# Left Inverse and Right Inverse is Inverse

## 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.

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

Its uniqueness comes from Inverse in Monoid is Unique.

$\blacksquare$