Left Inverse for All is Right Inverse

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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 $x \circ x_L = e_L$, 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.

$\blacksquare$


Also see


Sources