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

\(\displaystyle g * a\) \(=\) \(\displaystyle b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g^{-1} * \paren {g * a}\) \(=\) \(\displaystyle g^{-1} * b\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g^{-1} \circ g} * a\) \(=\) \(\displaystyle g^{-1} * b\) Group Action Axiom $GA \, 2$
\(\displaystyle \leadsto \ \ \) \(\displaystyle e * a\) \(=\) \(\displaystyle g^{-1} * b\) Group Axiom $G \, 3$: Inverses
\(\displaystyle \leadsto \ \ \) \(\displaystyle a\) \(=\) \(\displaystyle g^{-1} * b\) Group Action Axiom $GA \, 1$


and:

\(\displaystyle g^{-1} * b\) \(=\) \(\displaystyle a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle g * \paren {g^{-1} * b}\) \(=\) \(\displaystyle g * a\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {g \circ g^{-1} } * b\) \(=\) \(\displaystyle g * a\) Group Action Axiom $GA \, 2$
\(\displaystyle \leadsto \ \ \) \(\displaystyle e * b\) \(=\) \(\displaystyle g * a\) Group Axiom $G \, 3$: Inverses
\(\displaystyle \leadsto \ \ \) \(\displaystyle b\) \(=\) \(\displaystyle g * a\) Group Action Axiom $GA \, 1$

$\blacksquare$


Sources