Conditions for a regular semigroup to be a group. 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.
 A: 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.
A: 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).
A: 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.
