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:

\(\displaystyle x \circ e_L\) \(=\) \(\displaystyle x \circ \paren {x_L \circ x}\) $\quad$ Definition of Left Inverse Element $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left({x \circ x_L}\right) \circ x\) $\quad$ as $\circ$ is associative $\quad$
\(\displaystyle \) \(=\) \(\displaystyle e_L \circ x\) $\quad$ $x_L$ is a Right Inverse $\quad$
\(\displaystyle \) \(=\) \(\displaystyle x\) $\quad$ Definition of Left Identity $\quad$


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