Invertible Elements of Monoid form Subgroup

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Theorem

Let $\struct {S, \circ}$ be a monoid whose identity element is $e$.

Let $U \subseteq S$ be the subset of $S$ consisting of the invertible elements of $S$.


Then $\struct {U, \circ}$ forms a subgroup of $S$.


Proof

We have from Inverse of Identity Element is Itself that $e$ is invertible.

Hence $e \in U$ and so $U \ne \O$.


Let $x, y \in U$.

As $x$ and $y$ are invertible, it follows that $x^{-1}$ and $y^{-1}$ both exist in $S$.

Both $x^{-1}$ and $y^{-1}$ also have inverses $x$ and $y$ respectively, and so themselves are invertible.

Hence both $x^{-1} \in U$ and $y^{-1} \in U$.

Then we have:

\(\ds \paren {x \circ y} \circ \paren {y^{-1} \circ x^{-1} }\) \(=\) \(\ds x \circ \paren {y \circ y^{-1} } \circ x^{-1}\) $\circ$ is associative in $S$
\(\ds \) \(=\) \(\ds x \circ e \circ x^{-1}\) Definition of Inverse Element
\(\ds \) \(=\) \(\ds x \circ x^{-1}\) Definition of Identity Element
\(\ds \) \(=\) \(\ds e\) Definition of Inverse Element

and similarly for $\paren {y^{-1} \circ x^{-1} } \circ \paren {x \circ y}$

Hence $x \circ y$ is invertible.

The result follows by the Two-Step Subgroup Test.

$\blacksquare$


Sources

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