Compatible with trivial action iff image centralizes inner automorphisms

Statement

Suppose $G$ and $H$ are groups with homomorphisms $\alpha :G\to \operatorname {Aut} (H)$ and $\beta :H\to \operatorname {Aut} (G)$ such that $\alpha$ is the trivial homomorphism, i.e., it sends every element of $G$ to the identity automorphism of $H$.

Then, the following are equivalent:

1. $\alpha$ and $\beta$ form a compatible pair of actions.
2. The image of $\beta$ in $\operatorname {Aut} (G)$ lies inside the centralizer $C_{\operatorname {Aut} (G)}(\operatorname {Inn} (G))$, i.e., it commutes with all the inner automorphisms of $G$.

Related facts

• Trivial pair of actions is compatible

Proof

The actions are compatible if and only if the following two conditions hold, where $\cdot$ denotes the action $\alpha ,\beta$ or conjugation within a group, as is clear from context:

$g_{1}\cdot (h\cdot g_{2})=(g_{1}\cdot h)\cdot (g_{1}\cdot g_{2})\ \forall \ g_{1},g_{2}\in G,h\in H$

$h_{1}\cdot (g\cdot h_{2})=(h_{1}\cdot g)\cdot (h_{1}\cdot h_{2})\ \forall g\in G,h_{1},h_{2}\in H$

Using the triviality of $\alpha$, these conditions are equivalent to:

$g_{1}\cdot (h\cdot g_{2})=h\cdot (g_{1}\cdot g_{2})\ \forall \ g_{1},g_{2}\in G,h\in H$

$h_{1}\cdot h_{2}=h_{1}\cdot h_{2}\ \forall g\in G,h_{1},h_{2}\in H$

Note that the second condition is vacuously true, so it conveys no information. The first condition is equivalent to saying that the action of $h$ commutes with conjugation by $g_{1}$, which is equivalent to saying that $\beta (h)\in C_{\operatorname {Aut} (G)}(\operatorname {Inn} (G))$.