Subset Product Action is Group Action

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Theorem

Let $\struct {G, \circ}$ be a group whose identity is $e$.

Let $\powerset G$ be the power set of $\struct {G, \circ}$.


For any $S \in \powerset G$ and for any $g \in G$, the subset product action:

$\forall g \in G: \forall S \in \powerset G: g * S = g \circ S$

is a group action.


Proof

Let $g \in G$.

First we note that since $G$ is closed, and $g \circ S$ consists of products of elements of $G$, it follows that:

$g * S \subseteq G$


Next we note:

$e * S = e \circ S = \set {e \circ s: s \in S} = \set {s: s \in S} = S$

and so Group Action Axiom $\text {GA} 2$ is satisfied.


Now let $g, h \in G$.

We have:

\(\ds \paren {g \circ h} * S\) \(=\) \(\ds \paren {g \circ h} \circ S\)
\(\ds \) \(=\) \(\ds \set {\paren {g \circ h} \circ s: s \in S}\)
\(\ds \) \(=\) \(\ds \set {g \circ \paren {h \circ s}: s \in S}\)
\(\ds \) \(=\) \(\ds g * \set {h \circ s: s \in S}\)
\(\ds \) \(=\) \(\ds g * \paren {h \circ S}\)
\(\ds \) \(=\) \(\ds g * \paren {h * S}\)

and so Group Action Axiom $\text {GA} 1$ is satisfied.

Hence the result.

$\blacksquare$


Also see


Sources

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