Left Inverse and Right Inverse is Inverse
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Theorem
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 $x$
Proof
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.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Exercise $4.9 \ \text{(a)}$