Tensor product of groups is commutative up to natural isomorphism

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Statement

Suppose G and H are (not necessarily abelian) groups with a compatible pair of actions \alpha:G \to \operatorname{Aut}(H) and \beta:H \to \operatorname{Aut}(G). We can define the tensor product of groups G \otimes H and also the tensor product of groups H \otimes G. The claim is that these tensor products are isomorphic groups with a natural isomorphism G \otimes H \to H \otimes G given as follows on a generating set:

g \otimes h \mapsto (h \otimes g)^{-1}

Note that this natural isomorphism is self-inverse, i.e., the natural isomorphism from G \otimes H to H \otimes G is the inverse map to the similarly defined natural isomorphism from H \otimes G to G \otimes H.

Related facts

Proof

Proof idea

Consider the two axioms that define the tensor product:

  • (g_1g_2) \otimes h = ((g_1 \cdot g_2) \otimes (g_1 \cdot h))(g_1 \otimes h)
  • g \otimes (h_1h_2) = (g \otimes h_1)((h_1 \cdot g) \otimes (h_1 \cdot h_2))

Note that apart from interchanging the roles of G and H, the other key difference between the two axioms is that the order of multiplication on the respective right sides differs. Since the inverse map is involutive, this explains why g \otimes h \mapsto (h \otimes g)^{-1} works.