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Suppose we have a non-empty set $P$ equipped with an associative binary operation $\bullet$ such that for every $a \in P$ there exists a unique $b \in P$ with $aba=a$. How would we go about proving this is a group?

I have tried various things, and proved some smaller results such as for the element $b$, the corresponding unique element $c$ such that $bcb=b$ satisfies $c=a$, but every attempt to show this structure is in fact a group seems to rely on circular logic that either a unique identity exists, or each element has a unique inverse, both of which we obviously have to prove!

Any help would be much appreciated.

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marked as duplicate by Theo Bendit, nmasanta, Bill Dubuque, José Carlos Santos, Daniele Tampieri Sep 12 at 6:44

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This result is false unless one assumes closure.

For example, let $P$ consist of any two distinct reflections of a regular polygon. Then, under the usual composition of symmetries, all the required conditions are satisfied but $P$ is not a group.

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