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We know that a semigroup with a right identity and right inverse for all elements is a group (e.g. see here). Symmetrically, also a left identity together with a left inverse implies a group. We also know that a semigroup with a right identity and a left inverse is NOT necessarily a group (see here). My questions are:

  1. in a semigroup, is the existence of a UNIQUE right identity together with the existence of a left inverse enough to have a group?
  2. in a semigroup, is the existence of a right identity together with the existence of a UNIQUE left inverse enough to have a group?

I think both these claims are false, ut haven't found a counter-example so far.

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  • $\begingroup$ Related: mathoverflow.net/questions/27720/… $\endgroup$ Mar 26, 2019 at 16:21
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    $\begingroup$ The standard example in your link shows that (2) does not hold. Let $X$ be a set with more than one element, $e\in X$ one of them. Define $ab=a$ for all $a,b\in X$. Then $ae=a$ for all $a$, so $e$ is a right identity. And for any $a\in A$, the unique inverse of $a$ in $A$ relative to $e$ is $e$, for $xa=e$ if and only if $x=e$. But this is not a group. $\endgroup$ Mar 26, 2019 at 16:26
  • $\begingroup$ @ArturoMagidin It's not clear to me what "left inverse" means if the right identity is not unique. Is this a standard notion? If we interpret "$x$ is a left inverse of $a$" to mean "$xa$ is a right identity", then "every element has a unique left inverse" does imply "there is unique right identity". $\endgroup$
    – bof
    Sep 14, 2020 at 2:38
  • $\begingroup$ @bof: it would mean a left inverse relative to a given identity; if there are multiple identities, then you would need to have an inverse relative to each of them, which that example does have. Note that you are replying to a comment that is well over a year old. $\endgroup$ Sep 14, 2020 at 3:10

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As has been shown in a comment, (2) does not hold. However, (1) does.

Let $e$ be the unique right identity, and for any $x$, let $x'$ denote a left inverse.

For any $x$, $$ e = x''x' = x''ex' = x''x'xx' = ex x'. $$ Hence, for any $y$, $$ y = ye = yexx' = y xx', $$ which shows that $xx'$ is a right identity. Since it is unique, $xx' = e$. Hence every element $x$ has a two-sided inverse.

Finally, for any $x$, $$ ex = xx'x = xe = x, $$ so $e$ is a two-sided identity and the semigroup is a group.

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