# Construction of Inverse Completion/Invertible Elements in Quotient Structure

## Theorem

Let $\left({S, \circ}\right)$ be a commutative semigroup which has cancellable elements.

Let $\left({C, \circ {\restriction_C}}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.

Let $\left({S \times C, \oplus}\right)$ be the external direct product of $\left({S, \circ}\right)$ and $\left({C, \circ {\restriction_C}}\right)$, where $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.

Let $\boxtimes$ be the cross-relation on $S \times C$, defined as:

- $\left({x_1, y_1}\right) \boxtimes \left({x_2, y_2}\right) \iff x_1 \circ y_2 = x_2 \circ y_1$

This cross-relation is a congruence relation on $S \times C$.

Let the quotient structure defined by $\boxtimes$ be:

- $\left({T', \oplus'}\right) := \left({\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}\right)$

where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.

Every cancellable element of $S'$ is invertible in $T'$.

## Proof

From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.

Call this identity $e_{T'}$.

Let the mapping $\psi: S \to T'$ be defined as:

- $\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$

From Image of Cancellable Elements in Quotient Mapping:

- $C' = \psi \sqbrk C$

So:

\(\displaystyle x'\) | \(\in\) | \(\displaystyle C'\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \exists x \in C: x'\) | \(=\) | \(\displaystyle \map \psi x\) | as $\psi$ is a surjection | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall a \in C: x'\) | \(=\) | \(\displaystyle \eqclass {\tuple {x \circ a, a} } \boxtimes\) | Definition of $\psi$ |

The inverse of $x'$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$, as follows:

\(\displaystyle a \circ x\) | \(\in\) | \(\displaystyle C\) | Cancellable Elements of Semigroup form Subsemigroup | ||||||||||

\(\displaystyle a \circ a \circ x\) | \(\in\) | \(\displaystyle C\) | Cancellable Elements of Semigroup form Subsemigroup | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \eqclass {\tuple {x \circ a, a} } \boxtimes \oplus' \eqclass {\tuple {a, a \circ x} } \boxtimes\) | \(=\) | \(\displaystyle \eqclass {\tuple {x \circ a \circ a, a \circ a \circ x} } \boxtimes\) | by definition of $\oplus'$ | |||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \eqclass {\tuple {a \circ a \circ x, a \circ a \circ x} } \boxtimes\) | Commutativity of $\circ$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e_{T'}\) | Identity of Quotient Structure |

thus showing that the inverse of $\eqclass {\tuple {x \circ a, a} } \boxtimes$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 20$