Cancellable Elements of a Semigroup

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Theorem

The right cancellable elements of a semigroup $\left ({S, \circ}\right)$ form a subsemigroup of $\left ({S, \circ}\right)$.

Similarly, the left cancellable elements of a semigroup $\left ({S, \circ}\right)$ form a subsemigroup of $\left ({S, \circ}\right)$.

Consequently, the cancellable elements of a semigroup $\left ({S, \circ}\right)$ form a subsemigroup of $\left ({S, \circ}\right)$.

Let $C_\rho$ be the set of right cancellable elements of $\left ({S, \circ}\right)$, that is:

$C_\rho = \left\{{x \in S: \forall a, b \in S: a \circ x = b \circ x \implies a = b}\right\}$

Let $x, y \in C_\rho$. Then:

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle a \circ \left({x \circ y}\right)$	$=$	$\displaystyle b \circ \left({x \circ y}\right)$	$\displaystyle $	$\displaystyle $	$\displaystyle $
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle \left({a \circ x}\right) \circ y$	$=$	$\displaystyle \left({b \circ x}\right) \circ y$	$\displaystyle $	$\displaystyle $	$\displaystyle $	by associativity of $\circ$
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle a \circ x$	$=$	$\displaystyle b \circ x$	$\displaystyle $	$\displaystyle $	$\displaystyle $	as $y \in C_\rho$
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle a$	$=$	$\displaystyle b$	$\displaystyle $	$\displaystyle $	$\displaystyle $	as $x \in C_\rho$
$\displaystyle $	$\displaystyle \implies$	$\displaystyle $	$\displaystyle x \circ y$	$\in$	$\displaystyle C_\rho$	$\displaystyle $	$\displaystyle $	$\displaystyle $

Thus $\left({C_\rho, \circ}\right)$ is closed and is therefore by the Subsemigroup Closure Test is a semigroup of $\left ({S, \circ}\right)$.

A similar argument holds for the left cancellable elements.

Now let $C$ be the set of cancellable elements of $\left ({S, \circ}\right)$.

Let $x, y \in C$. Then $x$ and $y$ are both left cancellable and right cancellable.

Thus $x \circ y$ is both left cancellable and right cancellable, and therefore cancellable.

Thus $x \circ y \in C$.

Thus $\left ({C, \circ}\right)$ is closed and is therefore by the Subsemigroup Closure Test a semigroup of $\left ({S, \circ}\right)$.

$\blacksquare$

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a \circ \left({x \circ y}\right)\)	\(=\)	\(\displaystyle b \circ \left({x \circ y}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left({a \circ x}\right) \circ y\)	\(=\)	\(\displaystyle \left({b \circ x}\right) \circ y\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	by associativity of $\circ$
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a \circ x\)	\(=\)	\(\displaystyle b \circ x\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $y \in C_\rho$
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle a\)	\(=\)	\(\displaystyle b\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as $x \in C_\rho$
\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x \circ y\)	\(\in\)	\(\displaystyle C_\rho\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)