Invertible Elements of Monoid form Subgroup of Cancellable Elements

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.

Taking the group axioms in turn:

G0: Closure
By Inverse of Product, $x^{-1} \circ y^{-1} \in T$.

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

G1: Associativity
This is inherited from $S$, by Subsemigroup Closure Test.

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

G2: Identity
All the elements of $\left({C, \circ}\right)$ are by definition cancellable

So, by Identity of Cancellable Monoid is Identity of Submonoid:
 * $e_C \in T$

Thus $\left({T, \circ}\right)$ has an identity element.

G3: Inverses
By Inverse of Inverse: Monoid:


 * $\left({x^{-1} \circ y^{-1}}\right)^{-1} \in T$

Thus every element of $\left({T, \circ}\right)$ has an inverse.

All the group axioms are thus seen to be fulfilled, and so $\left({T, \circ}\right)$ is a group.