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:

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

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

$$a \circ \left({x \circ y}\right) = b \circ \left({x \circ y}\right)$$

$$\Longrightarrow \left({a \circ x}\right) \circ y = \left({b \circ x}\right) \circ y$$ (associativity of $$\circ$$)

$$\Longrightarrow a \circ x = b \circ x$$ (as $$y \in C_\rho$$)

$$\Longrightarrow a = b$$ (as $$x \in C_\rho$$)

$$\Longrightarrow x \circ y \in C_\rho$$

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