Subgroup Action is Group Action

From ProofWiki
Jump to: navigation, search

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