In contrast to the semilattice of groups case, an inverse semigroup S which is the union of strongly E-reflexive inverse subsemigroups need not be strongly E-reflexive. If, however, the union is saturated with respect to the Green's relation 6D, and in particular if the union is a disjoint one, then S is indeed strongly E-reflexive. This is established by showing that 6) -saturated inverse subsemigroups have certain pleasant properties. Finally, in contrast to the E-unitary case, it is shown that the class of strongly E-reflexive inverse semigroups is not closed under free inverse products. The reader is referred to [1], [2] for the basic theory of inverse semigroups, including the theory of free inverse products. Recall from [4], [5] that an inverse semigroup S is said to be strongly E-reflexive whenever S is a semilattice of E-unitary inverse semigroups, or alternatively, whenever there exists a semilattice of groups congruence on S such that only idempotents are linked to idempotents under -. In [4], [5] we studied this class of semigroups and showed that many of the properties of semilattices of groups and of E-unitary inverse semigroups generalise to this class, albeit sometimes in a weaker form. We continue this line of investigation here. In what is by now a classic theorem, Clifford showed that an inverse semigroup which is a union of groups is a semilattice of groups. We ask to what extent this is true for strongly E-reflexive inverse semigroups. It is already known that a semilattice of strongly E-reflexive inverse semigroups is again strongly E-reflexive [5]. The following simple example shows that we cannot hope for a full generalisation of Clifford's theorem. Consider the bisimple inverse o-semigroup S(G, a), where the endomorphism a of the group G is not injective. As noted in [4, p. 341], S(G, a) is not strongly E-reflexive. However, using [1, Lemma 1.31], it is easily seen that S(G, a) is a union of its maximal subgroups and copies of the bicyclic semigroup, and these are all E-unitary. The restriction we require will now be given, and the example just noted would seem to indicate that it is the weakest possible. Let S be an inverse semigroup with semilattice of idempotents E. Let U be an inverse subsemigroup of S which is 6D -saturated in the sense that x 6D y' E U implies x E U, where 6D denotes the usual Green's relation on S. The maximal group homomorphic image of U is denoted by U with udenoting the image of u (u E U). Let U' = {x E SIx > u for some u E U); note that U' may equal S. Received by the editors March 23, 1979. AMS (MOS) subject classifications (1970). Primary 20M10. ? 1980 American Mathematical Society 0002-9939/80/0000-0301 /$0 1.75 352 This content downloaded from 207.46.13.148 on Sun, 11 Sep 2016 04:20:48 UTC All use subject to http://about.jstor.org/terms STRONGLY E-REFLEXIVE INVERSE SEMIGROUPS 353 The first result shows that U' has some pleasant properties. PROPOSITION. (i) U' is an inverse subsemigroup of S which contains U, and xy E U'implies x E U'andy E U'. (ii) The rule: xp = uif x > u E U and x4 = 0 otherwise, gives a well-defined homomorphism 0: S U? such that k1 U is the canonical homomorphism onto U. PROOF. (i) xy > U E U = xx > xyy 1x uu X1 x > uu x R. u =X x E U', since U is 6D -saturated and 6R C 6D. Dually, y E U'. The remainder of the result is easily proven. (ii) Suppose x E U' with x > u E U and x > v E U. Then u = ex, v = fx where e=uuEUnE,f=vv-1EUnE. Hence efu=efv, and ef6EEn U, so that u= . It is then almost immediate that 0 is well-defined. The rest of the result involves a little routine calculation, using (i). REMARK. Taking S to be a semilattice with more than two elements, we see that U need not be an ideal of U' in Proposition 1. The proposition enables us to prove our main result. THEOREM. Let S be a union of 6D -saturated strongly E-reflexive inverse subsemigroups Si, i E I. Then S is strongly E-reflexive. PROOF. Each Si is a semilattice Ai of E-unitary inverse semigroups T7, X E Ai. It is easily shown that each T/x is 6D-saturated in S. Hence we may suppose without loss of generality that each Si is E-unitary. For each i E I, let Oi: S -> S?o be the homomorphism defined as in (ii) above, and let T be the direct product of the Si?. Then the pi induce a homomorphism 0: S -T with s0 having ith component s5ci, i E I. Now So is a semilattice of groups, since T is. Suppose that x4 = eo for some e E E, where x E Si say. Then x4i is the identity element of Si, and since S5 is E-unitary it follows that x E E; whence the result. COROLLARY. Let S be a disjoint union of strongly E-reflexive inverse subsemigroups. Then S itself is strongly E-reflexive. PROOF. Clearly each of the inverse subsemigroups in question is 6D -saturated in S. REMARK. The elementary theory of inverse semigroups shows that an inverse semigroup S which is a union of groups is a disjoint union of its maximal subgroups He, e E E, and that this is the 6D -decomposition of S. Hence S is the union of the 6D -saturated E-unitary inverse subsemigroups He. It is easy to show that the homomorphism 4 in the proof of the theorem is injective in this case. Hence S is a subdirect product of the He with zero added possibly. From this one can deduce, again by elementary means, that S is a semilattice of groups with the multiplication defined by linking homomorphisms. Thus, modulo some elementary results, our theory restricts to Clifford's classic theorems. Now let E be the semilattice { e, f, g} where e > g, f > g, and e, f are incomparable. Let S be the semilattice of groups Ge U Gf U Gg where Ge' Gg are trivial and Gf is the cyclic 2-group; let T be the semilattice of groups He U Hf U Hg This content downloaded from 207.46.13.148 on Sun, 11 Sep 2016 04:20:48 UTC All use subject to http://about.jstor.org/terms