Right Identity while exists Right Inverse for All is Identity

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:

\(\ds e_R \circ x\)

\(=\)

\(\ds \paren {x \circ x_R} \circ x\)

Definition of Right Inverse Element

\(\ds \)

\(=\)

\(\ds x \circ \paren {x_R \circ x}\)

Semigroup Axiom $\text S 1$: Associativity

\(\ds \)

\(=\)

\(\ds x \circ e_R\)

Right Inverse for All is Left Inverse

\(\ds \)

\(=\)

\(\ds x\)

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

• Left Identity while exists Left Inverse for All is Identity

Sources

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