Tensor product of groups is commutative up to natural isomorphism

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 $G \otimes H$.

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.