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In a set endowed with an associative operation with a left identity and where every element has a left inverse, I was wondering, is every element bound to have a right inverse too? Or is there at least one set with a respective operation and such properties that at least one element doesn't have a right inverse? Clearly, an easy proof shows that if there is a right and a left inverse and the aforementioned properties hold, they have to be equal.

Note: This is merely a question of mine and not somehow part of homework.

Henno Brandsma
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  • The answer is yes, having a left identity and left inverses also implies right identity and right inverses (whereas left identity and right inverse is not enough). It is a good exercise to show this. – Tobias Kildetoft Jul 27 '17 at 08:00
  • See also this related question – Henno Brandsma Jul 27 '17 at 08:08
  • @HennoBrandsma Both the question that mine has been marked as a duplicate of and the one you linked don't prove the existence of a right inverse (in my case, mutatis mutandis in their case) but instead take it for granted... –  Jul 27 '17 at 08:35
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    @Jason my linked one shows it is the same element in the body of the question. Did you read it? Once you show the same $e$ works for both sides, left inverses will give uniqueness, etc. – Henno Brandsma Jul 27 '17 at 08:39
  • @HennoBrandsma Now I understand it better. You can prove existence of a right-inverse of an element because the element is the left-inverse of it... –  Jul 27 '17 at 08:52
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    @Jason You get some $e$ that is a left identity for all $a$. Then you show for all $a$ that $ae=a$ as well. But any such $e$ is unique because we have left inverses for all $a$. Now for any $a$ we have $a^{-1}$ with $a^{-1}a = e$. Then the post shows that also $aa^{-1} = e$. – Henno Brandsma Jul 27 '17 at 08:59

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