# Invertible Element containing Identity in Power Structure

## Theorem

Let $\struct {S, \circ}$ be a magma.

Let identity element $e \in S$ be an identity element of $\struct {S, \circ}$.

Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.

Let $X \subseteq S$ such that:

$e \in X$
$X$ is invertible for $\circ_PP$.

Then $X = \set e$.

## Proof

Let $X \subseteq S$ be invertible for $\circ_PP$ and such that $e \in X$.

From Identity Element for Power Structure, $\struct {\powerset S, \circ_\PP}$ has an identity element $J = \set e$.

We have:

 $\text {(1)}: \quad$ $\ds \exists Y \in \powerset S: \,$ $\ds X \circ_\PP Y$ $=$ $\ds J$ Definition of Invertible Element: here $Y$ is the inverse of $X$ $\ds \leadsto \ \$ $\ds \set {x \circ y: x \in X, y \in Y}$ $=$ $\ds \set e$ Definition of Operation Induced on Power Set, Definition of $J$ $\ds \leadsto \ \$ $\ds \forall y \in Y: \,$ $\ds e \circ y$ $=$ $\ds e$ applying the above to $e$ as an element of $X$ $\ds \leadsto \ \$ $\ds \forall y \in Y: \,$ $\ds y$ $=$ $\ds e$ Definition of Identity Element $\ds \leadsto \ \$ $\ds Y$ $=$ $\ds \set e$ $\ds \leadsto \ \$ $\ds X \circ_\PP \set e$ $=$ $\ds \set e$ substituting for $Y$ and $J$ in $(1)$ $\ds \leadsto \ \$ $\ds \forall x \in X: \,$ $\ds x$ $=$ $\ds e$ Definition of Identity Element $\ds \leadsto \ \$ $\ds X$ $=$ $\ds \set e$

$\blacksquare$