# Is a semigroup with unique right identity and left inverse a group?

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.

• Mar 26, 2019 at 16:21
• 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. Mar 26, 2019 at 16:26
• @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".
– bof
Sep 14, 2020 at 2:38
• @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. Sep 14, 2020 at 3:10

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.