# Hopf's formula for Schur multiplier

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## 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.

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