Invertible Elements of Monoid form Subgroup

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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 $\left({C, \circ}\right)$.

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 Elements of Semigroup 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.

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

$\blacksquare$