Left inverse property implies two-sided inverses exist

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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.

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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.