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 $X = \set x$.
 * 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:

Similarly:

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.

$\exists z \in X$ such that $z \ne x$.

Then we have:


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

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 $\circ$.


 * $\circ$ is Cancellable

Let $\circ$ be a cancellable operation.

$\exists z \in X$ such that $z \ne x$.

Then we have:


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

Then:

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 $\circ$.

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:

and:

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