Left Identity while exists Left Inverse for All is Identity
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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:
\(\ds x \circ e_L\) | \(=\) | \(\ds x \circ \paren {x_L \circ x}\) | Definition of Left Inverse Element | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {x \circ x_L} \circ x\) | Semigroup Axiom $\text S 1$: Associativity | |||||||||||
\(\ds \) | \(=\) | \(\ds e_L \circ x\) | Left Inverse for All is Right Inverse | |||||||||||
\(\ds \) | \(=\) | \(\ds x\) | 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 $5$