# Action of Inverse of Group Element

## 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 $GA \, 1$

$\blacksquare$