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On the Endomorphism Semigroup of Simple Algebras*

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On the Endomorphism Semigroup of Simple Algebras* GRATZER, G. Math. Annalen 170,334-338 (1967) On the Endomorphism Semigroup of Simple Algebras* GEORGE GRATZER * The preparation of this paper was supported by the National Science Foundation under grant GP-4221. i. Introduction. Statement of results Let (A; F) be an algebra andletE(A ;F) denotethe set ofall endomorphisms of(A ;F). Ifep, 1.p EE(A ;F)thenwe define ep •1.p (orep1.p) asusual by x(ep1.p) = (Xlp)1.p. Then (E(A; F); -> is a semigroup, called the endomorphism semigroup of (A; F). The identity mapping e is the identity element of (E(A; F); .). Theorem i. A semigroup (S; -> is isomorphic to the endomorphism semigroup of some algebra (A; F) if and only if (S; -> has an identity element. Theorem 1 was found in 1961 by the author and was semi~published in [4J which had a limited circulation in 1962. (The first published reference to this is in [5J.) A. G.WATERMAN found independently the same result and had it semi~ published in the lecture notes of G. BIRKHOFF on Lattice Theory in 1963. The first published proofofTheorem 1 was in [1 Jby M.ARMBRUST and J. SCHMIDT. The special case when (S; .) is a group was considered by G. BIRKHOFF in [2J and [3J. His second proof, in [3J, provides a proof ofTheorem 1; namely, we define the left~multiplicationsfa(x) == a . x as unary operations on Sand then the endomorphisms are the right~multiplicationsxepa = xa. We get as a corollary to Theorem 1BIRKHOFF'S result. Furthermore, noting that if a (ES) is right regular then fa is 1-1, we conclude Corollary 2. A semigroup (S; .) is isomorphic to the semigroup of all 1-1 endomorphisms of an algebra if, and only if, it has an identity element and it is right regular. The problem of characterizing the semigroups of 1-1 endomorphisms and onto endomorphisms within the endomorphism semigroup was solved by M. MAKKAI [5]. In 1961 the author observed that there exist simple algebras 1 with a given automorphism group and there exist algebraswith oneelement automorphism group and arbitrary congruence lattice. The question was raised whether the automorphismgroupis independentofthecongruencelattice. Apositiveanswer was given by E. T. SCHMIDT [6]. 1 Let A be a set; the binary relations ro and I on A are defined by the rules: x == y(ro) iff x "" y ; X ==Y(I) for all x, yE A. An algebra <A; F> is simple if ro and I are the only congruence relations of <A; F>. The Endomorphism Semigroup 335 The next question is the independence of the congruence lattice and of the endomorphism semigroup. A negative answer to this question is given by the following result. Theorem 3. A semigroup (S,.·) is isomorphic to the endomorphism semigroup of some simple algebra (A,. F) if and only if (S,. .) has an identity element and every element in S is either right regular or a right annihilator. Let us assume that <S; .) satisfies the conditions of Theorem 3, and let R and N denote the set of right regular elements, and right annihilators, respec­ tively. Then (R; -) and <N; -) are subsemigroups of <S; -). Further, (R; -) is a right regular semigroup with identity element and <N; .) is a right annihila­ tor semigroup also with identity. For n E N, we can define a function fn(x) for R into N by fn(r) = nr. Then (a) fn(r) = fm(r) implies n = m; (b) if fn(r t ) = m, then fn(r t rz) = fm(r z). Theorem 4. Let <R,. .) be a right regular semigroup with identity element and (N,. .) be a right annihilator semigroup with identity such that Rand N have only one element, the identity, in common and for every n E N let fn be a mapping of R into N satisfying (a) and (b). Then <R v N,. -) can be made into a semigroup satisfying the conditions of Theorem 3 by keeping the operations in Rand N, and setting r' n = n, n' r = fn(r), for n E N, r E R. Conversely, every semigroup satisfying the conditions of Theorem 3, can be so constructed in one and only one way (up to isomorphism). Corollary 5. Let <R,. .) be a right regular semigroup with identity and (N,. .) be a right annihilator semigroup with identity. Then there exists a simple algebra (A,. F) such that <R,. -) and <N ,. -) are isomorphic to the semigroup of 1-1, and onto endomorphisms of <A,. F) respectively. By Theorems 3 and 4 all we have to do is to define the fn. Set fn(r) = n, for all r E R. Then both (a) and (b) are trivial, and the statement follows. It should be noted that in the proof of Theorem 1 we used only unary operations while in the proof of Theorem 3 we use a binary operation. The reason for this is that Theorem 3 fails to hold for unary algebras. Theorem 6. Let <A,. F) be a simple unary algebra and let (G,. .) be the automorphism group of (A,. F). Then (G ,. .) is a cyclic group of prime order p and if p '* 1 then A also has p elements. In conclusion, I mention a few open problems: Problem 1. Characterize the endomorphism semigroups of simple unary algebras. Problem 2. Characterize the endomorphism semigroups of algebras with a small congruence lattice (e.g. 3, 4, 5 element congruence lattices). Problem 3. Find the connection between the congruence lattice and the endomorphism semigroup of an algebra. 336 G. GRATZER: Problem 4. Let Iff be a semigroup ofmappings ofa set A into itself, containing the identical mapping. Under what conditions is Iff the semigroup of all endo­ morphisms of an algebra <A; F)? Remark. It is pointed out in [1] that some difficulties arise if A is infinite. However, the finite case is even more involved. Counter examples were found by J. HARTMANIS and M. MAKKAI (oral communications). 2. Proofofthe results To prove Theorem 3 let (A; F) be an algebra, IpEE(A; F). Let 8" be the relation on A defined by x == y(8,,) ifXlp = Yip. Then 8" is a congruence relation of <A; F). Thus if <A; F) is simple, we have 8,,=w or I. Obviously, 8,,=w iff Ip is 1- 1 and 8" = 1 iff A Ip consists of a single element. Let R consist of the 1 1 endomorphisms and let N consist of the endomorphisms with a one element range. Then E(A; F) = R v N and the elements in R are right regular, those in N are right annihilators. This proves the "only if" part of Theorem 3. To prove the converse, let S = R v N, where R consists of the right regular elements and N of the right annihilators of S. Let e denote the identity of S; then eE R. If S = {e}, the statement is trivial so we assume that S::j::: {e}. Set A = S v {a, 1}, °¢ S, 1¢ s, o::j::: 1. We define the following operations on A (my original construction contained two additional operations; one of them was eliminated by G. WENZEL, and the second, later on, by myself): p :p(x)=x for XES p(O) = 1 p(1)=0; fa' for aES :fa(x)=ax, for XES fa(x)=x, for x¢S; :xnx=x xny=O, ifx,yES and x::j:::y 1nx=xn1 =x, for XEA Onx=xnO=O, forxEA. Set F = Uala ES} V {n, p}. To show that <A; F) is simple, we first observe that if 8 is a congruence relation and°== 1(8)then for any x, YEA, x = x n 1 == x nO = 0(8)and similarly, y==0(8), thus x==y(8), i.e. 0=1. Now let 8 be any congruence relation ::j:::w, Le. X == y(8) for some x::j::: y. If {x, y} = {a, I} then 8 = 1 was proved above. So let us assume that x ¢ {O, 1}. If y ¢ {O, 1}, then x == xny(8) and xny = 0. Thus we may assume y E {O, 1}, e.g. y =0. Then x == O(8) implies x =p(x) == p(O) 1(8), thus 0== 1(8), completing the proof. Now we define mappings on A: Ipa for a E R: Xlpa=xa if XES; Xlpa=X ifx¢S. Ipa for aEN :Xlpa=a, for all xEA. The Endomorphism Semigroup 337 It should be noted that on S these definitions coincide with the ones given in the sketched proof of Theorem 1. This implies: (i) If qJ EE(A ; F) and qJ maps S into S then qJ restricted to S coincides with one of the qJa restricted to S. (ii) If qJ EE(A; F) and qJ is 1-1, then OqJ = 0 and 1qJ = 1. Indeed, 1qJ = a =F 1 would imply xqJ=(xn1)qJ=xqJn1qJ=xqJna, thus xqJ=a or 0, a contra­ diction. (iii) If qJ EE(A; F) and qJ is 1-1, then qJ = qJa for some a ER. Let XES; then XqJ is in S; otherwise XqJ = OqJ or XqJ = 1qJ. Thus <p maps S into S and by (i) qJ = qJa on S. Obviously, a ER, otherwise qJa would not be 1-1. By (ii), OqJ = 0 and 1qJ = 1, thus qJ = qJa on the whole A. (iv) If qJ EE(A; F) and qJ is not 1-1, then AqJ consists of one element. This follows from the fact that (A; F) is simple. (v) Let AqJ = {a}. Then a EN.
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