Left Inverse and Right Inverse is Inverse

From ProofWiki
Jump to navigation Jump to search


Let $\struct {S, \circ}$ be a monoid with identity element $e_S$.

Let $x \in S$ such that $x$ has both a left inverse and a right inverse. That is:

$\exists x_L \in S: x_L \circ x = e_S$
$\exists x_R \in S: x \circ x_R = e_S$

Then $x_L = x_R$, that is, $x$ has an inverse.

Furthermore, that element is the only inverse (both right and left) for $x$


We note that as $\struct {S, \circ}$ is a monoid, $\circ$ is associative by definition.

\(\ds x_L\) \(=\) \(\ds x_L \circ e_S\) Definition of Identity Element
\(\ds \) \(=\) \(\ds x_L \circ \paren {x \circ x_R}\) Definition of Right Inverse Element
\(\ds \) \(=\) \(\ds \paren {x_L \circ x} \circ x_R\) Definition of Associative Operation
\(\ds \) \(=\) \(\ds e_S \circ x_R\) Definition of Left Inverse Element
\(\ds \) \(=\) \(\ds x_R\) Definition of Identity Element

Its uniqueness comes from Inverse in Monoid is Unique.