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.

Related facts

• Equality of left and right inverses in monoid
• Two-sided inverse is unique if it exists in monoid
• Equivalence of definitions of inverse property loop
• Equivalence of definitions of gyrogroup

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.