Action of Inverse of Group Element
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Theorem
Let $\struct {G, \circ}$ be a group.
Let $S$ be a sets.
Let $*: G \times S \to S$ be a group action.
Then:
- $g * a = b \iff g^{-1} * b = a$
Proof
\(\ds g * a\) | \(=\) | \(\ds b\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds g^{-1} * \paren {g * a}\) | \(=\) | \(\ds g^{-1} * b\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {g^{-1} \circ g} * a\) | \(=\) | \(\ds g^{-1} * b\) | Group Action Axiom $\text {GA} 2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e * a\) | \(=\) | \(\ds g^{-1} * b\) | Group Axiom $\text G 3$: Existence of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a\) | \(=\) | \(\ds g^{-1} * b\) | Group Action Axiom $\text {GA} 1$ |
and:
\(\ds g^{-1} * b\) | \(=\) | \(\ds a\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds g * \paren {g^{-1} * b}\) | \(=\) | \(\ds g * a\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {g \circ g^{-1} } * b\) | \(=\) | \(\ds g * a\) | Group Action Axiom $\text {GA} 2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds e * b\) | \(=\) | \(\ds g * a\) | Group Axiom $\text G 3$: Existence of Inverse Element | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds b\) | \(=\) | \(\ds g * a\) | Group Action Axiom $\text {GA} 1$ |
$\blacksquare$
Sources
- 1965: J.A. Green: Sets and Groups ... (previous) ... (next): $\S 5.6$. Stabilizers: Lemma