Cancellable Elements of Monoid form Submonoid

Theorem
The cancellable elements of a monoid $$\left ({S, \circ}\right)$$ form a submonoid of $$\left ({S, \circ}\right)$$.

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

Obviously $$C \subseteq S$$.

From Cancellable Elements of a Semigroup, $$\left ({C, \circ}\right)$$ is a subsemigroup of $$S$$.

Let $$e_S$$ be the identity of $$\left ({S, \circ}\right)$$.

From Identity of Monoid is Cancellable, $$e_S$$ is cancellable, therefore $$e_S \in C$$.

As $$e_S$$ is the identity of $$S$$, $$\forall x \in C: x \circ e_S = x = e_S \circ x$$.

Thus $$e_S$$ is the identity of $$\left ({C, \circ}\right)$$, which is therefore a monoid. As $$C \subseteq S$$, the result follows.