1
$\begingroup$

An element $x$ of a semigroup $S$ is called regular provided that there exists $y\in S$ such that $xyx=x$. $S$ is called regular if all its elements are regular. Let $S$ be a monoid with identity element $1$. An element $a\in S$ is called invertible if there exists $b\in S$ such that $ab=ba=1$. The set of all inverse elements of the monoid $S$ is denoted by $S^{\star}$.

Are there examples of monoids that are non-regular and have more than one invertible element, particularly related to transformation semigroups?

$\endgroup$

2 Answers 2

5
$\begingroup$

The integers under multiplication has two invertible elements, yet it only has three regular elements.

The dyadic rationals (the rational numbers whose denominator is a natural power of $2$) under multiplication has infinitely many invertible elements (any integral power of $2$), yet any dyadic rational with an odd prime factor in its numerator is non-regular.

$\endgroup$
2
$\begingroup$

Minimal example. Let $M = \{1,a,b,0\}$ be the monoid in which $1$ is the identity, $0$ is a zero, $a^2 = 1$, $b^2 = 0$ and $ab = ba = b$. The group of units of $M$ is $\{1, a\}$, the cyclic group of order $2$ and $b$ is the unique non-regular element.

To obtain $M$ as a transformation monoid, just take its right representation. \begin{array}{c|c|c|c|c|} &1&2&3&4\\ \hline a&2&1&3&4\\ \hline b&3&3&4&4\\ \hline \end{array}

$\endgroup$

This site is temporarily in read-only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .