# Cancellable Elements of Monoid form Submonoid

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## 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 Semigroup form Subsemigroup, $\left ({C, \circ}\right)$ 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 $S$, we have that:

- $\forall x \in C: x \circ e_S = x = e_S \circ x$.

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

$\blacksquare$