# 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.

\(\displaystyle x_L\) | \(=\) | \(\displaystyle x_L \circ e_S\) | 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.

$\blacksquare$

## Sources

- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Exercise $4.9 \ \text{(a)}$