# Left Identity while exists Left Inverse for All is Identity

## 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 $e_L$ is also a right identity, that is, is an identity.

## Proof

From Left Inverse for All is Right Inverse we have that $x \circ x_L = e_L$.

Then:

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

\(\displaystyle \) | \(=\) | \(\displaystyle \left({x \circ x_L}\right) \circ x\) | as $\circ$ is associative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle e_L \circ x\) | $x_L$ is a Right Inverse | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x\) | Definition of Left Identity |

So $e_L$ behaves as a right identity as well as a left identity.

That is, by definition, $e_L$ is an identity element.

$\blacksquare$

## Also see

## 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 $5$