# 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 Group Action Axiom $GA \, 2$ is satisfied.

Now let $h_1, h_2 \in G$.

We have:

\(\displaystyle \paren {h_1 \circ h_2} * g\) | \(=\) | \(\displaystyle \paren {h_1 \circ h_2} \circ g\) | $\quad$ Definition of $*$ | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle h_1 \circ \paren {h_2 \circ g}\) | $\quad$ Group Axiom $G \, 1$: Associativity | $\quad$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle h_1 * \paren {h_2 * g}\) | $\quad$ Definition of $*$ | $\quad$ |

and so Group Action Axiom $GA \, 1$ is satisfied.

Hence the result.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 5.5$. Groups acting on sets: Example $106$ - 1971: Allan Clark:
*Elements of Abstract Algebra*... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 54 \alpha$