Cancellable Elements of Semigroup form Subsemigroup

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 $\left ({S, \circ}\right)$.

Proof
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:

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 and right cancellable.

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

Thus $x \circ y$ is both left 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 $\left ({S, \circ}\right)$.