Crossed module defines a compatible pair of actions

Statement
Suppose $$G$$ is a group and $$H$$ is a crossed module over $$G$$. Explicitly, there is a homomorphism $$\mu:H \to G$$ and a homomorphism $$\alpha:G \to \operatorname{Aut}(H)$$ (so that $$G$$ acts on $$H$$ by automorphisms) satisfying the following two compatibility conditions:


 * 1) The map $$\alpha$$ pushes forward via $$\mu$$ to the conjugation action of $$G$$ on itself: $$\mu(\alpha(g)(h)) = g\mu(h)g^{-1} \ \forall \ g \in G, h \in H$$
 * 2) The map $$\alpha$$ pulls back via $$\mu$$ to the conjugation action of $$H$$ on itself: $$\alpha(\mu(h_1))(h_2) = h_1h_2h_1^{-1} \ \forall h_1,h_2 \in H$$

The two conditions are more easily stated if we use $$\cdot$$ to denote all the conjugation actions within a group and to denote the action $$\alpha$$, so that $$\alpha(g)(h)$$ becomes $$g \cdot h$$. In that case, the conditions become:


 * 1) $$\mu(g \cdot h) = g \cdot \mu(h) \ \forall \ g \in G, h \in H$$
 * 2) $$\mu(h_1) \cdot h_2 = h_1 \cdot h_2 \ \forall \ h_1,h_2 \in H$$

The claim is that given such a crossed module structure, we can obtain a compatible pair of actions $$\alpha:G \to \operatorname{Aut}(H)$$ and $$\beta: H \to \operatorname{Aut}(G)$$ where, for all $$h \in H$$, $$\beta(h)$$ is defined as conjugation in $$G$$ by the element $$\mu(h)$$.

In other words, the claim is that the following are true:


 * $$g_1 \cdot (\mu(h) \cdot g_2) = \mu(g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H$$
 * $$\mu(h_1) \cdot (g \cdot h_2) = (\mu(h_1) \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H$$

Proof of first claim using $$\cdot$$ notation
To prove: $$g_1 \cdot (\mu(h) \cdot g_2) = \mu(g_1 \cdot h) \cdot (g_1 \cdot g_2) \ \forall \ g_1,g_2 \in G, h \in H$$

Proof:

Right side simplfies to left side: $$\mu(g_1 \cdot h) \cdot (g_1 \cdot g_2) = (g_1 \cdot (\mu(h)) \cdot (g_1 \cdot g_2) = (g_1\mu(h)g_1^{-1}) \cdot (g_1 \cdot g_2) = (g_1\mu(h)g_1^{-1}g_1) \cdot g_2 = (g_1\mu(h)) \cdot g_2 = g_1 \cdot (\mu(h) \cdot g_2)$$. Note that the first step of simplification uses one of the axioms for the crossed module definition.

Proof of second claim using $$\cdot$$ notation
To prove: $$\mu(h_1) \cdot (g \cdot h_2) = (\mu(h_1) \cdot g) \cdot (h_1 \cdot h_2) \ \forall g \in G, h_1,h_2 \in H$$

Proof:

Right side simplifies to left side: $$(\mu(h_1) \cdot g) \cdot (h_1 \cdot h_2) = (\mu(h_1) \cdot g) \cdot (\mu(h_1) \cdot h_2) = (\mu(h_1)g\mu(h_1)^{-1}) \cdot (\mu(h_1) \cdot h_2) = (\mu(h_1)g\mu(h_1)^{-1}\mu(h_1)) \cdot h_2 = (\mu(h_1)g) \cdot h_2 = \mu(h_1) \cdot (g \cdot h_2)$$ where the first step uses one of the axioms.