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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

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    $\begingroup$ If by regular you mean that for every $a\in M$ there exists $x\in M$ such that $axa=a$, then take $M$ to be a monoid obtained from a nontrivial left zero semigroup by adjoining a unit element. $\endgroup$ – Keith Kearnes Jun 29 '17 at 12:15
  • $\begingroup$ No, by regular I mean "every element is regular", that is ax=ay implies x=y and xa=ya implies x=y for all a. $\endgroup$ – Tig la Pomme Jul 11 '17 at 15:02
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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.

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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.

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  • $\begingroup$ Thank you, but it doesn't work : aa=ab so, if every element where regular, one should have a=b. $\endgroup$ – Tig la Pomme Jul 11 '17 at 15:03
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    $\begingroup$ I am afraid you have a wrong definition of a regular element in a semigroup. Please look at the definition of a Regular semigroup and reconsider your comment (and your vote). $\endgroup$ – J.-E. Pin Jul 11 '17 at 15:08
  • $\begingroup$ Sorry, this Wikipedia page seems to be erroneous! $\endgroup$ – Tig la Pomme Jul 11 '17 at 15:46
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    $\begingroup$ This wikipedia page is perfectly correct. Take any book of semigroup theory and you will find the same definition. $\endgroup$ – J.-E. Pin Jul 11 '17 at 16:20

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