Pure tensor involving identity element is identity element

Statement
Suppose $$G$$ and $$H$$ are groups with a compatible pair of actions $$\alpha:G \to \operatorname{H}$$ and $$\beta:H \to \operatorname{Aut}(G)$$. Denote by $$1_G$$ the identity element of $$G$$, by $$1_H$$ the identity element of $$H$$, and by $$1_{G \otimes H}$$ the identity element of $$G \otimes H$$. Then, the following are true:


 * $$1_G \otimes h = 1_{G \otimes H} \ \forall \ h \in H$$
 * $$g \otimes 1_H = 1_{G \otimes H} \ \forall \ g \in G$$

Proof idea
The statement is similar in spirit to the statement that in the tensor product of vector spaces, any pure tensor where either of the components is the zero vector must be the zero vector in the tensor product.

The main complication here is that the groups are acting on each other. However, if we concentrate on the behavior at and near the identity element, then the relevant actions are trivial.

Proof using the $$\cdot$$ notation for actions
Instead of using the letters $$\alpha$$ and $$\beta$$, we will use $$\cdot$$ to denote both $$\alpha$$ and $$\beta$$ as well as the actions of the groups on themselves by conjugation.

Proof of the statement involving identity element in $$G$$
To use: $$(g_1g_2) \otimes h = ((g_1 \cdot g_2) \otimes (g_1 \cdot h))(g_1 \otimes h)$$ for all $$g_1,g_2 \in G, h \in H$$ (this is one of the two axioms for the tensor product)

To prove: $$1_G \otimes h = 1_{G \otimes H}$$

Proof: We use the statement To use setting $$g_1 = g_2 = 1_G$$ and setting $$h = h$$. We get:

$$(1_G1_G) \otimes h = ((1_G \cdot 1_G) \otimes (1_G \cdot h))(1_G \otimes h)$$

This simplifies to:

$$1_G \otimes h = (1_G \otimes h)^2$$

Cancel one of the $$1_G \otimes h$$ factors from both sides and interchange the left and right side to get:

$$1_G \otimes h = 1_{G \otimes H}$$

Proof of the statement involving identity element in $$H$$
To use: $$g \otimes (h_1h_2) = (g \otimes h_1)((h_1 \cdot g) \otimes (h_1 \cdot h_2))$$ for all $$g \in G, h_1,h_2 \in H$$(this is one of the two axioms for the tensor product)

To prove: $$g \otimes 1_H = 1_{G \otimes H}$$ for all $$g \in G$$.

Proof: We use the statement To use setting $$h_1 = h_2 = 1_H$$ and setting $$g = g$$. We get:

$$g \otimes (1_H1_H) = (g \otimes 1_H)((1_H \cdot h) \otimes (1_H \cdot 1_H))$$

This simplifies to:

$$g \otimes 1_H = (g \otimes 1_H)^2$$

Cancel one of the g \otimes 1_H factors from both sides and interchange the left and right side to get:

$$g \otimes 1_H = 1_{G \otimes H}$$