Bol loops that are centrally nilpotent of class 2

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Bol loopsthat are centrally nilpotentof class 2
Orin Chein
(Edgar G Goodaire)
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• 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), 103-122.
• 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.

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Outline

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

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Why nilpotence class 2
Historical reasons Practical reasons
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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

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Practical reasons

• In a loop of nilpotence class 2, commutators and
  associators are central.

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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,
• A1 (x,z,y) (x,y,z)-1
• A2 (x, y, zyn) (x, y, z)
• (x, zyn, y) (x, z, y)
• A3 (xyn, y, z) (x, y, z)(y, y, z)n
• (xyn, z, y) (x, z, y)(y, z, y)n

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• 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,ym,zn) (x,y,z)mn
• A6 (yn, y, z) (y, y, z)n
• In particular, (y-1, y, z) (y, y, z)-1
• A7 (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

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• 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 (xm, yn) (x, y)mn(x, x, y)mn(m-1)(y, x,
  y)mn(n-1)
• C3 (xyn, 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)

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• Two-generator identities
• T1 (xmyn,xpyq,xrys) (x,x,y)m(ps-qr)(y,y,x)n(qr
  -ps)
• T2 (xmyn, xpyq) (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 (xmyn)(xpyq)
• xmpyqn(y,x)np(x,x,y)-np(p-1)-mp(q2n)(y,y,x)
  np(nq-1)

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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 xp yq 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.

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If associators square to 1

• S1 (x, z, y) (x, y, z).
• S2 x2 ? N(L).
• S3 (xm, yn, zr) (x,y,z)mnr
• S4 (xm,yn) (x,y)mn
• S5 (xmyn,xpyq)
• (x,y)mq-np(x, x, y)mp(nq)(y, y, x)nq(mp)
• S6 (xmyn)(xpyq)
• xmpyqn(y,x)np(x, x, y)mpq(y, y, x)npq
• S7 (xz,y) (x,y)(z,y)(y,x,z)

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If commutators also square to 1

• CS1 x2 ? Z(L).
• CS2 If L?2, then squares are central in L.

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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)21.
• 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.

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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 ? Cm by
• a, ib, j abzq, (ij),
• where (ij) is the least non-negative residue of
  ij 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).

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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 rm
  rn s2 t2 1.
• Let L B ? Cm ? Cn with multiplication defined
  by
• a, i, ?b, j, ? abrj?sij?tj??zpwq,(i
  j),(??)?,
• where, for any integer i, i and i? denote the
  least nonnegative residues of i modulo m and n,
  respectively,
• mp ij - (ij), and nq ? ? -(? ?)?.
• Then L is a right Bol loop.

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• 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).

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Explanation of notation

• We are thinking of the elements of our loop as
  being of the form (aui)v? (hence the notation
  a, i, ?), where a?B, u generates Cm and v
  generates Cn, and where um z?Z(B) and vn
  w?Z(B).
• Thus, for example, uij zpu(ij), where
• ij mp (ij).
• 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.

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• 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 vj.
• 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 (ui, uj, v?) and (v?, uj , v?),
  respectively.

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• We assume that rm rn s2 t2 1 so that we
  can ignore the impact of reducing modulo m and n
  in the second and third coordinates.
• Note that the conditions rm rn s2 t2 1
  are necessary.

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Bol loops of order 8

• The main construction produces all six Bol loops
  of order 8.
• L1 L(C2, 2, 2, a, a, a, a, a)
• L2 L(C2, 2, 2, a, a, 1, a, 1)
• L3 L(C2, 2, 2, a, a, 1, a, a)
• L4 L(C2, 2, 2, a, a, a, a, 1)
• L5 L(C2, 2, 2, a, a, a, 1, 1)
• L6 L(C2, 2, 2, a, a, 1, 1, 1)
• These are not isomorphic (by considering order
  structure and squares).

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Bol loops of order 16

• This could occur with
• B C2, m 2 and n 4,
• B C2, m 4 and n 2,
• B C4 and m n 2
• B C2 ? C2 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.

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

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Bol loops of order 24

• B must be C6 (since neither C3 nor S3 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 dont know how many of the 65 Bol loops of
  order 24 on Moorhouses web site arise.

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Properties of L(B,m,n,r,s,t,z,w)

• P1 For a, i, ? in L,
• a,i,?-1 a-1ri?si?ti? zpwq,(m-i),(n-?)
  ?,
• where p 0 if i 0 and p -1 otherwise, and q
  0 if ? 0 and q -1 otherwise.
• P2 Let B b,0,0 b ? B.
• Then B ? B and B ? L.
• P3 G b,i,0 b? B, i ?Cm , and let G be
  the group constructed in the preliminary
  construction.
• Then G ? G and G ? L.

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• P4 The commutator
• (a,i,?,b,j,?) (a,b)rj? i? sij(??)t(ij)??
  
• 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) lt r, s, t,
  Comm(B) gt.
• P5 The associator
• (a,i,?,b,j,?,c,k,?) (a,b,c)sij?ik?
  tj??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) lt s, t, Ass(B)
  gt.

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• P6 The commutator/associator subloop, L? is the
  subloop generated by r, s, t and B?. That is,
• L? lt r, s, t, B? gt.
• 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 ri si t? r?.

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• 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.

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Bol extensions of Moufang loops

• Let G be a Moufang loop which contains a normal
  subloop B such that G/B?Cm 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
• r2 t2 1, let L G ? Cn, with multiplication
  defined by
• (aui)v?(buj)v? abuijrj?sij?tj??wq,(??).
  
• Then L is a Bol loop that is not Moufang, and G
  is a normal subloop of L, with L/G ?Cn.

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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,? abrj? zpwq,(ij),(??)?,
• and L is just the direct product B?Cm?Cn modulo
  some subgroup of the center.

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• 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,? abzpwq,(ij),(??)?.
• And, again L is just the direct product B?Cm?Cn
  modulo some subgroup of the center.

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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), 557-564.
• E.G. Goodaire D.A. Robinson, A class of loops
  with right alternative loop rings, Comm. Algebra
  22 (1995), 5623-5634.

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• 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)zwx(yz)w and
  (xw)zyx(wz)y
• E(x,y,z,w) (xy)zwx(wz)y and
  (xw)zyx(yz)w
• F(x,y,z,w) (xy)zw(xw)zy and
  x(yz)wx(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)

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• 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.

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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(u2)
  (right multiplication by u2), then L is a group.
  Otherwise, if ?I or if ?R(u2), then L is SRAR.

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• 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)mn(m?)u
• (nu)(mu)n(m?).
• Then L is a loop.

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• If ?I, then L is a Bol loop if and only if both
• (1) (n2m)? n2(m?)
• (2) (n?)2m n2(m?2)
• for all n, m in N.
• If ?R(u2), right multiplication by u2, then L is
  a Bol loop if and only if both
• (3) (n2m)? (n?)2m
• (4) n2(m?)u2? n2(m?)u2
• for all n, m in N.
• If ?I and (1) and (2) hold or if ?R(u2) and (3)
  and (4) hold, but not both, then L is SRAR.

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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? gt 2.)

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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 n2 for any n in N,
  we have,
• (n2m)? n-2m-1 n2(m?), so (1) holds, and
• (n?)2m n-2m n2(m?2), so (2) holds too.
• Thus, L is SRAR.
• Note that u21?1, so that ? cannot be R(u2).
• (Again, L? gt 2.)

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• 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), 371-381 and denoted there
  by G(?xy ,?xy ,?xy ,?x-1y). Note,
  however,
• that our loops are the opposites of Petr's,
  whose Bol loops are left Bol.

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Example 3

• Let N be an abelian group with an element a of
  order 4. Let e a2 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 u2 1? 1, and so ? ?
  R(u2).
• 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 (n2m)? n2m n2(m?)
  if m ? S , and (n2m)? en2m n2(em) n2(m?),
  otherwise. Thus, equation (1) holds. Also n2
  (en)2 (n?)2 and n?2 n, so that equation (2)
  holds.
• Thus, L is SRAR (but here L? 2).

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Example 4

• Let N, S, and e be as in Example 3.
• Define ? by n? n if n ? S and n? en
  otherwise.
• Choose u2 ? S, and let ? R(u2).
• Note that, regardless of whether or not n is in
  S,
• n2? n2 (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,
• (n2m)? n2m (n?)2m (n?)2(m?), if m ? S
• and
• (n2m)? en2m e(n?)2m (n?)2(m?), if m ? S.

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• Thus, in either case, (3) holds.
• Also, it is easy to see that ?2 I.
• Therefore, since u2 ? S,
• n2(m?)u2? n2mu2 if m ? S and
• n2(m?)u2? en2mu2 n2e(m?)u2 n2mu2 if m ?
  S.
• Thus, again, in either case, (4) holds and so L
  is SRAR.
• Here, as in Example 3, L? 2.

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Example 5

• Let N be an abelian group of exponent 4 (but not
  of exponent 2), let u2 be any element of order 2
  in N, let ? R(u2) and let n? n-1 for all n in
  N. Then
• (n2m)? n-2m-1 (n-1)2m-1 (n?)2(m?), so
  equation (3) holds.
• Also, n2( m?)u2? n2m-1u2-1 n-2mu-2
  n2mu2
• 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-1y,?xy).
• (Again, our loops are the opposite of Petr's.)
• Often, L? gt 2.

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Example 6

• Let N lt a gt be a cyclic group of order 4k
• let u2 a2r for some integer r,
• let ? R(u2), and define ? by n? n2k1. Then
• (n2m)? (n2m)2k1 n4k2m2k1 (n2k1)2m2k1
  (n?)2(m?), so equation (3) holds.
• Also,
• n2(m?)u2? n2m2k1a2r? n4k2m(2k1)(2k1)a(
  4k2)r n2m4k4k1a2r n2mu2,
• so equation (4) holds too.
• Thus, L is SRAR. Here L? 2 .

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