Commutativity theorems for groups and semigroups
aa r X i v : . [ m a t h . G R ] N ov COMMUTATIVITY THEOREMS FOR GROUPS AND SEMIGROUPS
FRANCISCO ARA ´UJO AND MICHAEL KINYON ∗ Abstract.
In this note we prove a selection of commutativity theorems for various classesof semigroups. For instance, if in a separative or completely regular semigroup S we have x p y p = y p x p and x q y q = y q x q for all x, y ∈ S where p and q are relatively prime, then S is commutative. In a separative or inverse semigroup S , if there exist three consecutiveintegers i such that ( xy ) i = x i y i for all x, y ∈ S , then S is commutative. Finally, if S is aseparative or inverse semigroup satisfying ( xy ) = x y for all x, y ∈ S , and if the cubingmap x x is injective, then S is commutative. Introduction
Broadly speaking, a commutativity theorem in group theory is any result concluding thata group is commutative, i.e. abelian. Perhaps the best known example is the followingstandard exercise, usually given to students at the beginning of their study of group theory: If G is a group satisfying x = 1 for all x ∈ G , then G is commutative. Commutativity theorems can sometimes be extended to various classes of semigroupsproperly containing groups. For instance, a semigroup S is cancellative if it satisfies theconditions xy = xz = ⇒ y = z and yx = zx = ⇒ y = z for all x, y, z . Every finitecancellative semigroup is a group; the positive integers under addition provide an exampleof a cancellative semigroup which is not a group.The exercise above extends easily to cancellative semigroups once we reinterpret the con-dition “ x = 1”. Since we do not wish to assume the existence of an identity element, wereplace the condition with “ x = x ”, which is clearly equivalent to x = 1 in groups. Proposition 1.1. If S is a cancellative semigroup satisfying x = x for all x ∈ S , then S iscommutative (and in fact, is a group satisfying x = 1 for all x ∈ S ). The parenthetical part of the assertion suggests the proof: if x = x , then x y = xy andso cancelling gives x y = y for all x, y ∈ S . Dually, yx = y for all x, y ∈ S . Thus for all x, y ∈ S , x = x y = y . This constant, which we denote by 1, is an identity element.Hence x = x = x
1, and cancelling gives x = 1 for all x ∈ S . Therefore S is a group andwe have reduced the problem to the original exercise.Still using this elementary example to illustrate our point, further extensions of the resultare possible. A semigroup is separative if it satisfies the conditions xy = xx & yx = yy = ⇒ x = y and xy = yy & yx = xx = ⇒ x = y ([7], Def. II.6.2, p. 51). Every cancellativesemigroup is evidently separative.We also need the notion of a semilattice of semigroups. A semilattice is a partially orderedset ( I, ≤ ) such that every two elements x, y ∈ S have a greatest lower bound, denoted by ∗ Partially supported by Simons Foundation Collaboration Grant 359872 and by FCT project CEMAT-CIˆENCIAS UID/Multi/04621/2013. ∧ y . A semigroup S is a semilattice of semigroups if there exist a semilattice ( I, ≤ ) and a set Y = { S α } α ∈ I of pairwise disjoint subsemigroups S α ≤ S indexed by I such that S = ∪ α ∈ I S α ,and satisfying this property: for all α, β ∈ I and for all a ∈ S α , b ∈ S β , we have ab ∈ S α ∧ β .(For details, see [7], Def. II.1.4, p. 27).For our purposes, the following results are key ([7, Thm. II.6.4, p. 51], [6, Thm. 3.12, p.47]). Proposition 1.2.
Let S be a semigroup. (1) S is separative if and only if S is semilattice of cancellative semigroups. (2) S is commmutative and separative if and only if S is a semilattice of commutativecancellative semigroups. Now we can extend the original exercise even further.
Corollary 1.3.
Let S be a separative semigroup satisfying x = x for all x ∈ S . Then S iscommutative, and in fact, is a semilattice of abelian groups satisfying x = 1 . Indeed, by Proposition 1.2(1), S is a semilattice of cancellative semigroups S α , α ∈ I , suchthat each S α satisfies x = x . By Proposition 1.1, each S α is an abelian group satisfying x = 1. By Proposition 1.2(2), S is commutative.We can also view the generalizations of our exercise from a different perspective. Asemigroup S is regular if for each a ∈ S , there exists b ∈ S such that aba = a . This isequivalent to asserting that each a ∈ S has an inverse a ′ ∈ S satisfying aa ′ a = a and a ′ aa ′ = a ′ . If each a ∈ S has a unique inverse, then S is said to be an inverse semigroup . Equivalently,an inverse semigroup is precisely a regular semigroup in which all the idempotents commute.A semigroup S such that each element has a commuting inverse aa ′ = a ′ a is said to be completely regular . Equivalently, a completely regular semigroup is a union of groups. Acompletely regular, inverse semigroup is called a Clifford semigroup . A Clifford semigroupis characterized as a semilattice of groups. (For further details on regular semigroups, see,for instance, [3].) Note in particular that every Clifford semigroup is separative .This gives us a different way of viewing our exercise. A semigroup satisfying x = x forall x is completely regular, with the commuting inverse of each x being given by x ′ = x .It is easy to see that a regular, cancellative semigroup is a group by essentially the sameargument as above: xx ′ xy = xy and yxx ′ x = yx , so x ′ xy = y and yxx ′ = y for all x, y ∈ S .A semigroup (or more generally, any magma) with both a left identity element and a rightidentity element has a (necessarily unique) identity element 1, and we have xx ′ = x ′ x = 1for all x ∈ S . Thus we have another proof of Proposition 1.2.In this paper, we will extend three commutativity theorems from group theory to semi-groups. Our first result, which was our original motivation, is based on a recent preprint ofVenkataraman [8]. She proved that in a finite groups, if squares commute with squares andcubes commute with cubes then the group is commutative; she also proposed the problemof extending her result to infinite groups. More generally, in the same paper, she asked ifa group satisfying the conditions x p y p = y p x p and x q y q = y q x q for all x where p and q arerelatively prime is necessarily commutative. Although we did not know it when we began ourinvestigation, this is apparently a folk result in group theory [9], although we have not beenable to find a reference in the literature. (Note that the proofs given in the cited website do ot generalize directly to semigroups.) In the spirit of our discussion above, we prove thatVenkataraman’s desired result holds more generally. Theorem 1.4.
Let S be a separative or completely regular semigroup such that, for all x, y ∈ S , x p y p = y p x p and x q y q = y q x q where p and q are relatively prime positive integers.Then S is commutative. In Example 2.2, we note that this theorem cannot be extended from completely regularsemigroups to general regular semigroups.It is easy to see that a group, or more generally a cancellative semigroup, is commutativeif and only if ( xy ) = x y for all x, y . The direct implication is trivial; for the converse, xyxy = xxyy implies yx = xy after cancellation. In the same vein is the following well-knownexercise ([2], § G is a group such that ( ab ) i = a i b i for three consecutive integers i for all a, b ∈ G , show that G is abelian.The slightly awkward wording allows two interpretations: that the integers i depend onthe elements a, b , or that the same integers i work for all a, b . The proof for groups isessentially the same in either case. Our generalizations require both readings. Theorem 1.5.
Let S be a semigroup. (1) Suppose S is separative and suppose that for each a, b ∈ S , there exist three consecu-tive nonnegative integers i such that ( ab ) i = a i b i . Then S is commutative. (2) Suppose S is an inverse semigroup and suppose that there exist three consecutivenonnegative integers i such that ( ab ) i = a i b i for all a, b ∈ S . Then S is commutative. Part (2) of this theorem cannot be formulated in the same way as part (1) is; see Example3.2. In addition, part (2) cannot be generalized to other types of regular semigroups; seeExample 3.3.Another commutativity theorem for groups was motivated for us by another known exer-cise [2, Exer. 24, p. 48]:Let G be a finite group whose order is not divisible by 3. Suppose that( ab ) = a b for all a, b ∈ G . Prove that G must be abelian.The finiteness is not essential and the condition can be replaced with the assumption that G is a group with no elements of order 3, that is, G satisfies the condition x = 1 = ⇒ x = 1 (1)for all x ∈ G . More generally, a group G satisfying ( ab ) = a b for all a, b ∈ G can bedescribed by a more general theorem of Alperin [1] as a quotient of a subgroup of a directproduct of abelian groups and groups of exponent 3. The condition (1) rules out groups ofexponent 3, and so G is abelian. The existence of nonabelian groups of exponent 3, suchas the unique one of order 27, shows that some additional hypothesis like (1) is needed toconclude commutativity.There are two reasonable reformulations of (1) for semigroups. First, if a semigroup S satisfies ( ab ) = a b for all a, b ∈ S , then this just asserts that the cubing mapping S → S ; x x is an endomorphism. From this point of view, (1) asserts that in groups, thekernel of this endomorphism is trivial, or equivalently, that the endomorphism is injective. his latter formulation makes sense in any semigroup S : x = y = ⇒ x = y (2)for all x ∈ S .Another reformulation of (1) for groups which works for any semigroup S , possibly withoutan identity element, is weaker, but more straightforward: x = x = ⇒ x = x (3)for all x ∈ S . To see that this is weaker, suppose (2) holds and x = x . Then ( x ) = x = x ,and so applying (2) yields x = x . Theorem 1.6.
Let S be a semigroup satisfying ( xy ) = x y for all x ∈ S . (1) If S is separative and satisfies (2) , then S is commutative. (2) If S is an inverse semigroup and satisfies (3) , then S is commutative. The hypothesis of part (1) of Theorem 1.6 cannot be weakened to (3); see Example 4.2below. Also, neither part of the theorem extends to other types of regular semigroups; seeExample 4.3.All of our investigations were aided by the automated deduction tool
Prover9 createdby McCune [4]. Automated theorem provers are especially good at equational reasoning,being able to derive consequences of equational axioms much faster and more efficientlythan humans. Any currently available automated theorem prover would have sufficed forthis project, but
Prover9 has the advantage that its input and output are easily readableby mathematicians with no familiarity with such tools. For example, here are the axioms inan input file for the special case of Theorem 1.4 where S is a group, p = 2 and q = 3: % group axioms(x * y) * z = x * (y * z). % associativitye * x = x. x * e = x. % identity elementx’ * x = e. x * x’ = e. % inverses% x^2 y^2 = y^2 x^2(x * x) * (y * y) = (y * y) * (x * x).% x^3 y^3 = y^3 x^3(x * (x * x)) * (y * (y * y)) = (y * (y * y)) * (x * (x * x)). The goal is just x * y = y * x. % commutativity
Here each equation is interpreted by
Prover9 to be universally quantified in the variables.Everything written after a % symbol is a comment. Notice that the association of terms inany equation is made explicit; while there are settings in
Prover9 which allow one to avoidparenthesization of the input, they do not generally improve readability of proofs.Since we are working in semigroups, the associative law is part of any input file, and isheavily used throughout proofs, mostly as a rewrite rule. While this can lengthen proofs, itactually causes little to no trouble for a human reader, who can skip many lines where therule is being applied. unning Prover9 with its default settings on the above input gives a proof in less thanhalf a minute on a not particularly fast computer. The proof has 174 steps. Some of thesteps are long and unpleasant; for instance, the longest one has an equation with 67 symbolsin it (variables and occurrences of the operations ∗ and ′ . Here is a more typical one fromthe proof, split into two lines:
961 x * (y * (x’ * (y * y))) = y * (y * (x * (y * x’))).[para(864(a,2),51(a,1,2,2)),rewrite([14(7)])].
The number 961 is a clause identification number which is an internal index that
Prover9 usesto keep track of kept clauses. The part in square brackets is the justification for clause 961.Here “para” is short for paramodulation , which is the primary inference used in equationalreasoning. Paramodulation refers to the substitution of one side of an equation into a subtermof another equation. In this case, clause 864 was plugged into a subterm of clause 51. Thiswas followed by a rewrite of the resulting clause by clause 14.It is not terribly enlightening to show the details of this particular step in the proof norany other step, because it turns out not to be necessary for translation into humanly readableform. A proof of 167 steps is, by the standards of automated theorem provers, not very long,and so it is reasonable to try to obtain a human proof. For familiar associative structures suchas groups, lattices, rings, and so on, this is usually easy, albeit sometimes time-consuming.Given two equations and being told that under the axioms of group theory, the two yield athird is usually enough for someone familiar with groups to see how the proof goes.In addition, a human reader can take numerous shortcuts. For example, here is anotherstep in the proof, omitting the justification:
A human reader can immediately see that the proof is essentially finished: cancel x on eachside of the identity and then multiply on the right by y to get commutativity. Prover9 , onthe other hand, took 14 additional steps to reach commutativity. In other words, the “outof the box” proof that
Prover9 found was far from optimal.Experienced users can tweak
Prover9 ’s many parameters and use various specialized tech-niques to find proofs faster, to find shorter proofs, and so on. For example, changing the termordering from the default lexicographic path ordering (LPO) to the Knuth-Bendix orderinggets a different proof in just 10 seconds and the new proof is 19 steps shorter than the firstone.Certainly the most interesting use of
Prover9 was in our investigation of Theorem 1.4.Conditions such as x p y p = y p x p for all x, y where p is an arbitrary but fixed positive integercannot be directly encoded in Prover9 (or any other first-order theorem prover) because ithas no built-in description of the integers. Instead, we had to look at several special casessuch as the one above.It was only after examination of several special cases that we realized that many of thesteps in the proofs were similar. This enabled us to see the pattern of the proof of Theorem1.4 for groups. (Recall that at this point, we were not aware that the theorem was aknown folk result.) In particular, the special cases led us to our formulation of Lemma 2.1below as containing the essential idea of the proofs. It was also at this same point in ourinvestigations that we realized the theorem holds more generally in cancellative, and thenseparative semigroups. This motivated us to look at a sampling of other commutativitytheorems in various classes of semigroups. fter this paper was submitted, we became aware of the recently published paper ofMoghaddam and Padmanabhan [5]. The results contained therein are different from ours,but the spirit of the work is exactly the same: by extracting the essential features of syntacticproofs of commutativity theorems for groups, they were able to extend them to cancellativesemigroups. 2. Proof of Theorem 1.4
The goal of this section is to prove Theorem 1.4. We start with the following key lemma.
Lemma 2.1.
Let S be a cancellative semigroup and suppose that there exists a map g : S → S satisfying the following conditions: for all x, y ∈ S , (a) xg ( x ) = g ( x ) x ; (b) g ( x ) g ( y ) = g ( y ) g ( x ) ; (c) xg ( x ) · yg ( y ) = yg ( y ) · xg ( x ) .Then the semigroup S is commutative.Proof. We claim that the following identity holds: g ( g ( x ) y ) y = yg ( g ( x ) y ) . (4)In fact, g ( x ) y | {z } u g ( g ( x ) y ) | {z } g ( u ) ( a ) = g ( g ( x ) y ) | {z } g ( u ) g ( x ) y | {z } u = g ( g ( x ) y ) | {z } g ( u ) g ( x ) |{z} g ( x ) y ( b ) = g ( x ) |{z} g ( x ) g ( g ( x ) y ) | {z } g ( u ) y, which, eliminating g ( x ) by left cancellation, gives (4).Our next claim is that g ( x ) y = yg ( x ). We start by observing that yg ( y ) g ( x ) y | {z } u g ( g ( x ) y ) | {z } g ( u ) ( c ) = g ( x ) y | {z } u g ( g ( x ) y ) | {z } g ( u ) yg ( y ) . (5) ow, g ( x ) y g ( y ) g ( g ( x ) y ) | {z } g ( y ) g ( u ) y ( b ) = g ( x ) y g ( g ( x ) y ) g ( y ) | {z } g ( u ) g ( y ) y = g ( x ) yg ( g ( x ) y ) g ( y ) y | {z } ( a ) = g ( x ) yg ( g ( x ) y ) yg ( y ) | {z } ( ) = yg ( y ) g ( x ) yg ( g ( x ) y ) ( b ) = y g ( x ) g ( y ) | {z } yg ( g ( x ) y ) ( ) = yg ( x ) g ( y ) g ( g ( x ) y ) y | {z } , yielding g ( x ) y · (cid:16) g ( y ) g ( g ( x ) y ) y (cid:17) = yg ( x ) · (cid:16) g ( y ) g ( g ( x ) y ) y (cid:17) , which, by right cancellation, implies g ( x ) y = yg ( x ), as claimed.Now the proof that xy = yx is straightforward: x yg ( x ) | {z } g ( y ) = x g ( x ) y | {z } g ( y ) ( c ) = yg ( y ) xg ( x ) = y xg ( y ) | {z } g ( x ) ( b ) = yxg ( x ) g ( y ) , and xy = yx follows by right cancellation. This completes the proof of the lemma. (cid:3) The previous lemma opens the gate to the proof of our first main theorem.
Proof of Theorem 1.4.
We may assume without loss of generality that p, q > S is cancellative. Since p and q are relatively prime, by Bezout’s identitythere exist integers r, s such that pr + qs = 1. Since one of pr or qs must be negative, weassume without loss of generality that qs <
0; thus − qs > x − qs ∈ S for all x ∈ S .Since q >
0, we have s < − s >
0; thus x − s ∈ S for all x ∈ S . As pr > p > r > x pr , x r ∈ S for all x ∈ S .Let g ( x ) = x − qs . We claim that g ( x ) satisfies the three properties (a), (b) and (c) of theprevious lemma. By associativity, we have xg ( x ) = g ( x ) x so that (a) holds. Regarding (b)we have g ( x ) g ( y ) = x − qs y − qs = ( x − s ) q ( y − s ) q = ( y − s ) q ( x − s ) q = y − qs x − qs = g ( y ) g ( x ) . The second equality holds because by assumption the q th powers commute. Finally, regard-ing (c) we have xg ( x ) yg ( y ) = xx − qs yy − qs = x − qs y − qs = x pr y pr = ( x r ) p ( y r ) p = ( y r ) p ( x r ) p = y pr x pr = y − qs x − qs = yy − qs xx − qs = yg ( y ) xg ( x ) . The fourth equality holds because p th powers commute.We have proved that S admits a function g : S → S satisfying the three conditions of theprevious lemma. It follows that S is commutative. ext assume that S is separative. By Proposition 1.2, S is a semilattice of cancellativesemigroups S α , each of which satisfies the hypotheses of the theorem. It follows that each S α is commutative. By Proposition 1.2 again, S is commutative.Finally, assume that S is completely regular. If e, f ∈ S are idempotents, then ef = e p f p = f p e p = f e . It follows that S is an inverse semigroup. Since S is both completelyregular and inverse, it is a Clifford semigroup, hence is a semilattice of groups. In particular, S is separative and the desired result follows. (cid:3) Example . Theorem 1.4 does not generalize from completely regular semigroups to othertypes of regular semigroups. For example, let S be the Brandt semigroup of order 5. Thenfor every positive integer p and every x ∈ S , x p is an idempotent. Thus the hypotheses ofthe theorem are satisfied since idempotents commute in inverse semigroups, but S is notcommutative. 3. Proof of Theorems 1.5
We first need a lemma which will prove useful in both this section and the next.
Lemma 3.1.
Let S be an inverse semigroup and suppose there exists an integer k > suchthat ( xy ) k = x k y k for all x ∈ S . Then S is a Clifford semigroup.Proof. Denote the unique inverse of an element x ∈ S by x ′ . We will show that xx ′ = x ′ x forall x ∈ S . It will follow that S is completely regular, hence Clifford. First, since ( x ′ x ) k = x ′ x ,we have ( x ′ ) k x k = x ′ x (6)for all x ∈ S . Next, recalling that ( x ′ ) k − = ( x k − ) ′ in inverse semigroups, we compute( x ′ ) k − x k = ( x k − ) ′ x k = ( x ′ ) k − x k − · xx ′ | {z } x = xx ′ · ( x ′ ) k − x k − x = x ( x ′ ) k x k (6) = xx ′ x = x , where we used the fact that idempotents commute in the third equality. Thus( x ′ ) k − x k = x . (7)Next, we have x ′ xx (7) = x ′ x ( x ′ ) k − x k = ( x ′ ) k − x k (7) = x , where we used k > x ′ xx = x , (8) xx ′ x ′ = x ′ , (9)where (9) follows from (8) by replacing x with x ′ and using x ′′ = x . Finally, we compute xx ′ (8) = x ′ xxx ′ = xx ′ x ′ x (9) = x ′ x , where we used commuting idempotents in the second equality. This completes the proof ofthe lemma. (cid:3) Proof of Theorem 1.5.
We begin with part (1) and suppose first that S is cancellative. Oneof the standard proofs of Herstein’s exercise ([2], § i, i + 1 , i + 2. Then a i b i ab = ( ab ) i ab = ( ab ) i +1 = a i +1 b i +1 . Cancel a i on the left and b on the right to get i a = ab i . Repeating the same argument with i + 1 in place of i gives b i +1 a = ab i +1 . Thus b i · ab = ab i b = ab i +1 = b i +1 a = b i · ba . Cancelling gives ab = ba .Now suppose S is separative. By Proposition 1.2, S is a semilattice of cancellative semi-groups S α . Each S α satisfies the hypothesis of the theorem, hence is commutative. ByProposition 1.2, S is commutative.Now we turn to part (2) and assume that S is an inverse semigroup satisfying the hy-potheses of the theorem. Applying Lemma 3.1 with k = i + 1, we have that S is a Cliffordsemigroup. In particular, S is separative. Since the hypotheses of part (2) are stronger thanthose of part (1), we may now apply part (1) to conclude that S is commutative. (cid:3) Example . Let S be any nonClifford inverse semigroup with a zero 0, for instance, theBrandt semigroup of order 5 will suffice. Let b be any element such that bb ′ = b ′ b . Then(0 b ) k = 0 k b k for all k ≥
1. This shows that part (2) of Theorem 1.5 cannot be strengthenedto be like part (1).
Example . Let S = { e, f } be the 2-element left (say) zero semigroup. Then trivially( xy ) k = x k y k for all x, y ∈ S and all positive integers k , but S is not commutative. ThusTheorem 1.5 does not extend to arbitrary regular semigroups or even completely regularsemigroups. 4. Proof of Theorem 1.6
We start with a lemma of some independent interest.
Lemma 4.1.
Let S be a cancellative semigroup satisfying ( xy ) = x y for all x, y ∈ S .Then for all x, y ∈ S , x y = yx . (10) Proof.
First, cancellation on both sides of ( xy ) = x y gives ( yx ) = x y for all x, y ∈ S .Using this, we compute ( x · yx )( x · yx ) = ( yx ) x = x y x . Cancelling on both sides, weobtain xy x = yx x (11)for all x, y ∈ S . Next, x · y x · x = x ( xy ) x (11) = xy · x · xy . Cancelling xy on the left gives yx = x y for all x ∈ S , as desired. (cid:3) Finally, we prove our last main result.
Proof of Theorem 1.6.
For part (1), assume first that S is cancellative and satisfies (2).Then (10) shows that the image C = { x | x ∈ S } of the cubing map S → S ; x x iscommutative. The condition (2) asserts that this map is injective, hence S is isomorphic to C . In particular, S is commutative.Now assume S is separative. By Proposition 1.2, S is a semilattice of cancellative semi-groups S α , each of which satisfies both (2) and ( xy ) = x y for all x ∈ S α . By the argumentabove, each S α is commutative. Applying Proposition 1.2 again, we have that S is commu-tative.For (2), now let S be an inverse semigroup satisfying both (3) and ( xy ) = x y for all x ∈ S . By Lemma 3.1, S is a Clifford semigroup, hence a semilattice of groups S α . Eachgroup S α satisfies (3) as well, but in groups, (3) is equivalent to (2). In particular, S is aseparative semigroup satisfying the conditions of part (1), and so S is commutative. (cid:3) xample . The hypothesis of part (1) of Theorem 1.6 cannot be weakened to (3). Indeed,let S = (cid:26)(cid:20) a b (cid:21) | a, b ∈ Z + (cid:27) with matrix multiplication as the operation. Then S is acancellative semigroup without idempotents and thus trivially satisfies (3). However, S isnot commutative. Example . Let S be as in Example 3.2 and note once again that ( xy ) = x y is triviallysatisfied for all x, y ∈ S . Since S is idempotent, conditions (2) and (3) both hold. However S is not commutative. Thus neither part of Theorem 1.6 extends to other types of regularsemigroups. References [1] J. L. Alperin, The classification of n -abelian groups, Canad. J. Math. (1969), 1238–1244.[2] I. N. Herstein, Topics in Algebra , Second edition. Xerox College Publishing, Lexington, Mass.-Toronto,Ont., 1975.[3] J. M. Howie,
Fundamentals of semigroup theory , London Mathematical Society Monographs. New Series,12. Oxford Science Publications. The Clarendon Press, Oxford University Press, New York, 1995.[4] W. W. McCune,
Prover9 and Mace4 , version 2009-11A. [5] G. I. Moghaddam and R. Padmanabhan, Commutativity theorems for cancellative semigroups,
Semi-group Forum (2017), 448–454.[6] A. Nagy, Special classes of semigroups , Advances in Mathematics (Dordrecht) , Kluwer AcademicPublishers, Dordrecht, 2001.[7] M. Petrich, Introduction to Semigroups , Merrill Research and Lecture Series. Charles E. Merrill Pub-lishing Co., Columbus, Ohio, 1973.[8] G. Venkataraman, Groups in which squares and cubes commute, https://arxiv.org/abs/1605.05463 .[9] A group such that a m b m = b m a m and a n b n = b n a n ( m , n coprime) is abelian? https://math.stackexchange.com/questions/326702/a-group-such-that-am-bm-bm-am-and-an-bn-bn-an-m-n-coprime (Ara´ujo) Col´egio Planalto, R. Armindo Rodrigues 28, 1600–414 Lisboa, Portugal (Kinyon)
Department of Mathematics, University of Denver, Denver, CO 80208, USA (Kinyon)
CEMAT-CI ˆENCIAS, Departamento de Matem´atica, Faculdade de Ciˆencias, Uni-versidade de Lisboa, 1749-016, Lisboa, Portugal
E-mail address : [email protected]@du.edu