Left Inverse and Right Inverse is Inverse

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Theorem

Let $\left({S, \circ}\right)$ 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 $\left({S, \circ}\right)$ is a monoid, $\circ$ is associative by definition.

\(\displaystyle x_L\) \(=\) \(\displaystyle x_L \circ e_S\) $\quad$ Behaviour of identity $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x_L \circ \left({x \circ x_R}\right)\) $\quad$ Behaviour of right inverse $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x_L \circ x}\right) \circ x_R\) $\quad$ Associativity of $\circ$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e_S \circ x_R\) $\quad$ Behaviour of Left inverse $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x_R\) $\quad$ Behaviour of identity $\quad$

Its uniqueness comes from Inverse in Monoid is Unique.

$\blacksquare$


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