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 submonoid of $$\left({C, \circ}\right)$$, and $$\left({T, \circ}\right)$$ is a group.

Proof
From Cancellable Elements of a Monoid, $$\left({C, \circ}\right)$$ 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 also 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.


 * Closure: By Inverse of Product, $$x^{-1} \circ y^{-1} \in T$$ and therefore $$\left({T, \circ}\right)$$ is closed.


 * Identity: All the elements of $$\left({C, \circ}\right)$$ are by definition cancellable, so, by Cancellable Monoid Identity of Submonoid, $$e_C \in T$$.


 * Inverses: By Inverse of an Inverse, $$\left({x^{-1} \circ y^{-1}}\right)^{-1} \in T$$, thus every element has an inverse.


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

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