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Bol loops
that are centrally nilpotent
of class 2
Orin Chein
(Edgar G Goodaire)
1
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.
2
Outline
Why nilpotence class 2?
Properties of Bol loops of nilpotence class 2
Some constructions
Connections with strongly right alternative
ring (SRAR) loops
3
Why nilpotence class 2
Historical reasons
Practical reasons
4
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 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:
S2:
S3:
S4:
S5:
(x, z, y) = (x, y, z).
x2  N(L).
(xm, yn, zr) = (x,y,z)mnr
(xm,yn) =(x,y)mn
(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 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)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 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 = 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 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, 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 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 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-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(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,] | 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] = [abui+jrj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,(i+j)*,(+)],
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,(i+j)*,(+)].
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), 56235634.]
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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 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)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(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| > 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 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 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.
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Thus, L is SRAR (but here |L| = 2).
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  42S.
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| > 2.
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Example 6
Let 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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