Products of all elements in a loop and a framework for non-associative analogues of the Hall-Paige conjecture
aa r X i v : . [ m a t h . C O ] A ug Products of all elements in a loop and a framework fornon-associative analogues of the Hall-Paige conjecture
Kyle Pula
Department of MathematicsUniversity of Denver, Denver, CO, USA [email protected]
July 11, 2008
Mathematics Subject Classification: 05B15, 20N05
Abstract
For a finite loop Q , let P ( Q ) be the set of elements that can be represented as aproduct containing each element of Q precisely once. Motivated by the recent proof ofthe Hall-Paige conjecture, we prove several universal implications between the followingconditions:(A) Q has a complete mapping, i.e. the multiplication table of Q has a transversal,(B) there is no N E Q such that | N | is odd and Q/N ∼ = Z m for m ≥
1, and(C) P ( Q ) intersects the associator subloop of Q .We prove ( A ) = ⇒ ( C ) and ( B ) ⇐⇒ ( C ) and show that when Q is a group, theseconditions reduce to familiar statements related to the Hall-Paige conjecture (whichessentially says that in groups ( B ) = ⇒ ( A )). We also establish properties of P ( Q ),prove a generalization of the D´enes-Hermann theorem, and present an elementary proofof a weak form of the Hall-Paige conjecture. As the present work is motivated by the search for non-associative analogues of the well-known Hall-Paige conjecture in group theory, we begin with a brief historical sketch of thistopic. As the culmination of over 50 years of work by many mathematicians, the followingtheorem seems now to have been established.
Theorem 1. If G is a finite group, then the following are equivalent: (1) G has a complete mapping, i.e. the multiplication table of G has a transversal, (2) Sylow -subgroups of G are trivial or non-cyclic, and there is an ordering of the elements of G such that g · · · g n = 1 . The history of Theorem 1 dates back at least to a result of Paige from 1951 in which heproves (1) = ⇒ (3) [15]. In 1955 Hall and Paige proved that for all finite groups (1) = ⇒ (2)and that for solvable, symmetric, and alternating groups the converse holds as well. Theyconjectured that the converse holds in all finite groups, and this claim came to be known asthe Hall-Paige conjecture [13].In 1989 D`enes and Keedwell noted that condition (2) holds in all non-solvable groups andproved that condition (3) does as well [8]. Combining the above results, they observed thatthe Hall-Paige conjecture was thus equivalent to the statement that non-solvable groups havecomplete mappings, and many authors have since contributed to the effort of demonstratingsuch complete mappings. Evans provides an excellent survey of this progress up to 1992 in[11] and of the more recent progress in [12]. Jumping ahead to the most recent developments(consult the previous references for complete details), Wilcox improved on earlier resultsto show that a minimal counterexample to the Hall-Paige conjecture must be simple andreduced the number of candidates to the Tits group or a sporadic simple group (some ofwhich were already known to have complete mappings) [22]. The closing work on theseefforts was completed by Bray and Evans who demonstrated complete mappings in J andall remaining cases, respectively [1, 12].We would like to consider the Hall-Paige conjecture in more general varieties of loops,and this paper takes on the modest goal of making sense of conditions (1), (2), and (3) ofTheorem 1 in non-associative settings. While (1) translates directly, (2) and (3) presentdifficulties since it is not clear what a Sylow 2-subloop should be and any product in anon-associative loop requires the specification of some association.The primary contribution of this paper is to propose natural generalizations of theseconditions and establish several universal implications between them. As detailed in §
2, weprove a number of related results including a generalization of the D´enes-Hermann theoremand provide an elementary proof of a weak form of the Hall-Paige conjecture.We begin with a minimal background on the theories of latin squares and loops. A latin square of order n is an n × n array whose entries consist of n distinct symbols suchthat no symbol appears as an entry twice in a row or column. More formally, a Latin square L on a finite set of symbols S is a subset of S such that the projection along any pair ofcoordinates is a bijection. When the triple ( x, y, z ) ∈ L , we say that the symbol x appearsas a row, y as a column, and z as an entry.A k -plex is a subset of L in which each symbol appears as a row, column, and entryprecisely k times. A 1-plex is called a transversal and a k -plex with k odd is called anodd-plex. For the theory and history of k -plexes, consult [4, 10, 20, 21].We introduce the following generalizations of the concepts of transversals and k -plexes.A row k -plex of L is a collection of triples in L representing each row precisely k times. Wecall a subset C = { ( x i , y i , z i ) : 1 ≤ i ≤ m } ⊆ L column-entry regular , or just regular for2hort, if for each symbol s we have |{ i : y i = s }| = |{ i : z i = s }| . That is, s appears as anentry the same number of times it appears as a column. We denote by C r the multiset ofsymbols appearing as rows in C . For example, if C is a k -plex, then C r contains precisely k copies of each symbol. We will be primarily interested in regular row transversals, i.e.selections of a single cell from each row so that each symbol appears as a column the samenumber of times as an entry (Figure 1 depicts such a selection in the multiplication table ofa loop). A set with a binary operation, say ( Q, · ), is a loop if for each x, z ∈ Q , the equations x · y = z and y · x = z have unique solutions y , y ∈ Q and ( Q, · ) has a neutral element (which wealways denote 1). A group is a loop in which the associativity law holds. We assume in allcases that Q is finite and typically write Q rather than ( Q, · ).The left , right , and middle inner mappings of a loop are defined as L ( x, y ) = L − yx L y L x , R ( x, y ) = R − xy R y R x , and T ( x ) = R − x L x , respectively. A subloop S of a loop Q is said to be normal , written S E Q , if S is invariant under all inner mappings of Q . A loop Q is simple if it has no normal subloops except for { } and Q . This definition of normality is equivalentto what one would expect from the standard universal algebraic definition. In fact, just aswith groups, loops can be defined more formally as universal algebras though we have notelected to do so here.We write A ( Q ) for the associator subloop of Q , the smallest normal subloop of Q suchthat Q/A ( Q ) is a group. Likewise, we write Q ′ for the derived subloop of Q , the smallestnormal subloop of Q such that Q/Q ′ is an Abelian group. Note that A ( Q ) E Q ′ and thuscosets of Q ′ are partitioned by cosets of A ( Q ). When Q is a group, A ( Q ) = { } .The multiplication table of a loop Q is the set of triples { ( x, y, xy ) : x, y ∈ Q } . Multipli-cation tables of loops are latin squares and, up to the reordering of rows and columns, everylatin square is the multiplication table of some loop. It thus makes sense to say that a loophas a k -plex or more generally a regular row k -plex whenever its multiplication table does.Most literature on the Hall-Paige conjecture focuses on the concept of a complete mappingof a group rather than a transversal of its multiplication table, though the two are completelyequivalent [9, p. 7]. In the general loop setting, we prefer the latter concept as it emphasizesthe combinatorial nature of the problem and generalizes more naturally to the concepts of k -plexes and regular k -plexes.For H ⊆ Q and k ≥
1, let P k ( H ) be the set of elements in Q that admit factorizationscontaining every element of H precisely k times. We call these elements full k -products of H . When k = 1, we write just P ( H ) and refer to its elements as full products of H . We workprimarily in the case H = Q and simply refer to these elements as full k -products . While inthe group case, the set P ( G ) has been well-studied (see the commentary preceding Theorem4 for some background), to the best of our knowledge, the present article contains the firstinvestigation of the general loop case.Although we assume a basic familiarity with both loops and latin squares, we providereferences for any non-trivial results that we employ. For standard references on latin squares3onsult D´enes and Keedwell [7, 9] and for loops Bruck [2] and Pflugfelder [16]. We propose the following conditions as fruitful interpretations of (1), (2), and (3) of Theorem1 in varieties of loops in which associativity need not hold.
Definition 2 (HP-condition) . We say a class of loops Q satisfies the HP-condition if foreach Q ∈ Q the following are equivalent:(A) Q has a transversal,(B) there is no N E Q such that | N | is odd and Q/N ∼ = Z m for m ≥
1, and(C) A ( Q ) intersects P ( Q ).When Q is the variety of groups, satisfaction of the HP-condition reduces to Theorem1. The equivalence of (1) and (A) is clear; as is that of (3) and (C), given that when Q isa group, A ( Q ) = { } . We take an indirect approach to showing (2) ⇐⇒ (B) by showing(B) ⇐⇒ (C), a corollary of Theorems 5 and 6. In 2003, Vaughan-Lee and Wanless gavethe first elementary proof of (2) ⇐⇒ (3). Their paper also provides some background onthis result (whose initial proof invoked the Feit-Thompson theorem) [18]. As corollaries ofour main results, we show that in all loops (B) ⇐⇒ (C) and (A) = ⇒ (C).The following easy observation sets the context for our main results. It is well-known inthe group case and follows in the loop case for the same simple reasons. We include it aspart of Lemma 12 for completeness. Observation 3. P k ( Q ) is contained in a single coset of Q ′ . At least as far back as 1951, authors have asked whether, in the group case, this obser-vation can be extended to show that P k ( Q ) in fact coincides with this coset. For a historyof this line of investigation, see [7, p. 35] and [9, p. 40]. This result now bears the names ofD´enes and Hermann who first established the claim for all groups. Theorem 4 (D´enes, Hermann [6] 1982) . If G is a group, then P ( G ) is a coset of G ′ . Itfollows that P k ( G ) is also a coset of G ′ . An admittedly cumbersome but more general way to read this statement is that P ( G )intersects every coset of A ( G ) that is contained in the relevant coset of G ′ . Since A ( G ) = { } and these cosets partition cosets of G ′ , this technical phrasing reduces to the theorem asstated. We extend the D´enes-Hermann theorem to show that this more general phrasingholds in all loops. Although the result is more general, our proof of Theorem 5 reliesessentially upon the D´enes-Hermann theorem. In §
7, we discuss prospects of a furthergeneralization. 4 heorem 5. If P ( Q ) ⊆ xQ ′ , then P ( Q ) ∩ yA ( Q ) = ∅ for all y ∈ xQ ′ . That is, P ( Q ) intersects every coset of A ( Q ) contained in xQ ′ and, in particular, if P ( Q ) ⊆ Q ′ , then P ( Q ) intersects A ( Q ) . It follows that P k ( Q ) also intersects every coset of A ( Q ) in thecorresponding coset of Q ′ . Coupled with Theorem 5, our next result establishes (B) ⇐⇒ (C). Theorem 6. P ( Q ) ⊆ Q ′ if and only if ( B ) holds. In 1951, Paige showed that if a group G has a transversal, then 1 ∈ P ( G ) [15]. We extendthis result to a much wider class of structures. Theorem 7. If C is a regular subset of the multiplication table of Q , then P ( C r ) intersects A ( Q ) . In particular, if Q has a k -plex (or just a regular row k -plex), then P k ( Q ) intersects A ( Q ) . Applying these results, we establish that for all loops: • (A) = ⇒ (C) by Theorem 7 and • (B) ⇐⇒ (C) by Theorems 5 and 6.By 1779, Euler had shown that a cyclic group of even order has no transversal and in1894 Maillet extended his argument to show that all loops for which condition (B) fails lacktransversals [7, p. 445]. In 2002, Wanless showed that such loops lack not just transversals,i.e. 1-plexes, but contain no odd-plexes at all [20]. While their arguments are quite nice, ourproof of (A) = ⇒ (B) provides an alternative, more algebraic proof of these results. Corollary 8.
If a loop fails to satisfy ( B ) , then it has no regular row odd-plexes. It is not true in general that (B) ∧ (C) = ⇒ (A) (for a smallest possible counter-example,see Figure 1). In a separate paper still in preparation, we show that this implication holdsin several technical varieties of loops that include non-associative members and provide bothcomputational and theoretical evidence suggesting that it may hold in the well-known varietyof Moufang loops as well [17].The equivalent statements in Theorem 1 are typically stated with the additional claimthat G can be partitioned into n mutually disjoint transversals, i.e. G has an orthogonalmate. In the group case, it is easy to show that having an orthogonal mate is equivalentto having at least one transversal. While this equivalence may extend to other varieties ofloops (this question for Moufang loops is addressed directly in [17]), the argument seemsunrelated to the most difficult part of Theorem 1 (that (2) = ⇒ (1)).We do however introduce a weakening of the orthogonal mate condition in the followingtheorem. While this result follows directly from a combination of the Theorems 1 and 7,we provide an elementary proof (in particular, we avoid the classification of finite simplegroups). Theorem 9. If G is a group of order n , then the following are equivalent: Q with no transversal and yet P ( Q ) = Q ′ = Q . Q contains 168 regular rowtransversals, one of which has been bracketed. (i) G has a regular row transversal,(ii) G can be partitioned into n mutually disjoint regular row transversals,(iii) Sylow -subgroups of G are trivial or non-cyclic, and(iv) ∈ P ( G ) . Corollary 10.
Theorem 1 is equivalent to the claim that a group has a transversal if andonly if it has a regular row transversal.
We make the following two observations not to suggest that our methods may be usefulin tackling these important problems but rather to indicate their theoretical context.
Observation 11.
When Q is the class of odd ordered loops, condition ( B ) always holds andthus the implication (B) = ⇒ (A) is equivalent to Ryser’s conjecture, that every latin squareof odd order has a transversal [ ] .A natural question might be whether Q has a -plex if and only if A ( Q ) intersects P ( Q ) .Since this latter condition is satisfied in all loops, an affirmative answer to this question isequivalent a conjecture attributed to Rodney, that every latin square has a -plex [ , p. . P k ( Q ) We begin with a sequence of easy observations about the sets P k ( Q ). Lemma 12.
For i, j, k ≥ ,(i) ∈ P ( Q ) ,(ii) P i ( Q ) P j ( Q ) ⊆ P i + j ( Q ) and | P k ( Q ) | ≤ | P k +1 ( Q ) | ,(iii) P k ( Q ) is contained in a coset of Q ′ ,(iv) P k ( Q ) ⊆ P k +2 ( Q ) , v) P ( Q ) ⊆ Q ′ , and(vi) P ( Q ) ⊆ aQ ′ where a ∈ Q ′ .Proof. (i) Let q ρ be the right inverse of q . Then 1 = Q q ∈ Q qq ρ ∈ P ( Q ).(ii) Observe that P i ( Q ) P j ( Q ) = { ab : a ∈ P i ( Q ) , b ∈ P j ( Q ) } . Thus ab is a full ( i + j )-product. It then follows that | P k ( Q ) | ≤ | P k +1 ( Q ) | since for q ∈ P ( Q ), qP k ( Q ) ⊆ P k +1 ( Q )and | P k ( Q ) | = | qP k ( Q ) | .(iii) Any two elements of P k ( Q ) have factors that differ only in their order and association.In other words, if x, y ∈ P k ( Q ), then xQ ′ = yQ ′ .(iv) By (i), we have 1 ∈ P ( Q ); thus P k ( Q ) = P k ( Q ) · ⊆ P k ( Q ) P ( Q ). By (ii) we have P k ( Q ) P ( Q ) ⊆ P k +2 ( Q ). Thus P k ( Q ) ⊆ P k +2 ( Q ).(v) The claim follows immediately from (i) and (iii).(vi) By (iii), P ( Q ) ⊆ aQ ′ for some a ∈ Q and thus P ( Q ) ⊆ a Q ′ . By (ii), P ( Q ) ⊆ P ( Q )and by (v) P ( Q ) ⊆ Q ′ . It follows that a Q ′ = Q ′ and thus a ∈ Q ′ .Our next lemma uses the idea that Q/N is a set of cosets of N and thus P ( Q/N ) is asubset of these cosets.
Lemma 13. If N E Q , | N | = k , and a N, . . . , a k N ∈ P ( Q/N ) , then P ( Q ) ∩ ( a N · · · a k N ) = ∅ . That is, P ( Q ) intersects every member of P ( Q/N ) k .Proof. Let | Q | = mk . For any aN ∈ P ( Q/N ), we may select a system of coset representativesof N in Q , say { x , . . . , x m } , and some association of the left hand side such that x N · · · x m N = aN (1)and thus using the same association pattern x · · · x m ∈ aN . Furthermore, since (1) dependsonly on the order and association of the cosets of N (rather than the specific representativeschosen), we may select k disjoint sets of coset representatives of N in Q , say { x ( i, , . . . , x ( i,m ) :1 ≤ i ≤ k } , and corresponding association patterns such that( x (1 , · · · x (1 ,m ) ) · · · ( x ( k, · · · x ( k,m ) ) ∈ a N · · · a k N ∈ P ( Q/N ) k for any selection of a i N ∈ P ( Q/N ) for 1 ≤ i ≤ k . Having selected each element of Q as acoset representative precisely once, the left-hand side falls in P ( Q ) and we are done. For x ∈ Q , we write L x ( R x ) for the left (right) translation of Q by x . Our notation for theleft translation is not to be confused with the convention of using L for a latin square. The7atter usage appears only in our introduction. In any case, the meaning is always made clearby the context.The multiplication group of Q , written Mlt( Q ), is the subgroup of the symmetric group S Q generated by all left and right translations, i.e. h L x , R x : x ∈ Q i , while the left multiplicationgroup of Q , written LMlt( Q ), is generated by all left translations. If H E Q and ρ ∈ Mlt( Q ), we may define the map ρ H ( xH ) := ρ ( x ) H , which is said to be induced by ρ . It isstraightforward to verify that the map is well-defined and that ρ H ∈ Mlt(
Q/H ).To prove Theorem 7 in the group case one would like to use the fact that from an identitylike a ( a ( · · · ( a k x ) · · · ) = x we may conclude that a ( a ( · · · ( a k ) · · · ) = 1, which is trivial in the presence of associativitybut typically false otherwise. In the general loop case, the following lemma shows we can atleast conclude that a ( a ( · · · ( a k ) · · · ) ∈ A ( Q ) . Lemma 14. (i) If ρ ∈ LMlt ( Q ) , then ρ A ( Q ) = L ρ (1) A ( Q ) .(ii) If ρ ∈ Mlt ( Q ) , then ρ Q ′ = L ρ (1) Q ′ .(iii) Since they are left translations, ρ A ( Q ) and ρ Q ′ are constant if and only if they have fixedpoints.(iv) If a ( a ( · · · ( a k x ) · · · ) = x , then a ( a ( · · · ( a k ) · · · ) ∈ A ( Q ) .Proof. Set A := A ( Q ).(i) Let ρ = L a · · · L a k . Then ρ A ( qA ) = a ( a · · · ( a k q ) · · · ) A . Since Q/A is a group, wemay reassociate to get ρ A ( qA ) = a ( a · · · ( a k ) · · · ) A · qA = ρ (1) A · qA . Thus ρ A = L ρ (1) A .(ii) Let ρ = T ǫ a · · · T ǫ k a k where T ǫ i ∈ { L, R } . Since Q/Q ′ is an Abelian group, we mayreassociate and commute to get ρ Q ′ ( qQ ′ ) = a ( a · · · ( a k ) · · · ) Q ′ · qQ ′ = ρ (1) Q ′ · qQ ′ . Thus ρ Q ′ = L ρ (1) Q ′ .(iii) Since ρ A ( Q ) is a left translation, if it has a fixed point, it is constant. Likewise for ρ Q ′ .(iv) Let ρ ( z ) := a ( a ( · · · ( a k z ) · · · ). Since ρ A ( xA ) = ρ ( x ) A = xA , it is constant by (iii).Thus ρ A ( A ) = A and in particular ρ (1) = a ( a ( · · · ( a k ) · · · ) ∈ A .Lemma 14 is stated somewhat more generally then we actually need. If the translationnotation feels cumbersome, the idea is very basic. Given the product a ( a ( · · · ( a k x ) · · · ) = x ,we may reduce both sides mod A to get a Aa A · · · a k AxA = xAa Aa A · · · a k A = 1 Aa a · · · a k ∈ A. emma 15. If C = ∅ is regular, then there exists C ′ such that(i) ∅ 6 = C ′ ⊆ C ,(ii) P ( C ′ r ) intersects A ( Q ) , and(iii) C \ C ′ is regular (and possibly empty).It follows that P ( C r ) intersects A ( Q ) .Proof. Let [ k ] := { , . . . , k } . Suppose C = { ( x i , y i , z i ) : i ∈ [ k ] } is regular. Select i ∈ [ k ] atrandom. Having selected i , · · · , i m ∈ [ k ], pick i m +1 ∈ [ k ] such that y i m = z i m +1 . Since C isregular, such a selection can always be made. If i m +1
6∈ { i , . . . , i m } , continue.Otherwise, stop and consider the set { i j , . . . , i m } where i j = i m +1 . Reindex C such that h i j , . . . , i m i = h , · · · , s i and set C ′ := { ( x i , y i , z i ) : 1 ≤ i ≤ s } . Note that y s = z . Byconstruction, C ′ has the following form: C ′ = { ( x , y , y s ) , ( x , y , y ) , ( x , y , y ) , · · · ( x s − , y s − , y s − ) , ( x s , y s , y s − ) } .C ′ is clearly regular and thus so too is C \ C ′ . Furthermore, by construction we have x ( x ( · · · ( x s z ) · · · )) = z . By Lemma 14, x ( x ( · · · ( x s ) · · · )) ∈ A ( Q ). Since this product is in P ( C ′ r ) as well, P ( C ′ r ) ∩ A ( Q ) = ∅ . Iterating this construction we have P ( C r ) intersects A ( Q ). Proof of Theorem 7. If C is a k -plex, then C r consists of k copies of each element of Q andthus P ( C r ) = P k ( Q ). By Lemma 15, P k ( Q ) intersects A ( Q ). Lemma 16. If G is a group and g , . . . , g k ∈ G such that g · · · g k = 1 and no proper con-tiguous subsequence evaluates to , then G admits a regular set C such that C r = { g , . . . , g k } and no column (and thus no entry) is selected more than once.Proof. Set h i := g i +1 · · · g k for 1 ≤ i ≤ k − h = h k := 1. Note that we have g i h i = h i − for 1 ≤ i ≤ k . We claim that C := { ( g i , h i , h i − ) : 1 ≤ i ≤ k } is the desired regular set. It is9lear that C is regular and that C r = { g , . . . , g k } . To see that no column is selected morethan once, suppose that h i = h i + j for j ≥
1. That is, g i +1 · · · g k = g i + j +1 · · · g k . Cancelingon the right, we have g i +1 · · · g i + j = 1, a contradiction. Proof of Theorem 9. (i) = ⇒ (ii) In this case we may use the standard argument from the group case showingthat a single transversal extends to n disjoint transversals. Let T = { ( x i , y i , z i ) : 1 ≤ i ≤ n } be a regular row transversal of G . For each g ∈ G , form T g := { ( x, yg, zg ) : ( x, y, z ) ∈ T } .It is easy to check that the family { T g : g ∈ G } partitions the multiplication table of G intoregular row transversals.(i) ⇐ = (ii) If G admits a partition into regular row transversals, then it certainly has aregular row transversal.(i) = ⇒ (iv) Let T be a regular row transversal. By Theorem 7, P ( G ) ∩ A ( G ) = ∅ andthus 1 ∈ P ( G ).(i) ⇐ = (iv) Let g · · · g n = 1. We partition G as follows: • If no proper contiguous subsequence of g · · · g n evaluates to 1, stop. • Otherwise, extract the offending subsequence g i · · · g j = 1 and note that g · · · g i − g j +1 · · · g n = 1 . • Iterate this process with these shortened products.Suppose we have thus partitioned G into k disjoint sequences { g ( i, , . . . , g ( i,n i ):1 ≤ i ≤ k } suchthat g ( i, · · · g ( i,n i ) = 1 for 1 ≤ i ≤ k and no proper contiguous subsequence of g ( i, , . . . , g ( i,n i ) evaluates to 1. Now we apply Lemma 16 to each subsequence to get regular sets C i for1 ≤ i ≤ k . Then S ki =1 C i is a regular row transversal of G .(iii) ⇐⇒ (iv) As noted earlier, this is an established equivalence in the Hall-Paigeconjecture. Lemma 17. (2) ⇐⇒ (3) holds for Abelian groups.Proof. As mentioned above, Vaughan-Lee and Wanless give a direct, elementary proof ofthis result for all groups [18]. For an earlier though indirect proof, Paige showed that (1) ⇐⇒ (2) holds in Abelian groups [14] and Hall and Paige showed that (1) ⇐⇒ (3) insolvable groups [13]. Lemma 18.
If a group G has a cyclic Sylow -subgroup S , then there exists N E G suchthat G/N ∼ = S . roof. This is a direct application of Burnside’s Normal Complement theorem that can befound in most graduate level group theory texts (see [24] for example).
Proof of Theorem 6. ( ⇐ =) We show the contrapositive. Suppose P ( Q ) ⊆ aQ ′ = Q ′ . Since G := Q/Q ′ is an Abelian group, P ( G ) = { bQ ′ } such that b ∈ Q ′ . By Lemma 13, P ( Q )intersects every element of P ( G ) | Q ′ | = { b | Q ′ | Q ′ } and thus aQ ′ = b | Q ′ | Q ′ . Since aQ ′ = Q ′ and b ∈ Q ′ , it follows that aQ ′ = bQ ′ and | Q ′ | is odd.Since P ( G ) = { Q ′ } , by Lemmas 17 and 18 there is N E G such that | N | is odd and G/N ∼ = Z m . N is a collection of coset of Q ′ . Letting H be their union, we have Q/H ∼ = G/N ∼ = Z m and | H | = | N || Q ′ | is odd.(= ⇒ ) Again we argue the contrapositive. Suppose N E Q such that | N | = q is odd and Q/N ∼ = Z m for m ≥
1. Since
Q/N ∼ = Z m , P ( Q/N ) = { aN } 6 = { N } such that a ∈ N . ByLemma 13, P ( Q ) intersects every element of P ( Q/N ) | N | = { aN } | N | = { aN } .Given that Q/N is an Abelian group, Q ′ ⊆ N but since P ( Q ) intersects aN = N , it istherefore disjoint from Q ′ . An A -loop is a loop in which all inner mappings are automorphisms. The variety of A -loopsis larger than that of groups but is certainly not all loops. Bruck and Paige conducted theearliest extensive study of A -loops [3].Before proving Theorem 5, we make several additional observations about the sets P k ( Q ).While none of these results will be used directly in our proof, we hope they are of some interestin that they may suggest an alternative proof of the D´enes-Hermann theorem. Lemma 19.
Set P ω := S ∞ i =1 P i ( Q ) .(i) P ω ≤ Q ,(ii) P ω = P k ( Q ) ∪ P k +1 ( Q ) for sufficiently large k ,(iii) If P ω E Q , then P ω = Q ′ or P ω = Q ′ ∪ aQ ′ where a ∈ Q ′ , and(iv) P ω is fixed by all automorphisms of Q . Thus, if Q is an A -loop, then P ω E Q .Proof. (i) Since Q is finite, we need only verify that P ω is closed under multiplication. If x, y ∈ P ω , then x ∈ P i ( Q ) and y ∈ P j ( Q ) for some i, j ≥
1. Thus xy ∈ P i + j ( Q ) ⊆ P ω .(ii) Again, since Q is finite, the nested sequence ( P i ( Q ) : 1 ≤ i < ∞ ) must terminateat some step, say P k ( Q ). Likewise ( P i +1 ( Q ) : 1 ≤ i < ∞ ) must terminate at some step,say P k ( Q ). Thus letting k = max { k , k } , we have P ω = P k ( Q ) ∪ P k +1 ( Q ). (In fact, byLemma 12 part (ii), the sequences terminate at the same time.)(iii) Suppose P ω E Q . We show that Q/P ω is an Abelian group and thus Q ′ ⊆ P ω . Tosee that Q/P ω is a group, note that aP ω bP ω · cP ω = ( ab · c ) P ω . We would like to show that( ab · c ) P ω = ( a · bc ) P ω . 11o that end, let a ′ ∈ P ( Q \ { a } ) and likewise for b ′ and c ′ . We translate both ( ab · c ) P ω and ( a · bc ) P ω by ( a ′ b ′ · c ′ ) P ω on the left to get( a · bc ) P ω · ( a ′ b ′ · c ′ ) P ω = [( a · bc ) · ( a ′ b ′ · c ′ )] P ω ( ab · c ) P ω · ( a ′ b ′ · c ′ ) P ω = [( ab · c ) · ( a ′ b ′ · c ′ )] P ω Note that both ( a · bc ) · ( a ′ b ′ · c ′ ) and ( ab · c ) · ( a ′ b ′ · c ′ ) are elements of P ( Q ) and thus both right-hand sides reduce to P ω . Thus both ( ab · c ) P ω and ( a · bc ) P ω are left inverses of ( a ′ b ′ · c ′ ) P ω .Since left inverses are unique, we have ( ab · c ) P ω = ( a · bc ) P ω and Q/P ω is a group.To see that Q/P ω is Abelian, consider aP ω bP ω and bP ω aP ω . Again let a ′ ∈ P ( Q \ { a } )and b ′ ∈ P ( Q \ { b } ). We then have abP ω · a ′ b ′ P ω = ( ab · a ′ b ′ ) P ω baP ω · a ′ b ′ P ω = ( ba · a ′ b ′ ) P ω Since ( ab · a ′ b ′ ) and ( ba · a ′ b ′ ) are both members of P ( Q ), the right-hand sides reduce to P ω .As above, it follows that aP ω bP ω = bP ω aP ω .Since Q ′ is the smallest normal subloop of Q such that Q/Q ′ is an Abelian group, Q ′ ⊆ P ω .(iv) First note that for i ≥ P i ( Q ) is always fixed by automorphisms of Q and thus sois P ω . If Q is an A -loop, then P ω is fixed by every inner-mapping. That is, P ω E Q .In the spirit of the observations made in Lemma 19, Yff showed that when G is a group, P ( G ) coincides with a coset of G ′ [23, p. 269]. Although this fact is an easy applicationof the D´enes-Hermann theorem, his proof applies directly to all finite groups and avoids theuse of the Feit-Thompson theorem.To our knowledge, Theorem 5 is the first extension beyond groups of the D´enes-Hermanntheorem. Although our generalization is rather modest, we suspect the result extends almostcompletely. Question 20. If | Q | is sufficiently large, does it follow that P ( Q ) is a coset of Q ′ ? The D´enes-Hermann theorem is equivalent to the claim that for any finite group G |{ g · · · g n : ranging over all orderings }| = | G ′ | . To answer Question 20 affirmatively, it would suffice to show the perhaps stronger claim thatgiven any fixed ordering of the elements of Q , we have |{ q · · · q n : ranging over all associations }| = | A ( Q ) | . The restriction on | Q | in Question 20 is necessary as Vojt˘echovsk´y and Wanless haveobserved that for 3 of the 5 non-associative loops of order 5, P ( Q ) has order 4 whereas Q = Q ′ (see Figure 2 for one example) [19]. At the time of this writing, we are unaware ofany other loops with this property. 12 Q for which Q = Q ′ but P ( Q ) = Q \ { } . As such, Q demonstrates thenecessity of the cardinality restriction in Question 20.We recall the following special case of the correspondence and isomorphism theorems,proofs of which can be found in most standard universal algebra texts. Lemma 21. If N E Q and N ≤ H ≤ Q , then(i) H E Q if and only if H/N E Q/N and(ii) when H E Q , Q/H ∼ = ( Q/N ) / ( H/N ) . We employ the following lemma in our proof of Theorem 5.
Lemma 22. ( Q/A ( Q )) ′ = Q ′ /A ( Q ) .Proof. Set A := A ( Q ). By definition, ( Q/A ) ′ is the smallest normal subloop of Q/A such thatthe factor loop is an Abelian group. Since A ≤ Q ′ E Q , by the correspondence theorem wehave ( Q ′ /A ) E ( Q/A ) and (
Q/A ) / ( Q ′ /A ) ∼ = Q/Q ′ , an Abelian group. Thus ( Q/A ) ′ ≤ ( Q ′ /A ).We now show ( Q ′ /A ) ≤ ( Q/A ) ′ . Fix N/A E Q/A such that (
Q/A ) / ( N/A ) is an Abeliangroup. Again by the correspondence theorem, N E Q and Q/N ∼ = ( Q/A ) / ( N/A ). Since
Q/N is an Abelian group, Q ′ ≤ N and thus Q ′ /A ≤ N/A . It follows that ( Q ′ /A ) ≤ ( Q/A ) ′ Proof of Theorem 5.
Let A := A ( Q ) and k := | A | . Since P ( Q ) is contained in a single cosetof Q ′ , it suffices to show that P ( Q ) intersects at least [ Q ′ : A ] cosets of A (the maximumpossible).By Theorem 4, P ( Q/A ) = { xA ( Q/A ) ′ } such that x A ∈ ( Q/A ) ′ . By Lemma 22, xA ( Q/A ) ′ = xA ( Q ′ /A ) = ( xQ ′ ) A . Thus we have P ( Q/A ) = { ( xQ ′ ) A } . By Lemma 13, P ( Q )intersects each of the [ Q ′ : A ] elements of P ( Q/A ) k = { ( x k Q ′ ) A } = { qA : q ∈ x k Q ′ } . We have proposed the HP-condition as a possible framework for extensions of Theorem 1from groups into the larger world of non-associative loops. Having shown several universalimplications between the points of the HP-condition, we leave open the difficult problem ofidentifying interesting varieties of loops in which conditions (B) and (C) imply (A).13t would also be of interest to identify classes of loops in which the existence of a reg-ular row transversal implies the existence of a transversal. As noted in Corollary 10, thisimplication in groups is fully equivalent to Theorem 1.I thank Michael Kinyon and Petr Vojtˇechovsk´y for their helpful conversations regardingthis material and Anthony Evans and Ian Wanless for sharing several articles and preprintsrelated to the Hall-Paige conjecture. Lastly I thank the anonymous referee whose feedbackclarified the historical background and helped to focus the exposition.
References [1] J. Bray. Unpublished notes. 2008.[2] R. H. Bruck.
A survey of binary systems . Ergebnisse der Mathematik und ihrer Gren-zgebiete. Neue Folge, Heft 20. Reihe: Gruppentheorie. Springer Verlag, Berlin, 1958.[3] R. H. Bruck and L. J. Paige. Loops whose inner mappings are automorphisms.
Ann. ofMath. (2) , 63:308–323, 1956.[4] D. Bryant, J. Egan, B. Maenhaut, and I. M. Wanless. Indivisible plexes in latin squares.
Designs, Codes and Cryptography , 52(1), July 2009.[5] C. J. Colbourn and J. H. Dinitz.
The CRC Handbook of Combinatorial Designs . CRCPress, Boca Raton, FL, 1996.[6] J. D´enes and P. Hermann. On the product of all elements in a finite group. In
Algebraicand geometric combinatorics , volume 65 of
North-Holland Math. Stud. , pages 105–109.North-Holland, Amsterdam, 1982.[7] J. D´enes and A. D. Keedwell.
Latin squares and their applications . Academic Press,New York, 1974.[8] J. D´enes and A. D. Keedwell. A new conjecture concerning admissibility of groups.
European J. Combin. , 10(2):171–174, 1989.[9] J. D´enes and A. D. Keedwell.
Latin squares , volume 46 of
Annals of Discrete Mathe-matics . North-Holland Publishing Co., Amsterdam, 1991.[10] J. Egan and I. M. Wanless. Latin squares with no small odd plexes.
Journal of Combi-natorial Designs , 16(6):477–492, 2008.[11] A. B. Evans. The existence of complete mappings of finite groups. In
Proceedings of theTwenty-third Southeastern International Conference on Combinatorics, Graph Theory,and Computing (Boca Raton, FL, 1992) , volume 90, pages 65–75, 1992.[12] A. B. Evans. The admissibility of sporadic simple groups.
J. Algebra , 321:1407–1428,2009. 1413] M. Hall and L. J. Paige. Complete mappings of finite groups.
Pacific J. Math. , 5:541–549, 1955.[14] L. J. Paige. A note on finite Abelian groups.
Bull. Amer. Math. Soc. , 53:590–593, 1947.[15] L. J. Paige. Complete mappings of finite groups.
Pacific J. Math. , 1:111–116, 1951.[16] H. O. Pflugfelder.
Quasigroups and loops: introduction , volume 7 of
Sigma Series inPure Mathematics . Heldermann Verlag, Berlin, 1990.[17] K. Pula. The Hall-Paige conjecture for finite Moufang loops. In preparation, 2009.[18] M. Vaughan-Lee and I. M. Wanless. Latin squares and the Hall-Paige conjecture.
Bull.London Math. Soc. , 35(2):191–195, 2003.[19] Vojtˇechovsk´y and Wanless. Private correspondence. February 2009.[20] I. M. Wanless. A generalisation of transversals for Latin squares.
Electron. J. Combin. ,9(1):Research Paper 12, 15 pp. (electronic), 2002.[21] I. M. Wanless. Transversals in Latin squares.
Quasigroups Related Systems , 15(1):169–190, 2007.[22] S. Wilcox. Reduction of the hall-paige conjecture to sporadic simple groups.
J. Algebra ,321(5):1407–1428, March 2009.[23] P. Yff. On the D´enes-Hermann theorem: a different approach.
European J. Combin. ,12(3):267–270, 1991.[24] H. Zassenhaus.