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