Tensor square of a group

Arbitrary group
Suppose $$G$$ is a (not necessarily abelian) group. The tensor square of $$G$$ (sometimes termed the non-abelian tensor square) refers to the defining ingredient::tensor product of groups $$G \otimes G$$ where we take both the actions of $$G$$ on each other to be the action by conjugation.

Explicitly, it is the quotient of the free group on all formal symbols $$\{ g \otimes h \mid g,h \in G \}$$ by the following relations:


 * $$(g_1g_2) \otimes h = (g_1g_2g_1^{-1} \otimes g_1hg_1^{-1})(g_1 \otimes h)$$, one such relation for all $$g_1,g_2,h \in G$$
 * $$g \otimes (h_1h_2) = (g \otimes h_1)(h_1gh_1^{-1} \otimes h_1h_2h_1^{-1})$$, one such relation for all $$g,h_1,h_2 \in G$$

Abelian group
For $$G$$ an abelian group, the tensor square of $$G$$ is the same as its tensor square as a group. However, it can now also be thought of as follows: it is the tensor product of $$G$$ with itself in the sense of tensor product of abelian groups.

Related notions

 * Exterior square of a group
 * Tensor power of a group

Facts

 * There is a natural surjective homomorphism from the tensor square of a group to the exterior square of a group (namely, send $$x \otimes y$$ to $$x \wedge y$$; the kernel is the normal closure of elements of the form $$x \otimes x$$), and from that, via the commutator map, to the derived subgroup which sits inside the group.
 * Kernel of natural homomorphism from tensor square to group equals third homotopy group of suspension of classifying space
 * Exact sequence giving kernel of mapping from tensor square to exterior square
 * Perfect implies natural mapping from tensor square to exterior square is isomorphism
 * Equivalence of definitions of superperfect group: For a superperfect group, the commutator mapping defines an isomorphism from the tensor square to the whole group.