1 Bol loops Bol loops that are centrally nilpotent of class 2 Orin Chein (Edgar G Goodaire)

2 This talk is based on joint work with E. G. Goodaire, which may be found in the following articles: –Bol loops of nilpotence class 2, O. Chein and E.G. Goodaire, Canadian Journal of Mathematics, to appear. –A new construction of Bol loops of order 8m, O. Chein and Edgar G. Goodaire, J. Algebra, 287 (1) (2005), –A new construction of Bol loops: The "odd" case, O. Chein and Edgar G. Goodaire, submitted. –Bol loops with a unique nonidentity commutator/associator, O. Chein and Edgar G. Goodaire, submitted.

3 Outline Why nilpotence class 2? Properties of Bol loops of nilpotence class 2 Some constructions Connections with strongly right alternative ring (SRAR) loops

4 Why nilpotence class 2 Historical reasons Practical reasons

5 Historical reasons Unique nontrivial commutator/associator 1986: Loops Whose Loop Rings are Alternative 1990: Moufang Loops with a Unique Nonidentity Commutator (Associator, Square) Minimally nonassociative Moufang loops 2003: Minimally nonassociative nilpotent Moufang loops

6 Practical reasons In a loop of nilpotence class 2, commutators and associators are central.commutators and associatorscentral

7 Properties of Bol loops of nilpotence class 2 Associator identities: If L is a Bol loop that is centrally nilpotent of class 2, then, for all x, y, z, w in L and any integers m, n. p, q, r, s,centrally nilpotent of class 2 A1A1: (x,z,y) = (x,y,z) -1 A2A2: (x, y, zy n ) = (x, y, z) (x, zy n, y) = (x, z, y) A3:(xy n, y, z) = (x, y, z)(y, y, z) n (xy n, z, y) = (x, z, y)(y, z, y) n

8 A4: (x, y, zx) = (x, y, z)(x, yz, x)(x, x, z) (x, zx, y) = (x, z, y)(x, x, yz)(x, z, x) A5: (x,y m,z n ) =(x,y,z) mn A6: (y n, y, z) = (y, y, z) n [In particular, (y -1, y, z) = (y, y, z) -1 ] A7A7: (xy,z,w)(x,y,zw) = (x,y,z)(y,z,w)(x,yz,w) A8: (x, y, z) 2 (y, z, x) 2 (z, x, y) 2 = 1

9 Commutator identities: If L is a Bol loop that is centrally nilpotent of class 2, then, for all x, y, z, w in L C1: (x,y) -1 = (y,x) C2: (x m, y n ) = (x, y) mn (x, x, y) mn(m-1) (y, x, y) mn(n-1) C3: (xy n, y) = (x, y)(y, x, y) n C4: (xz, y) = (x, y)(z, y)(y, x, z)(x, z, y) 2, (x, yz) = (x, y)(x, z)(x, z, y)(y, x, z) 2 C5: (xz,y)(yx,z)(zy,x) = (x,z,y)(y,x,z)(z,y,x)

10 Two-generator identities: T1: (x m y n,x p y q,x r y s ) = (x,x,y) m(ps-qr) (y,y,x) n(qr-ps) T2: (x m y n, x p y q ) = (x, y) mq-np (x, x, y) r (y, y, x) s, where r = (mq - np)(m + p - 1) and s = (np - mq)(n + q - 1) T3: (x m y n )(x p y q ) = x m+p y q+n (y,x) np (x,x,y) -np(p-1)-mp(q+2n) (y,y,x) np(n+q-1)

11 Two-generator Bol loops If L is generated by x and y, then, by T1 and T2, all commutators and associators in L can be expressed in the form r  s  t , where r = (x,y), s = (x,x,y) and t = (y,y,x) are central. Thus every element in L can be expressed in the form x p y q r  s  t . T3 tells us how to multiply two such elements, so that multiplication in L is completely determined once we know |x|, |y|, |r|, |s| and |t|.

12 If associators square to 1 S1: (x, z, y) = (x, y, z). S2: x 2  N(L). S3: (x m, y n, z r ) = (x,y,z) mnr S4: (x m,y n ) =(x,y) mn S5: (x m y n,x p y q ) = (x,y) mq-np (x, x, y) mp(n+q) (y, y, x) nq(m+p) S6: (x m y n )(x p y q ) = x m+p y q+n (y,x) np (x, x, y) mpq (y, y, x) npq S7: (xz,y) = (x,y)(z,y)(y,x,z)

13 If commutators also square to 1 CS1: x 2  Z(L). CS2: If |L|=2, then squares are central in L.

14 Minimally non-Moufang Bol loops Let L be a finite Bol loop that is nilpotent of class two and in which, for all x, y in L, (x, x, y) 2 =1. Suppose that L is minimally non-Moufang in the sense that it is not Moufang, but every proper subloop of L is Moufang. Then every proper subloop of L is associative.

15 Preliminary construction Let B be a loop, and let z be any fixed element in the center of B. Define an operation on the set G = B  C m by [a, i][b, j] = [abz q, (i+j)*], where (i+j)* is the least non-negative residue of i+j modulo m, and where mq = i + j - (i + j)*. Then G is a loop. Furthermore, G is Bol (respectively Moufang, respectively associative, respectively commutative) if and only if B is Bol (respectively Moufang, respectively associative, respectively commutative).

16 The main construction Let m and n be even positive integers, let B be a loop satisfying the right Bol identity, and let r, s, t, z and w be (not necessarily distinct) elements in Z(B), the center of B, such that r m = r n = s 2 = t 2 = 1. Let L = B  C m  C n with multiplication defined by [a, i,  ][b, j,  ] = [abr j  s ij  t j  z p w q,(i + j)*,(  +  )], where, for any integer i, i* and i denote the least nonnegative residues of i modulo m and n, respectively, mp = i+j - (i+j)*, and nq =  +  -(  +  ). Then L is a right Bol loop.

17 Furthermore, L is Moufang if and only if B is Moufang and s = t = 1, L is a group if and only if B is a group (and s = t = 1), and L is commutative if and only if B is commutative and r = 1. We denote the loop constructed in this manner by L(B, m, n, r, s, t, z, w).

18 Explanation of notation We are thinking of the elements of our loop as being of the form (au i )v  (hence the notation [a, i,  ]), where a  B, u generates C m and v generates C n, and where u m = z  Z(B) and v n = w  Z(B). Thus, for example, u i+j = z p u (i+j)*, where i+j = mp + (i+j)*. We use a mix of Roman and Greek characters to indicate the source from which an exponent comes – Roman for the exponents of u, the second coordinate, and Greek for the exponents of v, the third coordinate.

19 The elements r, s and t represent commutators and associators. Specifically, r represents the commutator of u and v, and the exponent j  indicates that we are considering the commutator of u  and v j. Similarly, s and t, respectively, represent the associators (u, u, v) and (v, u, v), and the exponents ij  on s and j  on t indicate that we are associating (u i, u j, v  ) and (v , u j, v  ), respectively.

20 We assume that r m = r n = s 2 = t 2 = 1 so that we can ignore the impact of reducing modulo m and n in the second and third coordinates. [Note that the conditions r m = r n = s 2 = t 2 = 1 are necessary.]Note that the conditions r m = r n = s 2 = t 2 = 1 are necessary

21 Bol loops of order 8 The main construction produces all six Bol loops of order 8. L 1 = L(C 2, 2, 2, a, a, a, a, a) L 2 = L(C 2, 2, 2, a, a, 1, a, 1) L 3 = L(C 2, 2, 2, a, a, 1, a, a) L 4 = L(C 2, 2, 2, a, a, a, a, 1) L 5 = L(C 2, 2, 2, a, a, a, 1, 1) L 6 = L(C 2, 2, 2, a, a, 1, 1, 1) These are not isomorphic (by considering order structure and squares).

22 Bol loops of order 16 This could occur with B = C 2, m = 2 and n = 4, B = C 2, m = 4 and n = 2, B = C 4 and m = n = 2 B = C 2  C 2 and m = n = 2. By considering the possible choices for r,s,t,z,w in each case, eliminating any cases for which s = t = 1 (otherwise we would get a group), our construction gives rise to 1136 Bol loops of order 16.

23 Even if these were non-isomorphic (many are isomorphic), this would not account for all of the 2038 Bol loops of order 16 [Moorhouse].

24 Bol loops of order 24 B must be C 6 (since neither C 3 nor S 3 contains central elements of order 2). Also, m = n = 2. There are two choices each for r, s and t, six of these without s = t = 1. There are six choices each for z and w, so the construction produces 216 nonassociative Bol loops (many of which are isomorphic). We don’t know how many of the 65 Bol loops of order 24 on Moorhouse’s web site arise.

25 Properties of L(B,m,n,r,s,t,z,w) P1: For [a, i,  ] in L, [a,i,  ] -1 = [a -1 r i  s i  t i  z p w q,(m-i)*,(n-  ) ], where p = 0 if i = 0 and p = -1 otherwise, and q = 0 if  = 0 and q = -1 otherwise. BB P2: Let B = {[b,0,0] | b  B}. BB Then B  B and B  L. P3: G = {[b,i,0] | b  B, i  C m, and let G be the group constructed in the preliminary construction. Then G  G and G  L.

26 P4: The commutator ([a,i,  ],[b,j,  ]) = (a,b)r j  +i  s ij(  +  ) t (i+j)  is contained in the subloop generated by r, s, t and Comm(B), where Comm(B) denotes the commutator subloop of B. In fact, Comm(L)=. P5: The associator ([a,i,  ],[b,j,  ],[c,k,  ])= (a,b,c)s ij  +ik  t j  +k  is contained in the subloop generated by s, t, and Ass(B), where Ass(B) denotes the associator subloop of B. In fact, Ass(L) =.

27 P6: The commutator/associator subloop, L is the subloop generated by r, s, t and B. That is, L =. P7: The centrum C(L) of L, that is, the set of elements of L that commute with all elements of L is given by C(L) = {[a,i,  ] | a  C(B) and r i = s i = t  = r  }.

28 P8: The nucleus N(L) of L, that is, the set of elements that associate in all orders with every pair of elements of L is given by N(L) = {[a,i,  ] | a  N(B)} if s = t = 1 and N(L)={[a,i,  ] | a  N(B); i,  even} otherwise. P9: The center Z(L) of L is given by Z(L) = {[a,i,  ] | a  Z(B)} if r = s = t = 1} and Z(L) = {[a,i,  ] | a  Z(B), i,  even} otherwise.

29 Bol extensions of Moufang loops Let G be a Moufang loop which contains a normal subloop B such that G/B  C m for some even integer m, and such that B contains at least one nontrivial central element, s, of order 2. Suppose that we can select a central element u in G so that Bu generates G/B. Then, for any even integer n and any choice of r, t and w in Z(B), with r 2 = t 2 = 1, let L = G  C n, with multiplication defined by [(au i )v  ][(bu j )v  ] = [abu i+j r j  s ij  t j  w q, (  +  )’]. Then L is a Bol loop that is not Moufang, and G is a normal subloop of L, with L/G  C n.

30 The construction in the “odd” case Must m and n be even integers? If both m and n are odd, then s = t = 1, and so we only get a non-Moufang Bol loop if we start with a non-Moufang Bol loop B. In this case, the multiplication rule becomes [a,i,  ][b,j,  ] = [abr j  z p w q,(i+j)*,(  +  )], and L is just the direct product B  C m  C n modulo some subgroup of the center.

31 If m is odd and n is even or n is odd an m is even, then r = s = t = 1. The multiplication rule becomes [a,i,  ][b,j,  ] = [abz p w q,(i+j)*,(  +  )]. And, again L is just the direct product B  C m  C n modulo some subgroup of the center.

32 Strongly right alternative ring loops A nonassociative (necessarily Bol) loop L is a strongly right alternative ring (SRAR) loop if the loop ring RL is a right Bol loop for every ring R of characteristic 2. K. Kunen, Alternative loop rings, Comm. Algebra 26 (1998), [E.G. Goodaire & D.A. Robinson, A class of loops with right alternative loop rings, Comm. Algebra 22 (1995), ]

33 A Bol loop L is SRAR, if and only if, for every x, y, z, w in L, at least one of the following holds: D(x,y,z,w): [(xy)z]w=x[(yz)w] and [(xw)z]y=x[(wz)y] E(x,y,z,w): [(xy)z]w=x[(wz)y] and [(xw)z]y=x[(yz)w] F(x,y,z,w): [(xy)z]w=[(xw)z]y and x[(yz)w]=x[(wz)y] If L is SRAR, then for all x, y, z in L, at least one of the following holds: D(x,y,z): (xy)z = x(yz) and (xz)y = x(zy) E(x,y,z): (xy)z = x(zy) and (xz)y = x(yz) F(x,y,z): (xy)z = (xz)y and x(yz) = x(zy)

34 Chein & Goodaire: When is an L(B,m,n,r,s,t,z,w) loop SRAR? (submitted) The loop L(B,m,n,r,s,t,z,w) is SRAR if and only if |L|=2.

35 A construction of SRAR loops Chein & Goodaire : SRAR loops with more than two commutator/associators (submitted) Let L be a Bol loop whose left nucleus, N, is an abelian group which, as a subloop of L, has index 2. Then, for every element u not in N, L = N  Nu. Choose a fixed element u not in N. We can then define mappings  : N  N and  : N  N by un = (n  )u and n  = u(nu). If both  =I (the identity map) and  =R(u 2 ) (right multiplication by u 2), then L is a group. Otherwise, if  =I or if  =R(u 2 ), then L is SRAR.

36 Conversely, if N is an abelian group with bijections  : N  N and  : N  N, and if u is an indeterminate and L = N  Nu, we can define multiplication of L by n(mu)=(nm)u (nu)m=[n(m  )]u (nu)(mu)=n(m  ). Then L is a loop.

37 If  =I, then L is a Bol loop if and only if both (1) (n 2 m)  = n 2 (m  ) (2) (n  ) 2 m = n 2 (m  2 ) for all n, m in N. If  =R(u 2 ), right multiplication by u 2, then L is a Bol loop if and only if both (3) (n 2 m)  = (n  ) 2 m (4) [n 2 (m  )u 2 ]  = n 2 (m  )u 2 for all n, m in N. If  =I and (1) and (2) hold or if  =R(u 2 ) and (3) and (4) hold, but not both, then L is SRAR.

38 Example 1 Let N be an elementary abelian 2-group of order at least 4, let  =I, and let  be any nonidentity permutation on N such that  2 = I and  is not a right multiplication map. (For example, 1  =1, a  =b, b  =a, etc.) Since the square of any element of N is 1, equations (1) and (2) reduce respectively to the tautologies m  = m  and m = m. Thus L is an SRAR loop. (In most cases, |L| > 2.)

39 Example 2 Let N be an abelian group of exponent 4 (but not of exponent 2). Let  = I and define  : N  N by n  = n -1. Noting that  2 = I and n -2 = n 2 for any n in N, we have, (n 2 m)  = n -2 m -1 = n 2 (m  ), so (1) holds, and (n  ) 2 m = n -2 m = n 2 (m  2 ), so (2) holds too. Thus, L is SRAR. [Note that u 2 =1  =1, so that  cannot be R(u 2 ).] (Again, |L| > 2.)

40 The family of loops described in Example 2 coincides with a class of non-Moufang Bol loops containing an abelian group as a subloop of index 2 discussed by Petr in [A class of Bol loops with a subgroup of index two, Comment. Math. Univ. Carolin. 45 (2004), ] and denoted there by G(  xy,  xy,  xy,  x -1 y ). [Note, however, that our loops are the opposites of Petr's, whose Bol loops are left Bol.]

41 Example 3 Let N be an abelian group with an element a of order 4. Let e = a 2 and S be the set of squares in N. Note that e  S. Let  = I, the identity map on N, and define  by n  = n if n  S and n  = en otherwise. Then u 2 = 1  = 1, and so   R(u 2 ). Now the product of two elements in S is in S and the product of an element in S with an element not in S is not in S. Thus (n 2 m)  = n 2 m = n 2 (m  ) if m  S, and (n 2 m)  = en 2 m = n 2 (em) = n 2 (m  ), otherwise. Thus, equation (1) holds. Also n 2 = (en) 2 = (n  ) 2 and n  2 = n, so that equation (2) holds. Thus, L is SRAR (but here |L| = 2).

42 Example 4 Let N, S, and e be as in Example 3. Define  by n  = n if n  S and n  = en otherwise. Choose u 2  S, and let  = R(u 2 ). Note that, regardless of whether or not n is in S, n 2  = n 2 = (n  ) 2. Also, n   S if and only if n  S. Since the product of two elements of S is in S and the product of an element in S with an element not in S is not in S, (n 2 m)  = n 2 m = (n  ) 2 m = (n  ) 2 (m  ), if m  S; and (n 2 m)  = en 2 m = e(n  ) 2 m = (n  ) 2 (m  ), if m  S.

43 Thus, in either case, (3) holds. Also, it is easy to see that  2 = I. Therefore, since u 2  S, [n 2 (m  )u 2 ]  = n 2 mu 2 if m  S; and [n 2 (m  )u 2 ]  = en 2 mu 2 = n 2 e(m  )u 2 = n 2 mu 2 if m  S. Thus, again, in either case, (4) holds and so L is SRAR. Here, as in Example 3, |L| = 2.

44 Example 5 Let N be an abelian group of exponent 4 (but not of exponent 2), let u 2 be any element of order 2 in N, let  = R(u 2 ) and let n  = n -1 for all n in N. Then (n 2 m)  = n -2 m -1 = (n -1 ) 2 m -1 = (n  ) 2 (m  ), so equation (3) holds. Also, [n 2 ( m  )u 2 ]  = [n 2 m -1 u 2 ] -1 = n -2 mu -2 = n 2 mu 2 ; so equation (4) holds as well and L is SRAR. Again, we draw attention to the fact that the family of non-Moufang loops described in this example is one discussed by Petr in the article mentioned above, specifically the class he labels G(  xy,  xy,  x -1 y,  xy ). (Again, our loops are the opposite of Petr's.) Often, |L| > 2.

45 Example 6 Let N = be a cyclic group of order 4k; let u 2 = a 2r for some integer r, let  = R(u 2 ), and define  by n  = n 2k+1. Then (n 2 m)  = (n 2 m) 2k+1 = n 4k+2 m 2k+1 = (n 2k+1 ) 2 m 2k+1 = (n  ) 2 (m  ), so equation (3) holds. Also, [n 2 (m  )u 2 ]  = [n 2 m 2k+1 a 2r ]  = n 4k+2 m (2k+1)(2k+1) a (4k+2)r = n 2 m 4k+4k+1 a 2r = n 2 mu 2, so equation (4) holds too. Thus, L is SRAR. Here |L| = 2.

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