Trivial pair of actions is compatible

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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

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.