Existence theorems for weakly symmetric operations

Transcription

1 Algebra univers 59 (2008) /08/ , published online November 22, 2008 DOI /s Birkhäuser Verlag, Basel, 2008 AlgebraUniversalis Existene theorems for weakly symmetri operations Miklós Maróti and Ralph MKenzie Dediated to George Grätzer and E Tamás Shmidt on their 70th birthdays 1 Introdution A k-ary near-unanimity operation (or k-nu) on a set A is an operation that satisfies the equations f(y,x,,x) f(x,y,,x) f(x,x,,x,y) x A k-ary weak near-unanimity operation (or k-wnu) on A is an operation that satisfies the equations w(x,,x) x and w(y,x,,x) w(x,y,,x) w(x,x,,x,y) If an algebra A has a k-nu (or a k-wnu) term operation, we say that A satisfies NU(k) (or WNU(k), respetively) Likewise, a variety is said to satisfy NU(k) (or WNU(k), respetively), it it has a k-variable term satisfying these equations It has been onjetured that a finite idempotent algebra A has finite relational width if and only if V(A) (the variety generated by A) has meet semi-distributive ongruene latties The onept of finite relational width arises in the theory of omplexity of algorithms, in the algebrai study of onstraint-satisfation problems Atually, there are several different definitions of this onept and it is not known if they are equivalent One version of the onept and the onjeture mentioned above are due to B Larose and L Zádori [10] The important family of varieties with meet semi-distributive ongruene latties has various known haraterizations There is a haraterization by a ertain Maltsev ondition; also, it is known that a loally finite variety has this property iff it omits ongruene overs of types 1 and 2 (defined in the tame ongruene theory of D Hobby, R MKenzie [6]) Presented by P Pálfy Reeived September 19, 2006; aepted in final form May 12, Mathematis Subjet Classifiation: 08B05, 08A70 Key words and phrases: weak near-unanimity, Maltsev ondition, tame ongruene theory The first author was partially supported by the Hungarian National Foundation for Sientifi Researh (OTKA), grant nos T and K

2 464 M Maróti and R MKenzie Algebraunivers E Kiss showed that a finite idempotent algebra of relational width k must have an m-wnu term operation for every m k E Kiss and M Valeriote then observed that a finite algebra with a k-wnu term operation, k > 1, must omit ongruene overs of type 1 These observations led M Valeriote to make two onjetures: any loally finite variety omits ongruene overs of type 1 iff it satisfies WNU(k) for some k > 1; any loally finite variety has meet semi-distributive ongruene latties if and only if for some k, it satisfies WNU(m) for all m k In this paper, we prove both of these onjetures of M Valeriote The family of loally finite varieties omitting type 1 is the largest family of loally finite varieties defined by a nontrivial idempotent Maltsev ondition For this result, see D Hobby, R MKenzie [6, Theorem 96] Several equivalent haraterizations for this largest Maltsev family are given in this theorem of D Hobby and R MKenzie One that we will use states that a loally finite variety V omits type 1 if and only if V has a term p(x,y,z) with the property that whenever A is an algebra in V and θ is a loally solvable ongruene of A and (a,b) θ, then p(a,b,b) = a = p(b,b,a) An important orollary of this haraterization, whih we need, is that in suh a variety, every Abelian algebra is polynomially equivalent to a module over a ring The result we prove in this paper, that a loally finite variety V omits type 1 iff V = WNU(k) for some k > 1, is the simplest known Maltsev haraterization of this family of varieties In addition to proving the two onjetures of M Valeriote, we shall demonstrate here that every ongruene distributive variety satisfies WNU(k) for all k 3 (This result requires no assumption of loal finiteness) Thus, in partiular, it follows that NU(n) implies WNU(k) whenever k 3 and n 3 A onsequene of our haraterization of loally finite varieties omitting type 1 is that every loally finite, ongruene modular variety must satisfy WNU(k) for some k > 1 Our final ontribution will be a seond proof of this fat about ongruene modular varieties The hief results of this paper are listed below The remainder of the paper is devoted to presenting proofs of these results Theorem 11 Let V be a loally finite variety The following are equivalent: (1) V omits type 1 (2) There is an integer k > 1 suh that V = WNU(k) (3) There is an integer n > 1 suh that V = WNU(k) for all k > 1 with k 1 (mod n) Theorem 12 Let V be a loally finite variety The following are equivalent: (1) V omits types 1 and 2 (2) The ongruene latties of all algebras in V are meet semi-distributive (3) There is an integer m > 1 suh that V = WNU(k) for all k m

3 Vol 59, 2008 Existene theorems for weakly symmetri operations 465 (4) For every positive integer n there is a positive integer m suh that V = WNU(k) for all k with m + 1 k m + n Theorem 13 Every ongruene distributive variety satisfies WNU(k) for all integers k 3 Corollary 14 Let V be a loally finite, ongruene modular variety There is an integer n > 1 suh that V = WNU(k) for all k > 1 with k 1 (mod n) These results will be proved in the order listed Our proof of Theorem 13 has a different harater from our proofs of the first two theorems Corollary 14 is a orollary to Theorem 11 However, as we mentioned, we shall present, in the final setion of the paper, a different proof of Corollary 14 along the lines of our proof of Theorem 13 2 Easy observations and examples It is well known that if a loally finite variety V admits type 1 then V annot annot satisfy any non-trivial idempotent Maltsev ondition In partiular, it annot have a weak near-unanimity term This supplies one of two impliations needed to prove Theorem 11 We shall now review the argument that proves this well-known fat, and review a losely related argument that reveals some limitations on the existene of weak near-unanimity terms that our when V admits type 2 For the onepts and results used in these arguments, see D Hobby, R MKenzie [6] Suppose that V has a type 1 ongruene quotient Then there is a finite algebra F in V and a minimal ongruene µ in F suh that (0 F,µ) has type 1 Let U be a (0 F,µ)-minimal set and N be a (0 F,µ)-trae ontained in U and e(x) be a polynomial operation of F satisfying e(e(x)) = e(x) (for all x F) and e(f) = U Suppose that F has a k-wnu-term τ( x) Then the operation e(τ( x)) restrited to N is an operation on the set N that satisfies the WNU(k)-equations This operation is a polynomial of the algebra F N But sine the type is 1, every polynomial operation of F N depends on at most one of its variables Hene F N does not satisfy WNU(k) for any k > 1, implying that F, likewise does not satisfy WNU(k) for any k > 1 Next, we show that if V admits type 2 then there is a prime integer p suh that whenever p divides k then V admits no k-wnu term This and the observation above about type 1 quotients, give one of two impliations needed to prove Theorem 12 Suppose that V has a type 2 ongruene quotient Choose a finite algebra F in V with a minimal ongruene µ in F suh that (0 F,µ) has type 2 Let U be a (0 F,µ)-minimal set and N be a (0 F,µ)-trae ontained in U In this situation, F N is polynomially equivalent to a one-dimensional vetor spae over a

4 466 M Maróti and R MKenzie Algebraunivers finite field k Let p be the harateristi of k It is easily verified that if F N has a k-wnu operation among its polynomial operations, then this operation is unique, and defined as τ(x 1,,x k ) = r(x x k ), where r k and kr = 1 in k Suh r exists in k iff p does not divide k Now the same argument as above shows that F annot satisfy WNU(k) if p divides k Example 21 Let A n = Z n,x y + z where Z n,x+y is the group of integers modulo n The term operations of this algebra are simply those operations whih an be expressed in the form f(x 0,,x k 1 ) = i k i x i where k i are integers and their sum is 1 modulo n If this operation is a k-wnu, then all the k i are ongruent modulo n Thus A n = WNU(k) iff (k,n) = 1 For example, A 6 satisfies WNU(5) but not WNU(2), WNU(3), or WNU(4) Let A = Z,x y + z where Z,x + y is the group of integers This algebra does not satisfy WNU(k) for any k 3 Finally, we remark that if A is an Abelian algebra in a loally finite variety V that omits type 1, then A does have a k-wnu for some k > 1 In fat, by D Hobby, R MKenzie [6, Theorem 96], A has a term operation p(x, y, z) satisfying Maltsev s equations p(x,y,y) = x and p(x,x,y) = y It is well known (see [4]) that every Abelian algebra A with suh an operation is polynomially equivalent to a unitary module; and that all term operations of the module of the form f(x 0,,x k 1 ) = i k ix i, where k i are integers and their sum is 1, are term operations of A Hene A does have k-wnu term operations in this ase (in fat, for every k > 1 that is relatively prime to the ardinality of A) 3 Redution to idempotent algebras An operation f(x 1,,x n ) on a set A is said to be idempotent if f(a,,a) = a for all a A An algebra A = A, is alled idempotent iff all the operations of A are idempotent, equivalently, every one-element subset of A is a subalgebra of A By an idempotent term of a variety V we mean a term for whih t(x,x,,x) x is a law of V A variety is alled idempotent if all of its algebras are idempotent; equivalently, if all terms of V are idempotent If V is any variety, there is the idempotent redut of V, or V id The variety V id is idempotent It is formed in this way: for every idempotent term t = t( x) of V, there is a orresponding basi operation f t ( x) of V id with the same number of variables as t The signature of V id onsists just of the operation symbols f t with t

5 Vol 59, 2008 Existene theorems for weakly symmetri operations 467 ranging over the idempotent term operations of V For every algebra A V there is the algebra A id of the same signature as V id, A id = A, {t A : t an idempotent term of V} whose basi operations are the term operations over A indued by the terms that are idempotent over V Then V id is defined as the variety generated by all the algebras A id, with A ranging over V It an be shown that for any positive integer n, the free algebra of rank n in V id, or F V id(n), is isomorphi over the free generators, to the subalgebra of F V (n) id generated by the free generators of F V (n) We mention this onstrution beause it is an (easily verified) fat that eah of the properties of algebras and of varieties that we are onerned with in this paper is invariant under the onstrutions A A id, V V id An algebra A satisfies WNU(k) iff A id satisfies WNU(k) The same is true of a variety V We have that V is ongruene distributive, or ongruene modular, iff V id has the respetive property Moreover, if V is loally finite then V id is loally finite; and in ase V is loally finite, then V omits type 1 (respetively type 2) iff V id omits type 1 (respetively type 2) (This is not obvious, but is a onsequene of the Maltsev haraterizations of these properties given in D Hobby, R MKenzie [6], Chapter 9) Thus both the hypotheses, and the onlusions, in eah of our hief results, is invariant under passing to the idempotent redut Consequently, it will suffie to prove our results for idempotent varieties Heneforth, we work only with idempotent algebras, and idempotent varieties Idempotent algebras have several speial properties that will be useful in our proofs We mention that if θ is a ongruene on an idempotent algebra A, then every blok of θ is a subalgebra of A Other useful properties will be explained as the need for them arises 4 Totally symmetri relations From here to the end of this paper, all algebras are assumed to be idempotent, unless expliitly stated otherwise Definition 41 Let A be any algebra A subuniverse B of A n is totally symmetri iff B is invariant under all permutations of the oordinates If B is totally symmetri, then B is a subdiret power of a ertain subuniverse of A, whih we term D(B) and all the domain of B The projetion of B at eah oordinate is equal to D(B) To simplify notation, we shall write elements of A n as though they were semigroup words Thus, for example, ba n 1 will denote the vetor (b,a,,a) in A n where {a,b} A

6 468 M Maróti and R MKenzie Algebraunivers Definition 42 An algebra B A n will be alled a -subalgebra of A n iff for some {a,b} A, B is generated by the n vetors ba n 1,aba n 2,,a n 1 b Note that in this ase B (the universe of B) is totally symmetri, and D(B) is generated by {a,b} Definition 43 Let A be an algebra and k > 1 be an integer We say that A satisfies WNU(k) iff A has a k-ary term operation w( x) satisfying the weak near-unanimity equations w(y,x,,x) w(x,y,x,,x) w(x,x,,x,y) (and w(x,,x) x, but the idempotent equation is already assumed) We say that A satisfies ST(k) iff every -subalgebra of A k ontains a diagonal vetor a k We say that A satisfies TS(k) iff every non-empty totally symmetri subuniverse of A k ontains a diagonal vetor We begin a sequene of lemmas detailing useful properties of these onepts Lemma 44 For a fixed signature and a fixed integer k > 1, the lass of algebras of the given signature satisfying TS(k) is losed under subalgebras, homomorphi images, and finite produts Proof It is straightforward to show that this lass is losed under subalgebras and homomorphi images, and we leave this task to the reader Now suppose that C = A B and that A = TS(k) and B = TS(k) Let S be any non-empty totally symmetri subuniverse of C k Let π 0 : C A be the first projetion homomorphism Then π k 0(S) = S 0 is a non-empty totally symmetri subuniverse of A k We an hoose a diagonal element a k in S 0 Then S 1 = {(b 0,,b k 1 ) B k : ((a,b 0 ),,(a,b k 1 )) S} is a non-empty totally symmetri subuniverse of B k (Here, the idempoteny of A ensures that S 1 is a subuniverse) So there is b B with b k S 1 This means that k S, where = (a,b) Lemma 45 Let k > 1 be a positive integer (i) For any algebra A, the impliations WNU(k) ST(k) and TS(k) ST(k) are valid (ii) For any algebra A, A = WNU(k) iff A belongs to some variety V suh that the free algebra on two generators in V satisfies ST(k) (iii) For any finite algebra A, the impliations TS(k) WNU(k) ST(k) are valid Proof We regard (i) as obvious Just note that if w( x) is a k-variable weak-nu term for A, and if B A k is the subalgebra generated by the sequene of vetors (for

7 Vol 59, 2008 Existene theorems for weakly symmetri operations 469 some {a,b} A) q i = a i ba k i 1, 0 i < k, then w( q 0,, q k 1 ) (with w applied in B) is the vetor k where = w(b,a,,a) We remark that the truth of (ii) requires the assumed idempoteny of algebras To prove (ii), assume first that A satisfies WNU(k) and that w( x) is a k-variable weak-nu term for A Let V be the variety generated by A Then w( x) is a weak-nu term for every algebra in V (sine the weak-nu property is determined by satisfation of ertain equations) Let, then, F be the free algebra on two generators in V By (i), F = ST(k), sine F = WNU(k) Now onversely, suppose that F is the free algebra freely generated by a two-element set {x,y} in a (idempotent) variety V, and that F = ST(k) Let k be the subalgebra of F k generated by the sequene of vetors f i = x i yx k i 1, 0 i < k Sine F = ST(k), there is a diagonal element (,,) in k For some term w(x 0,,x k 1 ), we an write This means that for eah 0 i < k, (,,) = w( f 0, f 1,, f k 1 ) = w( f 0 (i), f 1 (i),, f k 1 (i)) = w(x,x,,x,y,x,,x) in whih the final appliation of w is to a sequene of i x s followed by y and then k i 1 x s Sine also w(x,x,,x) = x, then all the equations required for w( x) to be a weak-nu term for V are satisfied Thus w( x) is a weak-nu term for every algebra that belongs to V We now prove (iii) Let A be a finite algebra All that remains is to show that A = TS(k) implies A = WNU(k) So let us assume that A satisfies TS(k) Let F be the free algebra on two generators in the variety generated by A Now F is finite, and is isomorphi to a subalgebra of a finite diret power of A By Lemma 44, F = TS(k); a fortiori, F = ST(k) Finally, by (ii) of this lemma, A = WNU(k) The next three lemmas establish that if a finite algebra satisfies WNU(k) then it will satisfy TS(k ) for some k k Definition 46 By a speial weak near-unanimity operation on a set A we shall mean a weak-nu operation w( x) on A that satisfies the equation w(w(y,x,,x),x,,x) w(y,x,,x) Lemma 47 If a finite algebra of size n has a weak near-unanimity term of arity k, then it has a speial weak near-unanimity term of arity k n! Proof Let A be a finite algebra of size n and t be a weak near-unanimity term for A of arity k We define by indution a sequene t 1,t 2,,t n! of weak near-unanimity

8 470 M Maróti and R MKenzie Algebraunivers terms of arities k,k 2,,k n!, respetively Put t 1 = t, and for all integers i < n! define t i+1 (x 1,,x k i+1) = t i (t(x 1,,x k ),t(x k+1,,x 2k ),,t(x k(ki 1)+1,,x k i+1)) Sine t is idempotent, it is easy to see that t i is a weak near-unanimity term for all i Moreover, t i (y,x,,x) = τx(y) i (the ith iterate of τ x applied to y) where τ x is the unary operation on A defined by τ x (y) = t(y,x,,x) The n!-fold iteration of a unary operation on an n-element set yields an idempotent operation; that is, τx 2n! (y) = τx n! (y) for all x,y A This means that t n! is a speial weak nearunanimity term for A Lemma 48 If an algebra has a k +1-ary speial weak near-unanimity term, then it has an mk + 1-ary speial weak near-unanimity term for every integer m 1 Proof Let T be a speial weak k-nu for A We define by indution a sequene t 1,t 2, of terms of arities k + 1,2k + 1,, respetively Put t 1 = t If t m is defined, then define t m+1 (x 0,,x (m+1)k ) = t m (t(x 0,,x k ),x k+1,,x (m+1)k ) Clearly, by an indutive argument, t i is a speial weak-nu term for eah i 1 Lemma 49 Let A be a finite algebra of size n, with a speial weak near-unanimity term t of arity k + 1, and let m nk Then A satisfies TS(mk + 1) Proof Let B A mk+1 be a totally symmetri subalgebra Claim 1 Suppose that ab k x B for some a,b A and x A (m 1)k Then k+1 x B where = t(a,b,,b) For 0 i k let f i = b i ab k i x Sine B is totally symmetri, f i B for all i Then k+1 x = t(f 0,,f k ) B Claim 2 Suppose that a i b jk x B for some integers 1 i j where a,b A Then i+jk x B where = t(a,b,,b) By applying the previous laim to the oordinates that ontain ab k we get that k+1 a i 1 b (j 1)k x B By repeated appliations we obtain that i+ik b (j i)k x B Note, that t(,b,b) = beause t is speial Therefore, by applying the previous laim to the oordinates that ontain b k, we get that i+(i+1)k b (j i 1)k x B By repeated appliations we finally obtain that i+jk x B Claim 3 Suppose that a k2 b 0 b k x B for some elements a,b 0,,b k in A and tuple x Then k 0 k k d x B for some elements 0,, k,d in A

9 Vol 59, 2008 Existene theorems for weakly symmetri operations 471 For 1 i k and 0 j k put ȳ i,j = a i 1 b j a k i A k For 0 i k define f i = ȳ i,0 ȳ i,i 1 a k ȳ i+1,i+1 ȳ i+1,k b i x, whih is best desribed by the following matrix: f 0 f 1 f 2 f k 1 f k a a a b 1 a a b k a a b 0 x b 0 a a a a a a b k a b 1 x a b 0 a a b 1 a a a a b 2 x = a a a a a a a a a b k 1 x a a b 0 a a b 1 a a a b k x Clearly, f 0,,f k B, beause eah is obtained from a k2 b 0 b k x by permuting its oordinates By alulating f = t(f 0,,f k ) B we see that f = k 0 k k d x, where i = t(b i,a,,a) and d = t(b 0,,b k ) Claim 4 There exist elements a 1,a 2,a m,b A so that a k 1a k 2 a k mb B Let j m be the largest integer for whih there exist elements a 1,,a j, b 0,,b (m j)k A so that a k 1 a k j b 0 b (m j)k B To get a ontradition, assume that j < m Clearly m j < n, otherwise some element of A appears at least k-many times in b 0,,b nk by the pigeon-hole priniple, whih ontradits the maximality of j Consequently, n(k 1) m n < j Applying the pigeon-hole priniple again to the elements a 1,,a j we may assume that a 1 = a 2 = = a k Therefore, a k2 1 b 0 b k a k k+1 ak j b k+1 b (m j)k B Applying the previous laim to this tuple yields elements 0,, k,d A so that k 0 k k dak k+1 ak j b k+1 b (m j)k B But this ontradits the maximality of j Claim 5 There exist elements b, A so that mk b B Let j m be the largest integer for whih there exist elements b, A and a j+1,,a m A so that jk a k j+1 ak mb B Clearly, k j by applying the pigeon-hole priniple to the tuple a k 1 a k mb whose existene is guaranteed by the previous laim To get a ontradition, assume that j < m Using Claim 2 for the tuple a k j+1 jk a k j+2 ak mb we get that d (j+1)k a k j+2 ak mb B for some d A, whih is a ontradition Claim 6 (f) There exists an element a A so that a mk+1 B This follows immediately from the previous laim and Claim 2 Theorem 410 Let A 1,,A n be finite algebras of the same signature suh that for 1 i n, A i = WNU(k i ) for some k i > 1 There is an integer k > 0 and a term t( x) of k + 1 variables in the signature of these algebras so that for all 1 i n, t is a speial weak near-unanimity operation for A i and for all m 1, A i = TS(mk + 1)

10 472 M Maróti and R MKenzie Algebraunivers Proof It follows easily from Lemmas 47, 48 and 49 that there is an integer N > 1 suh that every algebra A i satisfies TS(N) Let F be the free algebra on two generators x,y in the variety generated by A 1,,A n By Lemma 44, F = TS(N), and by Lemma 45(iii), F = WNU(N) Then by Lemma 47, F has a speial weaknu term of K +1 variables for a ertain K > 0 This term is also a speial weak-nu term for eah A i Let M be the ardinality of the largest algebra among the A i Then by Lemma 49, when m MK then every algebra A i satisfies TS(mK + 1) We take k = MK 2 Thus for m 1, eah A i satisfies TS(mk + 1) Moreover, our proof of Lemma 48, starting with the speial weak-nu term of K + 1 variables for F, yields a term t( x) of k + 1 variables that is a speial weak-nu for eah of the algebras A i There is another set of hypotheses that fores satisfation of TS(k) Theorem 411 Suppose that A is a finite algebra Let m 3 and k (m 1) A If A satisfies ST(n) for all m n k then A satisfies TS(k) Proof Let T be a non-empty totally symmetri subuniverse of A k Choose x T Sine k (m 1) A, and by total symmetry, we an assume that x = m 1 x m x k = m 1 x Now let n be maximal, m 1 n k, so that T ontains an element of the form u n z, u A Let a n q T, a D(T) We an assume that n < k, else the proof is finished Write q = b p, b D(T) Put S = { x A n+1 : x p T } Obviously, S is a totally symmetri subuniverse of A n+1 (sine the operations are idempotent and T is totally symmetri) Moreover S ontains the vetor a n b Thus S ontains a -subalgebra of A n+1 Sine n m 1, then m n+1 k Then sine ST(n+1) is satisfied by A, we onlude that there is e A with e n+1 S This means that e n+1 p T But this ontradits the assumed maximality of n 5 Minimal star subalgebras If we are trying to prove that a given finite algebra satisfies ST(k), then we will be looking at a -subalgebra B A k, and attempting to show that B ontains a diagonal element Among all the -subalgebras inluded in B, there will be at least one that properly inludes no other -subalgebra of A k Every -subalgebra of A k ontains a diagonal vetor iff every one of these minimal -subalgebras of A k is a singleton, onsisting of a diagonal vetor Thus it is natural to fous our attention on minimal -subalgebras of A k These algebras, in general, have some very interesting properties

11 Vol 59, 2008 Existene theorems for weakly symmetri operations 473 Definition 51 Let T be a minimal -subalgebra of F k where F is a finite algebra and k 3 Thus D(T) (defined in Definition 41 as the projetion of T onto the first oordinate, where T is the universe of T), is a two-generated subuniverse of F For a,b D(T) we write b a iff ba k 1 T We say that a,b is a primitive pair for T when b a Notie that b a entails that {a, b} generates D(T) We all an element a D(T) generi with respet to T iff b a for some b D(T) We all an element b D(T) o-generi with respet to T iff b a for some a D(T) A two-generator of D(T) is an element u suh that for some v D(T), {u,v} generates D(T) Thus every element of D(T) that is generi or o-generi with respet to T is a two-generator of D(T) Lemma 52 Let T be a minimal -subalgebra of F k where F is a finite algebra and k 3 For a D(T) the following are equivalent (i) a is a generi element with respet to T (ii) There is some x = x 1 x 2 x k T suh that x 1 is o-generi with respet to T and x 2 = a (iii) For all w T there is ȳ T with w i = y i for 3 i k and w 1 = a Also, the following are equivalent (a) There are u,v,w D(T) with u v w (b) For all v D(T) there are u,w D(T) with u v w; ie, every element of D(T) is both generi and o-generi with respet to T () For all x D(T) k there is ȳ T with x i = y i for 1 i k 1 Proof First, we takle (iii) (i) Take any x in T Choose ȳ 1 T with y 1 1 = a Exhanging the first and third oordinates and applying (iii), we find some axa in T If 3 < k, then exhange the first and fourth o-ordinates in this vetor and apply (iii) to get some ayaa in T This proedure an be ontinued, and eventually produes aba k 2 T for some b D(T) This is equivalent to b a Thus a is generi with respet to T The impliation (i) (ii) is trivial, sine b a means that ba a T For (ii) (iii), suppose that d, x T and x 1 =,x 2 = a Sine T is a minimal -subalgebra, the vetors ε 1,,ε k defined by ε i (i) = and ε i (j) = d when i j onstitute a generating set for T Thus we an write x = t(ε 1,,ε k ) in T for some term t Then t(ε 3,ε 1,ε 2,,ε 2 ) = ȳ is a vetor in T of the form apd k 3 for some p D(T) Likewise t(ε 2,ε 1,ε 2,,ε 2 ) = z is a vetor of the form aqd k 2 where q D(T) We write τ 1 = z = aqd k 2, and for 3 i k we put τ i = apd i 3 d k i All these vetors an be obtained by permuting the oordinates in ȳ or z and so, they belong to T Now suppose that w is any vetor in T Write w = s(ε 1,,ε k ) for a term s, just as we wrote x above Consider w = s(τ 1,τ 1,τ 3,,τ k ) It is easily verified

12 474 M Maróti and R MKenzie Algebraunivers that w(i) = w (i) for 3 i k and that w (1) = a This onludes our proof that (ii) (iii) For proving the equivalene of (a), (b), (), we first note that () (a) is easy: Taking any D(T), and taking x = k 1, there is b k 1 T for some b, ie, b By the same token, there is some a D(T) with a b For (a) (b), hoose three elements of D(T) satisfying a b and let u be any element of D(T) Find x T with x 1 = u Write first x = s(τ 1,,τ k ) where τ i (i) = a and τ i (j) = b when i j Consider ȳ = s(τ 1,γ,,γ) where γ = b T We have ȳ = uv v where v = s(b,,,), ie, u v Now write x = t(γ 1,,γ k ) where γ i (i) = b and γ i (j) = when i j Consider z = t(τ k,γ k,,γ k ) We have z = u k 1 w where w = t(a,b,,b) Thus also wu k 1 T and w u We have shown, given u D(T), that there exist v,w D(T) with w u v The derivation of (a) (b) is finished To prove that (b) (), we use the impliation (i) (iii) Assume that (b) holds Then every element a D(T) is generi with respet to T Thus every element a D(T) satisfies (iii) Now, given x D(T) k, we an hoose a vetor x 1 ȳ T Exhanging the first and third oordinates in this vetor and applying (iii) (with a = x 2 ) we get a vetor x 2 px 1 z T, and then after a permutation of oordinates, we have a vetor pqx 1 x 2 w T We apply (iii) to this vetor with a = x 3 and obtain a vetor x 3 rx 1 x 2 w T Clearly we an ontinue in this fashion to eventually obtain x k 1 sx 1 x 2 x k 2 T for some s D(T) After a permutation of oordinates, we get x 1 x 2 x k 1 s T, as required Definition 53 A minimal -subalgebra of F k satisfying the equivalent onditions (a)-(b)-() of Lemma 52 will be alled k 1-omplete Lemma 54 Let F be a finite algebra and k 3 If F satisfies WNU(k 1), WNU(k) or WNU(k +1), then every minimal -subalgebra of F k is k 1-omplete Proof Let T be any minimal -subalgebra of F k If F satisfies WNU(k) then learly, T has a diagonal vetor a k, implying that T = {a k } and D(T) = {a} Clearly T is k 1-omplete in this ase Assume that F satisfies WNU(k 1) Let m(x 1,,x k 1 ) be a k 1-WNU term for F Choose any a b in D(T) Consider the vetor x = m(ab k 1,bab k 2,bbab k 3,,b k 2 ab) This vetor in T has the form k 1 b So we have a b Now by Lemma 52 and Definition 53 it follows that T is k 1-omplete Now assume that F satisfies WNU(k + 1) Let w(x 1,,x k+1 ) be a k + 1-WNU term for F Choose u,v D(T) suh that u v Put p = w(u,u,v,v), q = w(v,,v,u,u), v = w(u,v,,v) We find that pv v T by giving uv k 1,uv k 1,vuv k 2,,v k 1 u

13 Vol 59, 2008 Existene theorems for weakly symmetri operations 475 as arguments to w in T Thus p v To omplete the proof of this theorem, it will suffie, by Lemma 52, to show that p is generi with respet to T Analogously to the above, we have that q v Also, we have that pqv v v T, as an be demonstrated by giving w these arguments in T: uv k 1,uv k 1,vvuv k 3,vvvuv k 4,,v k 2 uv,vuv k 2,vuv k 2 Now sine pq T and p,q are both o-generi then, by Lemma 52 (equivalene of (i)-(ii)-(iii)), p, q are also both generi Thus T is k 1-omplete by Lemma 52 Lemma 55 Let F be a finite algebra and k 3 Let T be a minimal -subalgebra of F k Suppose that there is 1 2 k T where every i is generi with respet to T Then T is k 1-omplete Proof We laim that if y D(T) then y 2 3 k 1 u T for some generi element u Indeed, let 1 k 1 b T Then we an write y = s( 1,b) for some term s Now there is b 2 3 k 1 T for some sine 2, 3, are generi (using k 2 appliations of the equivalene (i) (iii) in Lemma 52) Here is generi by the equivalene (i) (ii) in Lemma 52 (sine b is o-generi) Now in T we have the equation s( 1 2 k,b 2 k 1 ) = y 2 3 k 1 s( k,) and s( k,) is generi, sine k, are generi (and the set of generi elements of D(T) is obviously a subalgebra) The laim obviously generalizes: if y D(T) and d 1 d k T where all d i are generi, then yd 2 d k 1 u T for some generi u Applying this laim to k and to y = 1, we find that k 1 u T for some generi element u Then applying the laim to k 1 u and with y = 1, we get that k 1 v T for some generi v Obviously, we an ontinue to replae 4, 5, by 1 and we eventually will obtain that k 1 1 q T for some generi element q Now q is both generi and o-generi Finally, it follows from the equivalene (a) () in Lemma 52 that T is k 1-omplete 6 Charateristi ongruene and two-generators for a minimal star subalgebra Definition 61 Let A be an algebra and k 2, and let T be a non-empty totally symmetri subalgebra of A k For x A k 1, put pr x (T) = {u a : u x T } Define ρ T to be the set of all pairs (u,v) A 2 suh that for some x A k 1,

14 476 M Maróti and R MKenzie Algebraunivers {u,v} pr x (T) For x A, and n 1 we put ρ n T (x) = {y A : (x,y) ρn T } where ρ n t is the n-fold relation omposition of ρ T with itself, so that ρ 1 T = ρ T, ρ 2 T = {(u,v) A 2 : (u,a),(a,v) ρ T for some a D(T)}, et Now, ρ n T is a reflexive and symmetri subuniverse of D(T)2, and ρ n T (x) is a subuniverse of D(T) ontaining x, whenever x D(T) Considering the algebra D(T) with universe D(T), let θ T denote the ongruene on D(T) generated by ρ T We all θ T the harateristi ongruene of T Note that if A is finite, then there is a positive integer n suh that θ T = ρ n T Lemma 62 Let A be a finite algebra, k 2, and T be a minimal -subalgebra of A k If T > 1 then for every x D(T) k 1, pr x (T) D(T) Proof Put W = { x D(T) k 1 : pr x (T) = D(T)} To work toward a ontradition, suppose that W Note that W is a symmetri subuniverse of D(T) k 1 Choose n maximal, 1 n k 1, suh that there is n x W for some x W k n 1 Sine T > 1, then n < k 1 (else k 1 W, implying k T, giving that { k } is a -subalgebra of A k ontained in T, ontraditing minimality of T) Note that sine k 1 W, then k 3 Now hoose any b D(T) that is o-generi with respet to T We have that b n x T By Lemma 52, is generi with respet to T, so with the right hoie of b, we an assume that b k 1 T, as well as b n x T and n x T (sine n x W) Now we laim that for any s T we have also s 1 s 2 s n+1 x T The laim obviously will imply that for any t W, we have also t 1 t 2 t n x W To prove the laim, letting s T, we hoose a term λ suh that λ(f 1,,f k ) = s where f i is the vetor with b at the i plae and everywhere else (Here we use the minimality of T) Now in the equation s = λ(f 1,,f k ), replae f n+2,,f k by the vetor n+1 x, whih belongs to T, and for 1 i n+1, replae f i by the vetor obtained from n+1 x by replaing at plae i by b These vetors also belong to T This yields a new equation: s 1 s 2 s n+1 x = λ(g 1,,g k ) T The laim is proved Now reall from above that we have n < k 1 Thus write x = dȳ for some d D(T) So we have n dȳ W, or equivalently, n 1 dȳ W By the laim, we get that n 1 ddȳ W (replaing ȳ by x = dȳ) Then again, n 2 ddȳ W; and this yields in the same fashion, n 2 dddȳ W Obviously, an indutive argument now gives that d n+1 ȳ W But this ontradits our hoie of n as maximal The ontradition finishes our proof that pr x (T) = D(T) is impossible Reall that we defined an element a of an algebra B to be a two-generator of B just in ase for some b B we have that {a,b} generates B

15 Vol 59, 2008 Existene theorems for weakly symmetri operations 477 Lemma 63 Let A be a finite algebra and k 2, and let T be a minimal - subalgebra of A k If T > 1 then for every x D(T), if ρ T (x) = D(T) then x is not a two-generator of D(T) Proof Suppose this fails Then we an hoose {a,b} D(T) so that {a,b} generates D(T) and ρ T (a) = D(T) Thus (a,b) ρ T There is x D(T) k 1 with {a,b} pr x (T) Sine pr x (T) is a subalgebra of D(T), then pr x (T) = D(T) This ontradits Lemma 62 Lemma 64 Let A be a finite algebra and k 2, and let T be a minimal - subalgebra of A k Suppose that {a,x,y} D(T) and a is generi with respet to T If a k 2 xy T then x is a two-generator of D(T) iff y is a two-generator of D(T) Proof Let a k 2 xy T and suppose that {x,z} generates D(T) We have to show that y is a two-generator Sine a is generi, there is a k 2 zw T for some w Write S = Sg({y,w}) We show that S = D(T), demonstrating that y is a two-generator Choose a term t so that t(x,z) = a Applying t in T to a k 2 xy, a k 2 zw, we obtain that a k 1 t(y,w) T Write g = t(y,w) Thus a,g is a primitive pair for T Choose a term s so that s(x,z) = g Then applying s in T to a k 2 xy and a k 2 zw, we get that a k 2 gs(y,w) T Write u = s(y,w), so that u is a generi element, sine g is o-generi and a k 2 gu T (by an appliation of Lemma 52) Now we have {u,g} S, a k 1 g T, a k 2 gu T Choose h so that u k 1 h T We laim that for any x T there is s S with x 1 x 2 x k 1 s T If this is true, then there must be s S with u k 1 s T Then u,s is a primitive pair with respet to T, and S ontains {u,s}, implying that S = D(T), whih is what we wished to prove To prove the laim, let x T and write x = σ(ρ 1,,ρ k ), where the vetor ρ i has a at all plaes but the ith, and g at the ith plae Define τ i (1 i < k) to agree with ρ i exept at the kth plae, where τ i (k) = u The vetors τ i belong to T, sine a k 2 gu T Thus {τ 1,τ 2,,τ k 1,ρ k } T Now σ(τ 1,τ 2,,τ k 1,ρ k ) = x, where x (i) = x(i) for 1 i < k, and x (k) = σ(u,,u,g) = s S This proves the laim Definition 65 An algebra A satisfies TST(k) iff every non-empty totally symmetri subalgebra of A k ontains a -subalgebra of A k Theorem 66 Assume that T is a minimal -subalgebra of A k where A is a finite algebra and k 3, and assume further that T > 1 Assume also that every proper subalgebra of A satisfies TST(k 1) Then θ T, the harateristi ongruene for T, is distint from D(T) D(T) Proof In searh of a ontradition, we assume that θ T = D(T) D(T) Thus there is a positive integer n suh that ρ n T = D(T) D(T) Choose n 0 to be the least suh positive integer

16 478 M Maróti and R MKenzie Algebraunivers We laim that n 0 = 1 To prove this, suppose that n 0 > 1 Choose a primitive pair a,b for T We an also hoose some u D(T) so that (a,u) ρ n0 1 T and (u,b) ρ T Our immediate aim is to prove that u must be a two-generator of D(T) By definition, we an hoose some x A k 1 so that u x,b x T Sine b x T, then by Lemma 52, x i is generi for 1 i k 1 Rewriting, we have xu T Write G for the set of all elements of D(T) generi with respet to T Then G is a subalgebra of D(T) and x G k 1 Define S = {ȳ G k 1 : ȳu T } Sine xu T, then S is a non-empty subalgebra of G k 1 Clearly, S is totally symmetri Now if G = A then u G, implying that u is a two-generator, as we hope If G A, then this algebra satisfies TST(k 1) We have then that S ontains a vetor a k 2 for some {a,} G At this point, we have that a k 2 u T and a, and are both generi with respet to T, implying that is a two-generator of D(T) Now Lemma 64 implies that u is a two-generator Thus, indeed, u must be a two-generator of D(T) Choose v so that {u,v} D(T) and {u,v} generates D(T) Reall that (u,a) ρ n0 1 T and (u,b) ρ T ρ n0 1 T Thus {a,b} ρ n0 1 T (u), implying that the algebra ρ n0 1 T (u) is idential with D(T) Consequently, (u,v) ρ n0 1 T But sine our algebras are idempotent, and {u,v} generates D(T), then the set of pairs {(u,u),(v,v),(u,v),(v,u)} generates D(T) D(T) Thus it must be that ρ n0 1 T = D(T) D(T) This ontradits the minimality of n 0 The ontradition establishes that, in fat, ρ T = D(T) D(T) Now we have (a,b) ρ T, implying that ρ T (a) = D(T) Sine a is a two-generator of D(T), this is a ontradition of Lemma 63 Theorem 67 Assume that T is a minimal -subalgebra of A k where A is a finite simple algebra and k 3 Assume also that D(T) = A and that every proper subalgebra of A satisfies TST(k 1) Then A is an Abelian algebra Proof Under the assumptions in this theorem, Theorem 66 implies that θ T is the identity relation on A Thus ρ T is the identity relation on A as well We now onsider three ongruenes on the algebra T They are the kernels of the projetions of T onto A at the first oordinate, and at the seond oordinate, and the kernel of the projetion of T into A k 2 over all remaining oordinates Label

17 Vol 59, 2008 Existene theorems for weakly symmetri operations 479 these ongruenes σ 1,σ 2 and σ 1,2 Thus for x,ȳ T we have ( x,ȳ) σ 1 x 1 = y 1 ( x,ȳ) σ 2 x 2 = y 2 ( x,ȳ) σ 1,2 x i = y i for all 3 i k The remainder of this argument relies on the entralizer theory and theory of the solvability ongruene on the ongruene lattie of a finite algebra, as detailed in D Hobby, R MKenzie [6, Chapters 3 and 7] Notie that ρ T = 0 A (the identity relation on A) is equivalent to σ 2 σ 1,2 = 0 T and sine T is totally symmetri, also equivalent to σ 1 σ 1,2 = 0 T Let us put σ = σ 1 σ 2, and α = σ σ 1,2 The fat that σ i is disjoint from σ 1,2 (for i {1,2}) implies that σ i entralizes σ 1,2 This implies that σ entralizes σ 1,2, and thus α entralizes α The ongruene α on T is Abelian Thus σ 1 and σ 1 α are solvably equivalent Choose a primitive pair a,b for T It is easy to see that ba k 1 and aba k 2 are ongruent modulo α Thus a and b are ongruent modulo the image of σ 1 α under the projetion map D(T) A through the first oordinate Thus this image is a non-identity ongruene of A; it an only be A A sine it ontains the pair (a,b), and A is simple But sine σ 1 and σ 1 α are solvably equivalent, this image is a solvable ongruene of A This means that A A is a solvable ongruene Sine A is simple, we are fored to the onlusion that A is Abelian 7 Proof of Theorem 11 Of the three onditions whose equivalene Theorem 11 asserts, (3) trivially implies (2), and we saw in Setion 2 that (2) implies (1) The five Lemmas in Setion 4 together with Theorem 410 show that (2) and (3) are equivalent To omplete a proof of the theorem, what remains is to show that in a loally finite variety that omits type 1, every finite algebra (or just the free algebra on two generators) satisfies TS(k) for some integer k > 1 So let V be a loally finite variety that omits type 1 and assume that V has a finite algebra A that does not satisfy TS(k) for any integer k > 1 We shall derive a ontradition We assume that the ardinality of A is as small as possible This means that whenever B V and B < A then B satisfies TS(k) for some k > 1 We hoose a finite sequene of algebras A 1,A 2,,A n so that A i V and A i < A for 1 i n, and every algebra B V with B < A is isomorphi to some A i With an appliation of Theorem 410, we hoose an integer k 0 > 1 and a term t 0 ( x) of k variables in the signature of V so that for all 1 i n, t 0 is a speial weak near-unanimity operation for A i and for all m 1, A i = TS(mk 0 + 1)

18 480 M Maróti and R MKenzie Algebraunivers Lemma 71 A is a simple non-abelian algebra Proof First we prove that A is simple Clearly A > 1 Suppose that A has a ongruene θ different from A A and the identity relation We shall derive a ontradition by showing that A = TS(k 0 + 1) Let T be any non-empty totally symmetri subalgebra of A k0+1 Where π : A A/θ is the anonial epimorphism, we have that π k0+1 (T) is a non-empty totally symmetri subalgebra T θ of (A/θ) k0+1 Here A/θ = TS(k 0 + 1); thus T θ has a diagonal vetor This means that there is a vetor x T with the property that for some θ-equivalene lass E, x E k0+1 Now E is a subalgebra of A (sine A is idempotent) and T E k0+1 is a non-empty totally symmetri subalgebra of E k0+1 Sine θ A A, then E < A The algebra E thus satisfies TS(k 0 +1), implying that T E k0+1 ontains a diagonal vetor We have shown that every non-empty totally symmetri subalgebra of A k0+1 ontains a diagonal vetor This ontradits our assumption that A satisfies TS(k) for no k > 1 The ontradition onludes our proof that A is simple Now suppose that A is Abelian Then, as we saw in Setion 2, sine A V and V omits type 1, then A = WNU(k) for some k > 1 and as we have seen, it follows that A = TS(k) for some k > 1 This is another ontradition Reall that the property TST(m) was defined in Definition 65 We put M = max( A,k 0 + 1) k 0 Lemma 72 For m M and for 1 i n, A i = TST(m) Proof Suppose that R is a non-empty totally symmetri subalgebra of A m i where m > A i k 0 and m k0 2 + k 0 We must prove that R ontains the vetor m 1 b for some b, A i If m 1 (mod k 0 ) then A i = TS(m) implying that A i = TST(m) There is nothing to prove in this ase In the ontrary ase, we write m = lk r where l > 0 and 0 < r < k 0 Choose any vetor x R Sine m > A i k 0 > A i r, some member of A i ours r times in x After a permutation of oordinates in x, we get, say, a r ū R where a A i and ū A lk0+1 i Thus S = {ȳ A lk0+1 i : a r ȳ R} is a non-empty totally symmetri subalgebra of A lk0+1 i It follows that there is b A i with a r b lk0+1 R Rewrite this as a r b lk0 b R and note that the assumed inequality for m implies that l k 0 > r Now the argument proving Claim 2 in the proof of Lemma 49 shows that there is A i with r+lk0 b R, ie, m 1 b R Definition 73 An algebra C satisfies IST(k) iff for all a,b C there exists an element C so that the totally symmetri subalgebras of C k generated by a k 1 b and a k 1 ontain the vetor k

19 Vol 59, 2008 Existene theorems for weakly symmetri operations 481 Lemma 74 Let C be a finite algebra If C satisfies ST(k), then it satisfies IST(k C! ) Proof Sine C satisfies ST(k), we an hoose, for eah pair (a,b) C 2 a term t a,b of arity k so that t a,b (b,a,,a) = t a,b (a,b,a,,a) = = t a,b (a,,a,b) For eah a C we define a unary operation f a on C as f a (b) = t a,b (b,a,,a) Consider the term s(x 0,,x k2 1) = t a,fa(b)(t a,b (x 0,,x k 1 ),,t a,b (x k(k 1),,x k2 1) Sine C is idempotent, s(b,a,,a) = = s(a,,a,b) = f a (f a (b)) By iterating this argument, for eah a,b C we get a term s a,b of arity k C! so that Put = f C! a that is f C! a s a,b (b,a,,a) = = s a,b (a,,a,b) = f a C! (b) (b) We know that the C!-th power of the funtion f a is idempotent; () = f a C! (f a C! (b)) = f a C! (b) = This implies that s a, (,a,,a) = = s a, (a,,a,) = f C! a () =, so we have the required terms for IST(k C! ) Definition 75 An algebra C satisfies RST(k) iff for every minimal -subalgebra T C k, D(T) C Lemma 76 For m > M, we have that A satisfies RST(m) For m > M and m 1 (mod k 0 ), we have that A satisfies ST(m) Proof Let m > M and let T be a minimal -subalgebra of A m Clearly we have m 3 By Lemma 72, every proper subalgebra of A satisfies TST(m 1) Thus it follows from Lemma 71 and Theorem 67 that D(T) A This argument shows that A = RST(m) Next, let m > M, m = lk 0 + 1, and let T be a minimal -subalgebra of A m Sine A = RST(m), then T is atually a minimal -subalgebra of B m for some proper subalgebra B < A But B = A i for some 1 i n and B = ST(m) Thus T is a singleton, and ontains a diagonal vetor This shows that A = ST(m) Now let k 1 be the least multiple of k 0 that is not less than M and not less than A, and let k be the multiple of k 0 suh that k+1 = (k 1 +1) A! Let t be a term of k + 1 variables whih is a speial weak near-unanimity term for all A i, 1 i n The proof of Lemma 48, starting with t 0, will produe suh a term

20 482 M Maróti and R MKenzie Algebraunivers The next lemma summarizes what we know about A Lemma 77 Let A be the algebra from the previous lemma (1) A is a finite, simple, non-abelian algebra that satisfies TS(m) for no m > 1 (2) t is a k + 1-ary term whih is a speial weak near-unanimity term for every proper subalgebra of A (3) Eah proper subalgebra of A satisfies TS(mk + 1) for all m 1 (4) Eah proper subalgebra of A satisfies TST(m) for all m k (5) A satisfies RST(m) for all m k (6) A satisfies ST(mk + 1) for all m > 0 (7) A satisfies IST(k + 1) Proof These statements are easily justified with the help of the preeding lemmas of this setion The next lemma ontradits statement (1) of the above lemma Its proof will onlude our proof by ontradition, of Theorem 11 Lemma 78 Let A be the algebra and k be the integer from the previous lemma Then A satisfies TS(N) for some large integer N Proof Let N be large and N 1 (mod k), and let B A N be an arbitrary totally symmetri subalgebra We argue that B ontains a diagonal element Claim 1 Suppose that a i b jk x B for some integers 1 i j and a,b, x Then i+jk x B for some A Use an analogous argument to that of the proof of the first two laims in our proof of Lemma 49, using the fat that A satisfies IST(k + 1) as a replaement for the existene of a speial weak near-unanimity term for A Claim 2 If a k 1 a k p x B and p > k A, then b pk x B for some b A Let i p be a maximal integer so that b ik k k 1 p i x B for some elements b, 1,, p i A By the pigeon-hole priniple i k If i < p, then we an use Claim 1 for b ik k 1 to get that d ik+k k k 2 p i x B This ontradition shows that i = p Claim 3 a i b N i 1 B for some a,b, A and integer i < k Take an arbitrary tuple from B and write it in the form a k 1 a k pb q 1 r Choose suh a tuple where kp + q is maximal and 0 q < k If q = 0, then r = 0 by the maximality of kp + q If r = 0, then the laim holds by Claim 2 (Here we require only that N 1 > k 2 A ) Therefore, we may assume that q > 0 and r > 0 Clearly, r < k A, otherwise kp+q was not maximal by the pigeon-hole priniple Thus by Claim 2 (assuming that N > (k 2 + k) A + k), we may assume that a 1 =

21 Vol 59, 2008 Existene theorems for weakly symmetri operations 483 = a p Using Claim 1 for 1 and a pk 1 we get a tuple of the form d pk b q d 2 r If {d, 2,, r } generates A as a subalgebra, then kp + q was not maximal, beause the totally symmetri subalgebra of A r generated by d 2 r ontains a tuple of the form b 2 r 1 Now we use TST(N q) for the oordinates where we have elements from {d, 2,, r }, whih yields a tuple of the form b q e N q 1 f Claim 4 a N 2 b B for some a,b, A From the previous laim we know that a i b N i 1 B for some elements a,b, A and integer i < k Sine A satisfies RST(k) we an apply this to ab k 1 to get de k 1 suh that d,e generate a proper subalgebra of A We do this i-many times, one to eah a, and get that d i e ki i b N 1 ki B As d,e generate a proper subalgebra, we an apply TST(ki) and get that fg ki 1 b N 1 ki B for some elements f,g A Sine N 1 (mod k), N 1 ki 0 (mod k) Moreover, by hosing N large enough, we an make N 1 ki arbitrary large Therefore, we an apply Claim 1 to the tuple g ki 1 b N 1 ki, and get that fh N 2 B Claim 5 ab i N i 1 B for some a,b, A and integer i A suh that {a,} and {b,} generate proper subalgebras of A From Claim 4 we have a tuple ab N 2 B By RST(N 1) we an assume that {b,} generates a proper subalgebra of A We define a sequene a 1,a 2, A of elements suh that a i b i N i 1 B Put a 1 = a If a i is defined and {a i,} generates A, then there is a binary term t suh that t(a i,) = b In this ase we put a i+1 = t(,a i ) By the onstrution we have that a i+1 b i+1 N i 1 B If for some i A we arrive to an element a i so that {a i,} does not generate A then we are done Otherwise, we have a repetition in a 1,,a A,a A +1, so we an ontinue with the repeated sequene to get that a N 1 b N 1 B From this by ST(N) we get a onstant tuple Claim 6 ab k 0 b k k N 1 k k2 B for some a,b 1,,b k+1, A From the previous laim we have a tuple ab i N i 1 B It is easy to see that we have k > A i Thus we an permute the oordinates to get k2 b 0 b k N k2 k 1 B where b 0 b k = ab i k i Now {b i,} generates a proper subalgebra of A for 0 i k; hene t satisfies the weak-nu equations on inputs from {b i,} for eah i Now the argument used to prove Claim 3 in the proof of Lemma 49 works in this situation, using the term t; and it yields the desired result

22 484 M Maróti and R MKenzie Algebraunivers Claim 7 B ontains a diagonal element Observe, that N 1 k k 2 0 (mod k), so we an apply Claim 2 to the tuple ab k 0 b k k N 1 k k2 B and get that ad N 1 B for some d A Now using ST(N) we get a diagonal tuple in B 8 Proof of Theorem 12 D Hobby, R MKenzie [6, Theorem 910] gives the equivalene of statements (1) and (2) in Theorem 12 We showed in Setion 2 that (4) implies (1), and (3) trivially implies (4) The proof of Theorem 12 will therefore be ompleted with the proof of the next theorem, whih shows that (1) implies (3) Theorem 81 Let V be a loally finite variety that omits types 1 and 2 Let A be any finite algebra in V with n = A > 1 For k > 2(n 1)!, A satisfies ST(k), and for k 2n!, A satisfies TS(k) and WNU(k) Proof First, note that sine V omits types 1 and 2, it has no Abelian algebra with more than one element We prove this theorem by indution on n If A V and n = A = 2 then A is simple and non-abelian Let k > 2 (= 2(n 1)!) Now, proper subalgebras of A have only one element and thus satisfy TS(k), and TST(k) Thus it is immediate from Theorem 67 that every minimal -subalgebra T of A k has D(T) A, implying that T is a singleton This means that A satisfies ST(k) To onlude the proof for n = 2, notie that sine A = ST(k) for k 3, then by Theorem 411, A = TS(k) for k 4 = 2n!, and then by Lemma 45, A = WNU(k) for k 4 as well Now suppose that n > 2 and the onlusions of this theorem are valid for all finite algebras A V with 2 A < n Let A V with A = n, and let k > 2(n 1)! By our indution assumption, every proper subalgebra and every proper homomorphi image of A satisfies TS(k) and TS(k 1), and thus also satisfies TST(k 1) Thus the same proof used in the proof of Lemma 71 gives us the onlusion that A satisfies ST(k), unless A is simple Suppose that A is simple Let T be any minimal -subalgebra of A k We have that all the hypotheses of Theorem 67 exept possibly the hypothesis that D(T) = A are true The onlusion of Theorem 67 is false (as A is non-abelian) Thus we are fored to the onlusion that D(T) A In this ase, T is a minimal -subalgebra of B k, B = D(T), and B = ST(k) Thus T is a singleton So we onlude that A = ST(k) This holds for all k m = 2(n 1)!+1 By Theorem 411, we have that A = TS(k) for all k (m 1) A = 2n! By Lemma 45, A = WNU(k) for the same values of k This ompletes our indutive proof of Theorem 81