Trivial pair of actions is compatible

Statement
Suppose $$G$$ and $$H$$ are groups and $$\alpha:G \to \operatorname{Aut}(H)$$ and $$\beta:H \to \operatorname{Aut}(G)$$ are group homomorphisms. Then, if $$\alpha$$ and $$\beta$$ are both trivial, they form a compatible pair of actions.

Related facts

 * Compatible with trivial action iff image centralizes inner automorphisms
 * Trivial pair of Lie ring actions is compatible

Proof
We want to show that the trivial pair of actions is a compatible pair.

Want to show: If both the actions of the groups on each other are trivial, then the following holds, where $$\cdot$$ is interpreted from context as the action $$\alpha$$, the action $$\beta$$, or the action of a group on itself by conjugation:

$$g_1 \cdot (h \cdot g_2) = (g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H$$

$$h_1 \cdot (g \cdot h_2) = (h_1 \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H$$

Proof:

First equality, left side: $$g_1 \cdot (h \cdot g_2) = g_1 \cdot g_2$$ where we use the triviality of the action of $$H$$ on $$G$$.

First equality, right side: $$(g_1 \cdot h) \cdot (g_1 \cdot g_2) = h \cdot (g_1 \cdot g_2) = g_1 \cdot g_2$$ where the first step uses the triviality of the action of $$G$$ on $$H$$ and the second step uses the triviality of the action of $$H$$ on $$G$$.

Thus, the first equality holds.

Second equality, left side: $$h_1 \cdot (g \cdot h_2) = h_1 \cdot h_2$$ using the triviality of the action of $$G$$ on $$H$$.

Second equality, right side: $$(h_1 \cdot g) \cdot (h_1 \cdot h_2) = g \cdot (h_1 \cdot h_2) = h_1 \cdot h_2$$ where the first step uses the triviality of the action of $$H$$ on $$G$$ and the second step uses the triviality of the action of $$G$$ on $$H$$.