Subgroup Action is Group Action

Theorem
Let $\struct {G, \circ}$ be a group.

Let $\struct {H, \circ}$ be a subgroup of $G$.

Let $*: H \times G \to G$ be the subgroup action defined for all $h \in H, g \in G$ as:
 * $\forall h \in H, g \in G: h * g := h \circ g$

Then $*$ is a group action.

Proof
Let $g \in G$.

First we note that since $G$ is closed, and $h \circ g \in G$, it follows that $h * g \in G$.

Next we note:
 * $e * g = e \circ g = g$

and so is satisfied.

Now let $h_1, h_2 \in G$.

We have:

and so is satisfied.

Hence the result.