# Right Identity while exists Right Inverse for All is Identity

## Contents

## Theorem

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

- $\forall x \in S: \exists x_R: x \circ x_R = e_R$

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

Then $e_R$ is also a left identity, that is, is an identity.

## Proof

Let $x \in S$ be any element of $S$.

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

Then:

\(\displaystyle e_R \circ x\) | \(=\) | \(\displaystyle \paren {x \circ x_R} \circ x\) | Definition of Right Inverse Element | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \circ \paren {x_R \circ x}\) | as $\circ$ is associative | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle x \circ e_R\) | $x_R$ is a Right Inverse | ||||||||||

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

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

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

$\blacksquare$

## Also see

## Sources

- 1966: Richard A. Dean:
*Elements of Abstract Algebra*... (previous) ... (next): $\S 1.4$: Lemma $4$