Commutator map is homomorphism from exterior product to commutator subgroup

Statement
Suppose $$G$$ and $$H$$ are subgroups of a group $$Q$$ such that $$G$$ and $$H$$ normalize each other. The commutator map defines a homomorphism from the exterior product $$G \wedge H$$ to the commutator subgroup $$[G,H]$$. Explicitly, the homomorphism:

$$G \wedge H \to [G,H]$$

is defined on a generating set as follows:

$$g \wedge h \mapsto [g,h]$$

Related facts

 * Tensor product of groups maps to both groups: This is one route to the proof.
 * Commutator map is homomorphism from exterior square to derived subgroup: This is the special case $$G = H$$.
 * Commutator map is homomorphism from exterior square to derived subgroup of central extension
 * Schur multiplier is kernel of commutator map homomorphism from exterior square to derived subgroup