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

Proof

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: