Left Inverse for All is Right Inverse

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 $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity.


Proof

Let $y = x \circ x_L$. Then:

\(\ds e_L \circ y\) \(=\) \(\ds \paren {y_L \circ y} \circ y\) Definition of Left Inverse Element
\(\ds \) \(=\) \(\ds y_L \circ \paren {y \circ y}\) Semigroup Axiom $\text S 1$: Associativity
\(\ds \) \(=\) \(\ds y_L \circ y\) Product of Semigroup Element with Left Inverse is Idempotent: $y = x \circ x_L$
\(\ds \) \(=\) \(\ds e_L\) Definition of Left Inverse Element


So $x \circ x_L = e_L$, and $x_L$ behaves as a right inverse as well as a left inverse with respect to the left identity.

$\blacksquare$


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 $5$