C/C} = \frac{[F/C,F/C] \cap S/C}{[F/C,S/C]} \cong \frac{[\bar{G},\bar{G}] \cap \bar{H}}{[\bar{G},\bar{H}]}$$


