Left Inverse for All is Right Inverse

Jump to: navigation, search

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:

 $\displaystyle e_L \circ y$ $=$ $\displaystyle \paren {y_L \circ y} \circ y$ Definition of Left Inverse Element $\displaystyle$ $=$ $\displaystyle y_L \circ \paren {y \circ y}$ as $\circ$ is associative $\displaystyle$ $=$ $\displaystyle y_L \circ y$ Product of Semigroup Element with Left Inverse is Idempotent: $y = x \circ x_L$ $\displaystyle$ $=$ $\displaystyle e_L$ 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.

$\blacksquare$