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Bol loopsthat are centrally nilpotent

of 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) =

xm+pyq+n(y,x)np(x,x,y)-np(p-1)-mp(q+2n)(y,y,x)np(n+q-1)

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Two-generator Bol loopsIf L is generated by x and y, then, by T1 and T2, all

commutators and associators in L can be expressed in the form rst ,

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 1S1: (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(n+q)(y, y, x)nq(m+p)

S6: (xmyn)(xpyq)

= xm+pyq+n(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 1CS1: 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)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.

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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, i][b, j] = [abzq, (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).

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The main constructionLet 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, ] = [abrjsijtjzpwq,(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.

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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 notationWe are thinking of the elements of our loop as being of

the form (aui)v (hence the notation [a, i, ]), where aB, u generates Cm and v generates Cn, and where um = zZ(B) and vn = wZ(B).

Thus, for example, ui+j = zpu(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.

•

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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 8The 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 24B 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 don’t know how many of the 65 Bol loops of order 24 on Moorhouse’s 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-1risiti 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 BB = {[b,0,0] | b BB}. Then BB B and BB 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(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)= < r, s, t, Comm(B) >.

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) = < s, t, Ass(B) >.

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P6: The commutator/associator subloop, L is the subloop generated by r, s, t and B. That is,

L = < r, s, t, B >.

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,] | aC(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,] | aN(B)} if s = t = 1

and

N(L)={[a,i,] | aN(B); i, even} otherwise.

P9: The center Z(L) of L is given by

Z(L) = {[a,i,] | aZ(B)} if r = s = t = 1}

and

Z(L) = {[a,i,] | aZ(B), i, even} otherwise.

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Bol extensions of Moufang loopsLet G be a Moufang loop which contains a normal

subloop B such that G/BCm 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] = [abui+jrjsijtjwq,(+)’]. 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,(i+j)*,(+)],and L is just the direct product BCmCn 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,(i+j)*,(+)].And, again L is just the direct product

BCmCn 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)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)

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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 loopsChein & 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)m=[n(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(m2)

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 1Let 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.)

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Example 2Let N be an abelian group of exponent 4 (but not of

exponent 2). Let = I and define : NN 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(m2), so (2) holds too.

Thus, L is SRAR.[Note that u2=1=1, so that cannot be R(u2).](Again, |L| > 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 3Let 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 n2 = n, so that equation (2) holds.

Thus, L is SRAR (but here |L| = 2).

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Example 4Let 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 5Let 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| > 2.

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Example 6Let N = < a > be a cyclic group of order 4k;

let u2 = a2r for some integer r,

let = R(u2), and define by n = n2k+1. Then

(n2m) = (n2m)2k+1 = n4k+2m2k+1 = (n2k+1)2m2k+1 = (n)2(m), so equation (3) holds.

Also,

[n2(m)u2] = [n2m2k+1a2r] = n4k+2m(2k+1)(2k+1)a(4k+2)r = n2m4k+4k+1a2r = n2mu2,

so equation (4) holds too.

Thus, L is SRAR. Here |L| = 2 .

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