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 $g_{2}\otimes h$:

$g_{2}\otimes h\mapsto g_{1}g_{2}g_{1}^{-1}\otimes \alpha (g_{1})h$

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 $g\otimes h_{2}$:

$g\otimes h_{2}\mapsto \beta (h_{1})g\otimes h_{1}h_{2}h_{1}^{-1}$

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