Kernel of natural homomorphism from tensor square to group equals third homotopy group of suspension of classifying space

Statement
Suppose $$G$$ is a group. Let $$G \otimes G$$ denote the tensor square of $$G$$. There is a natural homomorphism from $$G \otimes G \to [G,G]$$ given by $$x \otimes y \mapsto [x,y]$$ (note that this is the composite of the natural map $$x \otimes y \mapsto x \wedge y$$ and the commutator map $$x \wedge y \to [x,y]$$.

The kernel of the homomorphism:

$$G \otimes G \to [G,G]$$

is canonically isomorphic to the third homotopy group of the suspension of the classifying space of $$G$$ (viewed as a discrete group), i.e., its first Eilenberg-Maclane space. Explicitly:

$$\operatorname{Ker}(G \otimes G \to [G,G]) = \pi_3SK(G,1)$$

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