# Left Inverse for All is Right Inverse

## Contents

## Theorem

Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that:

- $\forall x \in S: \exists x_L: x_L \circ x = e_L$

That is, every element of $S$ has a left inverse with respect to the left identity.

Then $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity.

## Proof

Let $y = x \circ x_L$. Then:

\(\displaystyle e_L \circ y\) | \(=\) | \(\displaystyle \paren {y_L \circ y} \circ y\) | Definition of Left Inverse Element | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y_L \circ \paren {y \circ y}\) | as $\circ$ is associative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle y_L \circ y\) | Product of Semigroup Element with Left Inverse is Idempotent: $y = x \circ x_L$ | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e_L\) | Definition of Left Inverse Element |

So $x \circ x_L = e_L$, and $x_L$ behaves as a right inverse as well as a left inverse with respect to the left identity.

$\blacksquare$

## Also see

- Left Identity while exists Left Inverse for All is Identity
- Right Identity while exists Right Inverse for All is Identity

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.4$ - 1978: Thomas A. Whitelaw:
*An Introduction to Abstract Algebra*... (previous) ... (next): Chapter $5$: Semigroups: Exercise $5$