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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 of $G$ and if

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

is a universal cover 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 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.

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    $\begingroup$ That is an elegant formula which I do not recognise in that form. I'll think before I write more. $\endgroup$
    – Geoff Robinson
    Commented May 3, 2019 at 16:21
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    $\begingroup$ $K\leqslant \bar{H}\trianglelefteq \bar{G}$ and $\bar{G}/\bar{H}\cong G/H$. Now since $H_2(\bar{G})=0$, the usual proof using Stallings exact sequence carries to this setting, see the bottom of groupprops.subwiki.org/wiki/… $\endgroup$
    – Andrei Smolensky
    Commented May 3, 2019 at 17:58
  • $\begingroup$ @AndreiSmolensky, many thanks, but why is $H_2(\overline{G},\mathbb{Z})=0$? For $G=V_4$ we can have $\overline{G} \cong D_4$, which does not have trivial Schur multiplier... $\endgroup$
    – André Macedo
    Commented May 3, 2019 at 18:09
  • $\begingroup$ @AndréMacedo Ah, indeed, this works only if $G$ is perfect (in which case the resulting formula also simplifies). $\endgroup$
    – Andrei Smolensky
    Commented May 3, 2019 at 18:14

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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

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

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  • $\begingroup$ Many thanks, Derek Holt! A quick observation: in the case where $K \subset [\overline{H},\overline{G}]$ (this happens for instance when $\operatorname{Cor}:\operatorname{H}_2(H,\mathbb{Z}) \to \operatorname{H}_2(G,\mathbb{Z})$ is surjective) the resulting quotient is also isomorphic to $\frac{H \cap [G,G]}{[H,G]}$. $\endgroup$
    – André Macedo
    Commented May 6, 2019 at 14:00

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