Group acts naturally on its tensor product with any group

Statement
Suppose $$G$$ and $$H$$ are groups and $$\alpha:G \to \operatorname{Aut}(H)$$ and $$\beta: H \to \operatorname{Aut}(G)$$ form a compatible pair of actions of $$G$$ and $$H$$ on each other. Let $$G \otimes H$$ be the tensor product of groups corresponding to this compatible pair of actions. Then, we have a natural action of $$G$$ on $$G \otimes H$$, i.e., a map:

$$G \to \operatorname{Aut}(G \otimes H)$$

defined as follows. The automorphism induced by $$g_1 \in G$$ is as follows on a generator of the form $$g_2 \otimes h$$:

$$g_2 \otimes h \mapsto g_1g_2g_1^{-1} \otimes \alpha(g_1)h$$

If we use $$\cdot$$ to denote the action of each group on itself by conjugation, the action $$\alpha$$, and the newly defined action on the tensor product, then the action can be written as:

$$g_1 \cdot (g_2 \otimes h) = (g_1 \cdot g_2) \otimes (g_1 \cdot h)$$

Similarly, we have a natural action of $$H$$ on $$G \otimes H$$, i.e., a map:

$$H \to \operatorname{Aut}(G \otimes H)$$

defined as follows. The automorphism induced by $$h_1 \in H$$ is as follows on a generator of the form $$g \otimes h_2$$:

$$g \otimes h_2 \mapsto \beta(h_1)g \otimes h_1h_2h_1^{-1}$$

If we use $$\cdot$$ to denote the action of each group on itself by conjugation, the action $$\beta$$, and the newly defined action on th etensor product, then the action can be written as:

$$h_1 \cdot (g \otimes h_2) = (h_1 \cdot g) \otimes (h_1 \cdot h_2)$$

Combining both of these, we get a homomorphism:

$$G * H \to \operatorname{Aut}(G \otimes H)$$

Related facts

 * Tensor product of groups maps to both groups