
Epigroups 347
epigroups is negative. Moreover, unipotently partitionable epigroups cannot be characterized
in any way in terms of forbidden epidivisors or Rees epidivisors. The fact is that the class of
such epigroups (even the finite ones) is not closed under Rees quotient semigroups. This is
demonstrated by the following example.
3.13 Example Let S = c, d |cdc = c, dcd = d, c
2
d = cd
2
= c
2
,d
2
c = dc
2
= d
2
.Itis
easy to see that S has six elements and four torsion classes, all of which are subsemigroups:
{c, c
2
}, {d, d
2
}, {cd},and{dc}. The subset I = {c
2
,d
2
} is an ideal of S,andS/I
∼
=
B
2
.
The largest homomorphically closed subclass of the class of unipotently partitionable
epigroups, and only this subclass, has the divisor characterization just mentioned; it can be
also transparently characterized from slightly different natural points of view. This is done
in the following theorem proved in [104, §2]; the equivalence of conditions (1), (2a)–(2c),
(3a)–(3c) was already stated in [96].
3.14 Theorem The following conditions on an epigroup S are equivalent:
(1) all homomorphic images of S are unipotently partitionable;
(2a) in any homomorphic image S
with zero, the set Nil S
is a subsemigroup;
(2b) for any ideal I of S,theset
√
I is a subsemigroup;
(2c) in any Rees epidivisor H
of S,thesetNil H
is a subsemigroup;
(3a) there is no semigroup B
2
among the epidivisors of S;
(3b) there is no semigroup B
2
among the Rees epidivisors of S;
(3c) there is no subsemigroup B
2
in any homomorphic image of S;
(3d) for any ideal I of S, there are no subsemigroup B
2
in S/I;
(4) for any subepigroup H of S, Reg H ⊆Gr H;
(5) for any e, f, i ∈ E
S
, e |i and f |i imply ef |i or fe|i.
The fact that just epidivisors but not arbitrary divisors appear in conditions (3a), (3b)
is essential. For (3a) this is obvious: groups trivially satisfy the conditions of Theorem 3.14,
but any semigroup is a divisor of a free group of suitable cardinality. The following example
shows that also in condition (3b) epidivisors cannot be replaced by arbitrary divisors.
3.15 Example Let S be the Rees matrix semigroup M
2
G;2, 2;
0
eg
−1
g
−1
e
13
,whereG =
g is the infinite cyclic group with identity element e.Putc =(1,g,1), d =(2,g,2),
Q = {(i, g
n
,λ) |i, λ ∈{1, 2},n>1}. It can be shown by a direct calculation that Q is an
ideal of the subsemigroup c, d, cdc = c, dcd = d, c
2
,d
2
∈ Q, c, d\Q = {c, d, cd, dc},and
therefore c, d/Q
∼
=
B
2
.ThusB
2
occurs among the Rees divisors of S.ButS satisfies the
conditions of Theorem 3.14, which can be seen at once, for instance, from condition (2b),
since S, being a completely simple semigroup, contains no proper ideals.