# Left inverse property implies two-sided inverses exist

## Statement

Suppose $(L,*)$ is a loop with neutral element $e$. Suppose $L$ is a left inverse property loop, i.e., there is a bijection $\lambda:L \to L$ such that for every $a,b \in L$, we have: $\! \lambda(a) * (a * b) = b$

Then, $\lambda(x)$ is the unique two-sided inverse of $x$ (in a weak sense) for all $x \in L$: $\! \lambda(x) * x = x * \lambda(x) = e$

Note that it is not necessary that the loop be a right-inverse property loop, so it is not necessary that $\lambda(x)$ be a right inverse for $x$ in the strong sense.

## Proof

Given: A left-inverse property loop $L$ with left inverse map $\lambda$.

To prove: $\! \lambda(x) * x = x * \lambda(x) = e$, where $e$ is the neutral element.

Proof: Putting $a = x, b = e$ in the left inverse property condition, we obtain that $\lambda(x) * x = e$.

Next, putting $a = x, b = \lambda(x)$, we obtain that: $\! \lambda(x) * (x * \lambda(x)) = \lambda(x)$

Writing the $\lambda(x)$ on the right as $\lambda(x) * e$ and using cancellation, we obtain that: $\! x * \lambda(x) = e$

This completes the proof.