# Condition for Invertibility in Power Structure on Associative or Cancellable Operation

## Theorem

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

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

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

Let $\circ$ be either:

an associative operation
a cancellable operation.

Let $X \subseteq S$ be a subset of $S$.

Then:

$X$ is invertible for $\circ_PP$
there exists an element $x \in S$ which is invertible for $\circ$ such that $X = \set x$.

## Proof

First we note that from Identity Element for Power Structure, the algebraic structure $\struct {\powerset S, \circ_\PP}$ has an identity element $J = \set e$.

### Sufficient Condition

Let $X$ be invertible for $\circ_PP$.

Then:

 $\ds \exists Y \in \powerset S: \,$ $\ds X \circ_\PP Y$ $=$ $\ds J$ Definition of Invertible Element $\ds \leadsto \ \$ $\ds \set {x \circ y: x \in X, y \in Y}$ $=$ $\ds \set e$ Definition of Operation Induced on Power Set $\ds \leadsto \ \$ $\ds \exists x \in X: \exists y \in S: \,$ $\ds x \circ y$ $=$ $\ds e$

Similarly:

 $\ds \exists Y \in \powerset S: \,$ $\ds Y \circ_\PP X$ $=$ $\ds J$ Definition of Invertible Element $\ds \leadsto \ \$ $\ds \set {y \circ x: x \in X, y \in Y}$ $=$ $\ds \set e$ Definition of Operation Induced on Power Set $\ds \leadsto \ \$ $\ds \exists x \in X: \exists y \in S: \,$ $\ds y \circ x$ $=$ $\ds e$

That is, there exists $x \in X$ such that:

$x \circ y = e = y \circ x$

That is:

$x$ is invertible for $\circ$
$y$ is the inverse of $x$.

Hence we have $Y \in \powerset S$ such that:

$\forall y \in Y: x \circ y = e = y \circ x$

$\circ$ is Associative

Let $\circ$ be an associative operation.

Aiming for a contradiction, suppose $\exists z \in X$ such that $z \ne x$.

Then we have:

$\forall y \in Y: z \circ y = e = y \circ z$

 $\ds z \circ e$ $=$ $\ds z$ Definition of Identity Element $\ds \leadsto \ \$ $\ds z \circ \paren {y \circ x}$ $=$ $\ds z$ $\ds \leadsto \ \$ $\ds \paren {z \circ y} \circ x$ $=$ $\ds z$ as $\circ$ is associative $\ds \leadsto \ \$ $\ds e \circ x$ $=$ $\ds z$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds z$

This contradicts our supposition that $z \ne x$.

Hence there can be no elements in $X$ apart from $x$.

That is:

$X = \set x$

where $x$ is invertible for $\circ$.

$\Box$

$\circ$ is Cancellable

Let $\circ$ be a cancellable operation.

Aiming for a contradiction, suppose $\exists z \in X$ such that $z \ne x$.

Then we have:

$\forall y \in Y: z \circ y = e = y \circ z$

Then:

 $\ds x \circ y$ $=$ $\ds z \circ y$ $\ds \leadsto \ \$ $\ds x$ $=$ $\ds z$ as $\circ$ is cancellable

This contradicts our supposition that $z \ne x$.

Hence there can be no elements in $X$ apart from $x$.

That is:

$X = \set x$

where $x$ is invertible for $\circ$.

$\Box$

### Necessary Condition

Let there exist an element $x \in S$ which is invertible for $\circ$.

Hence there exist $y \in S$ such that:

$x \circ y = e = y \circ x$

Let $X = \set x$.

We have:

 $\ds X \circ \set y$ $=$ $\ds \set {x \circ y}$ $\ds$ $=$ $\ds \set e$ $\ds$ $=$ $\ds J$

and:

 $\ds \set y \circ X$ $=$ $\ds \set {y \circ x}$ $\ds$ $=$ $\ds \set e$ $\ds$ $=$ $\ds J$

That is, $X$ is invertible for $\circ_\PP$.

$\blacksquare$