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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.

Shaun
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1 Answers1

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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.

  • Maybe I am misinterpreting your meaning, but I don't see that such a $P$ necessarily satisfies the required conditions. For instance if $P$ consists of reflections of a square about each of its diagonals, then $aba = b$, not $a$. – Paul Sinclair Sep 09 '21 at 14:00
  • @PaulSinclair. If a is one of the reflections then the unique element x such that axa=a is the element a itself. Therefore the condition is satisfied. –  Sep 10 '21 at 15:39
  • I see. I hadn't misinterpreted your meaning, but also hadn't thought deeply enough. Thanks for the explanation. – Paul Sinclair Sep 10 '21 at 15:48