Non-commutative regular monoid Good afternoon,
Does anybody have an example of a non-commutative regular monoid which cannot be embeded in any group, please? I cannot find one by myself.
Thanks
 A: The following terminology is standard in semigroup theory. An element $a$ of a  semigroup or monoid $M$ is left cancellative if $ax=ay$ implies $x=y$ for all $x, y$, and is right cancellative if $xa=ya$ implies $x=y$ for all $x, y$. It is cancellative if it is both left and right cancellative. $M$ is cancellative if all of its elements are. An element $a\in M$ is regular if there exists an $x\in M$ such that $axa=a$, and $M$ is regular is all of its elements are.
The following terminology is standard in ring theory. An element $a\in R$ is regular if it is cancellative in the multiplicative monoid of $R$. 
It is important to agree on a standard terminology, especially for this problem, since it is known that a monoid is a group iff it is cancellative + regular in the semigroup sense. [If you are willing to mix terminology, a monoid is a group iff it is regular + regular!]

The original question appears to be this: Is there a (noncommutative) cancellative monoid that is not embeddable in a group?
The answer to this question is ``yes''. The famous paper where the answer appeared is 
Malcev, A.
Uber die Einbettung von assoziativen Systemen in Gruppen. Rec. Math. N.S. 6 (48), (1939). 331–336.
Malcev's solution is summarized on Wikipedia.
The Wikipedia page does not describe a particular example, but here is one: the monoid $M$ with the presentation $\langle a, b, c, d, a', b', c', d'\;|\;ab=a'b', cb=c'b', cd=c'd'\rangle$. $M$ is cancellative but not embeddable in a group. It takes some work to show (i) $M$ is cancellative, (ii) $ad\neq a'd'$ in $M$, but (iii) any semigroup homomorphism of $M$ into a group must map $ad$ and $a'd'$ to the same group element.
It may be interesting to note that the variety generated by a cancellative monoid that is not embeddable in a group is the variety of all monoids. Said another way, if $S$ is a cancellative monoid that satisfies a proper monoid identity, then $S$ is embeddable in a group.
A: There are two minimal examples, both defined on the three-element set $\{1, a, b\}$. The product on the first one is defined by the rules $aa = ab = a$ and $bb = ba = b$. The second one is its dual version, defined by $aa = ba = a$, $bb = ab = b$. Note that both examples are not only regular, but also idempotent.
