Tensor product of groups maps to both groups

Statement
Suppose $$G$$ and $$H$$ are groups with a compatible pair of actions $$\alpha:G \to \operatorname{Aut}(H)$$ and $$\beta:H \to \operatorname{Aut}(G)$$ on each other. Suppose $$G \otimes H$$ is the tensor product of groups for this compatible pair of actions. Then, there are natural homomorphisms:

$$\lambda_G: G \otimes H \to G, \lambda_H: G \otimes H \to H$$

given by extending the following mapping from the generating set:

$$\lambda_G(g \otimes h) = g (\beta(h)(g^{-1}))$$

$$\lambda_H(g \otimes h) = (\alpha(g)(h))h^{-1}$$

If we use $$\cdot$$ to denote both the actions $$\alpha$$ and $$\beta$$, then this becomes:

$$\lambda_G(g \otimes h) = g(h \cdot g^{-1})$$

$$\lambda_H(g \otimes h) = (g \cdot h)h^{-1}$$

In particular, combining these, we get a natural homomorphism to the external direct product:

$$G \otimes H \to G \times H$$

Particular cases

 * In case that both $$G$$ and $$H$$ are abelian groups and the actions are trivial, both the homomorphisms are trivial, hence the homomorphism to the direct product is also trivial.
 * In case that $$G = H$$ and we are using the action by conjugation, so that we are mapping from the tensor square of a group, both maps are identical to each other, and they are both the same as the commutator map viewed as a homomorphism (which factors through the exterior square).
 * More generally, if they are both subgroups inside some big group and normalize each other, then both mappings are equal and are given by a commutator mapping in the big group.

Proof
We will give below the proof for the map $$G \otimes H \to G$$. The proof for the map $$G \otimes H \to H$$ is perfectly analogous.

To do the proof, we need to demonstrate that the set map:

$$G \times H \to G$$

given by:

$$(g,h) \mapsto g(\beta(h))g^{-1}$$

is a crossed pairing for the compatible pair of actions $$\alpha,\beta$$.

The checking of all the conditions is relatively straightforward.

Checking the condition with respect to $$\alpha$$ (the action of $$G$$ on $$H$$): framed the easy way using the $$\cdot$$
The condition for a general function $$f$$ would read:

$$f(g_1g_2,h) = f(g_1 \cdot g_2,g_1 \cdot h)f(g_1,h)$$ for all $$g_1,g_2 \in G, h \in H$$

For our particular function, what we need to show is below:

To prove: $$(g_1g_2)(h \cdot (g_1g_2)^{-1}) = (g_1 \cdot g_2)((g_1 \cdot h) \cdot (g_1 \cdot g_2^{-1}))(g_1)(h \cdot g_1^{-1})$$

Proof: We start from the right side and perform

Checking the condition with respect to $$\beta$$ (the action of $$H$$ on $$G$$): framed the easy way using the $$\cdot$$
The condition for a general function $$f$$ to satisfy this are:

$$f(g,h_1h_2) = f(g,h_1)f(h_1 \cdot g, h_1 \cdot h_2)$$ for all $$g \in G, h_1,h_2 \in H$$

For our particular function, what we need to prove is:

To prove: $$g((h_1h_2) \cdot g^{-1}) = g(h_1 \cdot g^{-1})((h_1 \cdot g)((h_1 \cdot h_2) \cdot (h_1 \cdot g)^{-1}))$$

Proof: We will simplify the right side to the left side.