# Show that a semigroup with $aS \cup \{a\} = bS \cup \{b\}$ and $Sa \cup \{a\} = Sb \cup \{b\}$ is a group

Let $S$ be a semigroup such that for all $a,b \in S$ $$aS \cup \{a\} = bS \cup \{b\} \quad \text{and} \quad Sa \cup \{a\} = Sb \cup \{b\}.$$ where $aS = \{as : s \in S \}$ and similarly $Sa$. I want to show that $S$ is a group.

The above conditions imply $$a \in bS$$ for $a \ne b$, hence $S \setminus\{b\} \subseteq bS$ for all $b \in S$. If $S$ contains an idempotent $e$, then as $e \in eS$ we have $eS = S = Se$ and $e$ is the identity. With an identity element it is easy to see that $aS = S$ and $S = Sa$ for every $a \in S$, hence that $S$ is a group.

So, if I could show that $S$ contains an idempotent then it follows that $S$ is a group. But here I am stuck, so could anyone provide a hint how to proceed?

By the way, this is an exercise from J. Howie Fundamentals of Semigroup Theory (page 61).

• Why is this downvoted? – Maria Mazur Jul 31 '18 at 12:51
• Before downvoting please leave a comment what I could improve (or add a comment to your downvote at least)... I dont know either why this was downvoted! – StefanH Jul 31 '18 at 12:55
• @DerekHolt That being said, if the purpose of actively downvoting is, as far as I can tell, to make this site a better place, then leaving a comment is the only way to teach the poster what they can do better. So downvoting without a comment serves little purpose except possibly scare new users away from the site (which, if I'm allowed to be cynical, might be a goal to some). – Arthur Jul 31 '18 at 13:33
• I don't know much about semigroups, but i had a quick look at the book. Lemma 2.2.1 seems to say that multiplying by $s$ on the right is a bijection of $S$, and similarly on the left. So, for $a \in S$, there exists $e$ with $ae=a$, and then $ae^2=ae$, so $e^2=e$, and $e$ is an idempotent. – Derek Holt Jul 31 '18 at 14:01
• @DerekHolt Okay, yes by my assummption there is just one $\mathcal L$-class! So this is fine, and also by this cancellability follows. – StefanH Jul 31 '18 at 14:06

As pointed out in the comments by DerekHolt, the key is Green's Lemma (as cited in the mentioned book on page 49). To formulate it, we start with an arbitrary semigroup $S$, and we define equivalence relations $\mathcal R$ and $\mathcal L$ on $S$ by setting \begin{align*} a \mathcal R b & :\Leftrightarrow \{a\} \cup aS = \{ b \} \cup bS, \\ a \mathcal L b & :\Leftrightarrow \{a\} \cup Sa = \{ b \} \cup Sb \end{align*} for all $a, b \in S$. These are part of what is known as Green's relations. Then we have:

(Green's Lemma) If $a \ne b$ and $a \mathcal R b$ with $as = b, a = bt$, then the maps $$x \mapsto xs \quad \mbox{and} \quad x \mapsto xt$$ are mutually inverse bijections between the $\mathcal L$-classes of $a$ and $b$, which preserve $\mathcal R$-classes. A similar (dual) relation holds if $a\mathcal L b$.

Proof: Its easy to see that $\mathcal L$ is a left congruence and $\mathcal R$ is a right congruence. Hence if $x\mathcal L a$ then $xs \mathcal L as$, so $xs \mathcal L b$ because $b = as$. Similarly, right multiplication by $t$ maps the $\mathcal L$-class of $b$ into the one of $a$. So the two maps between the $\mathcal L$-classes are well-defined. Also, if $x = ua$ then $$xst = uast = ubt = ua = x$$ and similarly $yts = y$ for all $y \mathcal L b$; hence they are mutually inverse on the corresponding $\mathcal L$-classes. Also, as $\mathcal R$ is a right congruence, the images of $\mathcal R$-related elements stay $\mathcal R$-related, but further as $x = (xs)t$ for $x \mathcal L a$ we have $x \mathcal R xs$ and so the mapping stays in the same $\mathcal R$-class for all elements $\mathcal L$-equivalent to $a$. $\square$

The assumption of the question gives that we just have a single $\mathcal R$-class and a single $\mathcal L$-class. So pick arbitrary $a \ne b$ and $as = b$ and $a = bt$, then for every $x \in S$ by the above we have $$xst = x$$ giving that $e := st$ is idempotent (set $x := e$ above) and a right-identity. Further as the equations $ux = e$ and $xu = e$ are solvable for each $u \in S$ it must be a group.

Remark 1: With further theory from the book it is shown that every non-empty intersection of an $\mathcal L$-class and an $\mathcal R$-class is a group.

Remark 2: In the paper Intersections of maximal ideals in semigroups by P. Grillet I found

An ideal $I$ is maximal iff $S - I$ is a $\mathcal J$-class.

And the same proof works in case of right-ideals and $\mathcal R$-classes and left-ideals and $\mathcal L$-classes. So if $aS = S \setminus \{a\}$, then as $aS$ is a right ideal it is maximal and $\{a\}$ would be a single $\mathcal R$-class, which is excluded. Hence $aS = S$ for all $a \in S$. Similar $Sa = S$ for all $a \in S$ so that $S$ is a group. And the result follows without Green's lemma.

Your question is indeed related to Green's relations. The hypothesis tells you that all the elements of your semigroup are $\mathcal{R}$-equivalent and $\mathcal{L}$-equivalent. Therefore, there are all $\mathcal{H}$-equivalent. It now suffices to apply the following standard result (see for instance [1, Proposition 1.4]).

Let $H$ be an $\mathcal{H}$-class of a semigroup. The following conditions are equivalent:

1. $H$ contains an idempotent,
2. there exist $s$, $t \in H$ such that $st \in H$,
3. $H$ is a group.

[1] P.A. Grillet, Semigroups, an introduction to the structure theory, Marcel Dekker, Inc., New York, 1995