Hopf's formula for Schur multiplier

Statement
Let $$G$$ be a group isomorphic to the quotient group $$F/R$$, where $$F$$ is a free group and $$R$$ is a normal subgroup of $$F$$. Then, the Schur multiplier of $$G$$, denoted $$M(G)$$, which is the same as the second homology group for the trivial group action of $$G$$ on the integers, denoted $$H_2(G,\mathbb{Z})$$, is an abelian group given by the formula:

$$M(G) = H_2(G; \mathbb{Z}) \cong (R \cap [F,F])/[R,F]$$.

In other words, it equals the quotient of the intersection of $$R$$ with the commutator subgroup of $$F$$ by the focal subgroup of $$R$$ in $$F$$. ($$[R,F]$$ equals the focal subgroup because $$R$$ is a normal subgroup of $$F$$).

Note that any choice of generating set for $$G$$ gives a choice of $$F$$ and $$R$$ for which the theorem can be applied: $$F$$ is the free group on those generators with the natural surjection, and $$R$$ is the kernel of the surjection.

Related facts

 * The definition of Baer invariant generalizes this formula
 * Hopf's formula for nilpotent multiplier
 * Variant of Hopf's formula for Schur multiplier for nilpotent group that uses the free nilpotent group of class one more

Facts used

 * 1) uses::Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup
 * 2) uses::Schur multiplier of free group is trivial

Direct proof
Consider the short exact sequence:

$$1 \to R \to F \to G \to 1$$

This gives a related short exact sequence:

$$1 \to R/[F,R] \to F/[F,R] \to G \to 1$$

Proof of being an initial object in the category of central extensions with homoclinisms
On account of $$F$$ being free, this short exact sequence gives an initial object in the category of central extensions with quotient group $$G$$. Explicitly, for any central extension:

$$1 \to A \to E \to G \to 1$$

there is a (unique) homomorphism $$(F/[F,R])' \to E'$$ that, composed with the quotient map $$E' \to G'$$, gives the quotient map $$(F/[F,R])' \to G'$$. Here is a sketch of the process:


 * Consider a freely generating set for $$F$$. Take its image in $$G$$.
 * For each element in the generating set for $$F$$, pick an element of $$E$$ that maps to its image in $$G$$.
 * Consider the homomorphism $$\theta:F \to E$$ obtained from the set map given from the generating set of $$F$$ above. Note that this exists because $$F$$ is free.
 * Verify that $$[F,R]$$ is in the kernel of $$\theta$$, so $$\theta$$ descends to a map $$\overline{\theta}: F/[F,R] \to E$$, which restricts to a homomorphism $$(F/[F,R])' \to E'$$.

Note that the second step of the construction introduces non-uniqueness. However, from general considerations, the map has to be unique, so this is a non-issue.

Consequence for the exterior square and Schur multiplier
This immediately implies that $$G \wedge G \cong (F/[F,R])'$$ and $$M(G)$$ (the Schur multiplier of $$G$$) is the kernel of the quotient map $$(F/[F,R])' \to G'$$. Let's calculate both:


 * $$(F/[F,R])'$$ simplifies to $$[F,F]/[F,R]$$. Thus, $$G \wedge G \cong [F,F]/[F,R]$$.
 * The kernel of the map from $$F$$ to $$G$$ is $$R$$, hence the kernel of $$F/[F,R] \to G$$ is $$R/[F,R]$$. The kernel of the map $$G \wedge G \to G'$$ is the intersection $$([F,F]/[F,R]) \cap (R/[F,R]) = (R \cap [F,F])/[F,R]$$. This, then, is the Schur multiplier.

Proof using the Stallings exact sequence
Given: A group $$G$$ written as a quotient group of a free group $$F$$ by a normal subgroup $$R$$ of $$F$$.

To prove: Let $$F' = [F,F]$$ and $$G' = [G,G]$$. The Schur multiplier $$M(G)$$, defined as the kernel of the commutator map homomorphism $$G \wedge G \to [G,G]$$, is isomorphic to $$(R \cap F')/[R,F]$$.

Proof: Consider the short exact sequence:

$$1 \to R \to F \to G \to 1$$

The corresponding Stallings exact sequence is:

$$H_2(F;\mathbb{Z}) \stackrel{\alpha}{\to} H_2(G;\mathbb{Z}) \stackrel{\beta}{\to} R/[F,R] \stackrel{\sigma}{\to} H_1(F;\mathbb{Z}) \stackrel{\tau}{\to} H_1(G;\mathbb{Z})$$

Since $$F$$ is free, $$H_2(F;\mathbb{Z})$$ is trivial. Thus, the sequence simplifies to:

$$1 \stackrel{\alpha}{\to} H_2(G;\mathbb{Z}) \stackrel{\beta}{\to} R/[F,R] \stackrel{\sigma}{\to} H_1(F;\mathbb{Z}) \stackrel{\tau}{\to} H_1(G;\mathbb{Z})$$

The homomorphism $$\sigma: R/[F,R] \to F/[F,F] = H_1(F;\mathbb{Z})$$ has kernel equal to $$(R \cap [F,F])/[F,R]$$. Thus, by exactness, that is also the image of $$\beta$$. Since $$\alpha$$ is trivial, $$\beta$$ is an isomorphism to its image. Thus, we get:

$$H_2(G;\mathbb{Z}) \cong (R \cap [F,F])/[F,R]$$