Right Identity while exists Right Inverse for All is Identity
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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
Sources
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.4$: Lemma $4$