Conditions for a regular semigroup to be a group.

Question

This is Exercise 2.6.11 of Howie's "Fundamentals of Semigroup Theory".

The Details:

Let $S$ be a semigroup.

Definition 1: We say $S$ is regular if for all $a\in S$ there exists an $x$ in $S$ such that $a=axa$.

Definition 2: We say $S$ is cancellative if for all $a, b, c\in S$, $$ac=bc\implies a=b$$ and $$ca=cb\implies a=b.$$

The Question:

Let $S$ be a regular semigroup. Show that the following are equivalent:

(a) $S$ has exactly one idempotent.

(b) $S$ is cancellative.

(c) $S$ is a group.

My Attempt:

Suppose $S$ is regular.

Showing (a) implies (b):

Suppose (a). Suppose $ac=bc$ for $a, b, c\in S$. We have $c=cxc$ for some $x$, so that $cx=cxcx=e$ and similarly $xc=e$. Then $acx=bcx$ implies $ae=be$.

How do I show that $a=b$? The proof of the left cancellative property should be dual to that of the right.

Showing (b) implies (c):

Suppose (b). Then what?

Showing (c) implies (a):

Suppose (c). Suppose $a^2=a$. Then $a=(aa)a^{-1}=aa^{-1}=1_S$. Thus the only idempotent in $S$ is the identity $1_S$. Thus (a). $\square$

Please help.

Asinomás · Answer 1 · 2017-05-15T14:37:01.527

2

Let $e$ be the unique idempotent.

For every $c$ there is an $x$ so that $cxc=c$, this means $cx=cxcx$, we conclude $cx=e$, and so $c=cxc=ec$.

Hence $e$ is a left identity, you can show analogously that $e$ is right identity, so $e$ is identity.

We also showed that $cx=e$ and analogously $xc=e$. So in fact we showed $a) \implies c)$ directly.

edited May 15 '17 at 14:37

answered May 15 '17 at 14:30

Asinomás

107,565

This only shows that (a) implies (c). – Shaun May 15 '17 at 14:48
So now I have that (a) if and only if (c), what about (b)? – Shaun May 15 '17 at 14:56
Oh sorry, I thought it was trivial how to adapt it. Let me add it in a minute. – Asinomás May 15 '17 at 15:02
(c) implies (b) is very easy! – Derek Holt May 15 '17 at 15:16
If I'm honest b implies c is making me struggle – Asinomás May 15 '17 at 15:21
Ok, I think I got it. – Asinomás May 15 '17 at 15:38
nevermind.${}{}{}$\ – Asinomás May 15 '17 at 16:14

score 1 · Accepted Answer · answered May 15 '17 at 21:09

OK, so Jorge has proved that (a) implies (c), and (c) implies (b) is easy, so it remains to prove (b) implies (a).

Actually, for this I think we need also to assume that $S$ is nonempty, because the empty semigroup satiisfies (b), but not (a) or (c). Assuming that, there exists $a \in S$ and $x \in S$ with $a = axa$, so $ax$ is an idempotent, and hence there is at least one such.

Now, if $e$ and $f$ are both idempotents, then we have $ef= eeff$, so cancelling the $f$ on the right gives $e = eef=ef$, whereas cancelling the $e$ on the left gives $f = eff = ef$, so $e=f$, giving (a).

score 1 · Answer 3 · answered Mar 09 '21 at 15:36

A "direct" proof for (b) $\rightarrow$ (c) (using quasigroup + semigroup $\leftrightarrow$ group):

This is to show that $\forall a, b\in S (\exists! x$ s.t. $ax=b)$ for a given semigroup $S$.

It suffices to show that $\forall a, b\in S (\exists x$ s.t. $ax=b)$ since the uniqueness is already given by $S$ being cancellative.

Let $p$ be an inner inverse of $a$: $apa = a$, since the semigroup $S$ is regular, such $p$ always exists. Also note that $ap$ is idempotent.

Now we have $(ap)b = (apap)b = (ap)(apb) \rightarrow b = apb = a(pb)$ so $x=pb$ is a desired solution.

Conditions for a regular semigroup to be a group.

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