# Invertible Elements of Monoid form Subgroup

## Theorem

Let $\left({S, \circ}\right)$ be an monoid whose identity is $e_S$.

Let $C$ be the set of all cancellable elements of $S$.

Let $T$ be the set of all invertible elements of $S$.

Then $\left({T, \circ}\right)$ is a subgroup of $\left({C, \circ}\right)$.

## Proof

From Cancellable Elements of Monoid form Submonoid, $\left({C, \circ}\right)$ is a submonoid of $\left({S, \circ}\right)$.

Let its identity be $e_C$ (which may or may not be the same as $e_S$).

Let $T$ be the set of all invertible elements of $S$.

From Invertible Element of Monoid is Cancellable, all the invertible elements of $S$ are also all cancellable, so $T \subseteq C$.

Let $x, y \in T$.

Clearly $x^{-1}, y^{-1} \in T$, as if $x, y$ are invertible, then so are their inverses.

Thus $\left({T, \circ}\right)$ is a group.

$\blacksquare$