Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup

Statement
Suppose $$G$$ is a group. The Schur multiplier of $$G$$, denoted $$M(G)$$, which is also described as the second homology group for trivial group action $$H^2(G;\mathbb{Z})$$, can be computed as follows: it is the kernel of the homomorphism from the exterior square to the derived subgroup given by the commutator map. Explicitly:

$$M(G) = \operatorname{ker}(G \wedge G \stackrel{\partial}{\to} [G,G])$$

where $$\partial$$ sends $$x \wedge y$$ to $$[x,y]$$.

Similar facts

 * Perfect implies natural mapping from tensor square to exterior square is isomorphism
 * Exact sequence giving kernel of mapping from tensor square to exterior square
 * Kernel of natural homomorphism from tensor square to group equals third homotopy group of suspension of classifying space

Applications

 * Schur muliplier of abelian group is its exterior square: This follows easily from the result of this page as follows. For an abelian group $$G$$, the commutator map homomorphism is trivial, so its kernel is the entire group $$G \wedge G $$.