ON THE COMMUTATIVE CENTER OF MOUFANG LOOPS

arXiv:1711.07001v1 [] 19 Nov 2017

ALEXANDER N. GRISHKOV AND ANDREI V. ZAVARNITSINE
Abstract. We construct two inﬁnite series of Moufang loops of exponent 3
whose commutative center (i. e. the set of elements that commute with all elements of the loop) is not a normal subloop. In particular, we obtain examples
of such loops of orders 38 and 311 one of which can be deﬁned as the Moufang
triplication of the free Burnside group B(3, 3).
Keywords: Moufang loop, commutative center, normal subloop, Burnside
group
MSC2010: 20N05

1. Introduction
A Moufang loop is a loop in which the identity
(xy)(zx) = (x(yz))x

(1)

holds. Moufang loops are known to be diassociative, i. e. their subloops generated by
a pair of elements are groups. In particular, elements of Moufang loops have unique
inverses and (xy)−1 = y −1 x−1 for all x, y.
The commutative center (also known as the Moufang center or the commutant 1)
of a Moufang loop M is the set
C(M ) = {c ∈ M | cx = xc for all x ∈ M }.
It is known [1, Theorem IV.3.10] that C(M ) is a subloop of M which is characteristic.
The following problem had been open for quite a while and oﬃcially raised by A. Rajah
at Loops ’03 conference in Prague.
Problem. Is C(M ) a normal subloop of M ?
It was stated in [2] that the answer to this question is aﬃrmative. Here we show
that the answer is in fact generally negative by constructing two series of examples of
Moufang loops whose commutative center is not normal. The ﬁrst series is given by an
explicit multiplication formula over any ﬁeld of characteristic 3. This series contains
a ﬁnite loop of order 311 . The second series is a particular case of the triplication of
Moufang loops of exponent 3. We show that, for any Moufang loop M of exponent 3
that does not satisfy the identity [[x, y], z] = 1, there is a Moufang loop M(M, 3) whose
commutative center is not normal. In particular, we have the example M(B(3, 3), 3)
of order 38 , where B(3, 3) is the free 3-generator Burnside group of exponent 3.
Recall that, for a Moufang loop M ,
Nuc(M ) = {a ∈ M | (ax)y = a(xy) for all x, y ∈ M }
is the nucleus of M . Our results reopen the following conjecture by S. Doro [3]:
Supported by FAPESP (proc. 2017/14489-2).
1We avoid using the term ‘commutant’ due to the unfortunate collision with the term ‘derived
subgroup’ as both are translated into Russian as ‘коммутант’.
1

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ALEXANDER N. GRISHKOV AND ANDREI V. ZAVARNITSINE

Conjecture. If Nuc(M ) = 1 then C(M ) is normal in M .
2. Motivation
For elements x, y, z of a Moufang loop M , we denote by [x, y] the unique element
s ∈ M such that xy = (yx)s and by (x, y, z) the unique element t ∈ M such that
(xy)z = (x(yz))t.
The idea behind the construction is the following straightforward observation.
Lemma 1. If C(M ) is normal in M then, for every a ∈ C(M ) and every b, c, d ∈ M ,
we have [(a, b, c), d] = 1.
Proof. Since C(M ) is normal and a ∈ C(M ), we have (a, b, c) ∈ C(M ).



Therefore, for any Moufang loop L, if there is a ∈ C(L) such that [(a, b, c), d] 6= 1
for some b, c, d ∈ L then C(L) is not a normal subloop in view of Lemma 1. Examples
of such loops are constructed in the next sections.
3. Algebraic loop
The underlying set of the loop is an 11-dimensional vector space V over a ﬁeld F
of characteristic 3. An element x ∈ V will be written as a tuple x = (x1 , x2 , . . . , x11 ),
xi ∈ F . We introduce a new operation ’◦’ on V which is given, for x, y ∈ V , by
x ◦ y = x + y + f,

(2)

where f = (f1 , . . . , f11 ) and fk are polynomials in xi , yj explicitly given below.
f1 = f2 = f3 = f4 = 0,
f5 = −x3 y2 , f6 = −x4 y2 , f7 = −x4 y3 ,
f8 =

x1 x3 y 2 − x1 y 2 y 3 − x2 x3 y 1 + x2 y 1 y 3 ,

f9 =

x1 x4 y 2 − x1 y 2 y 4 − x2 x4 y 1 + x2 y 1 y 4 ,

f10 =

x1 x4 y 3 − x1 y 3 y 4 − x3 x4 y 1 + x3 y 1 y 4 ,

f11 = −x1 x2 x4 y3 + x1 x2 y3 y4 + x1 x3 y2 y4 + x1 x4 y2 y3 + x2 x3 y1 y4 + x2 x4 y1 y3
− x2 y 1 y 3 y 4 + x3 x4 y 1 y 2 − x4 y 1 y 2 y 3 − x1 x5 y 4 + x1 x6 y 3 − x1 x7 y 2 + x1 y 2 y 7
− x1 y 3 y 6 + x1 y 4 y 5 + x2 x7 y 1 − x2 y 1 y 7 − x3 x6 y 1 + x3 y 1 y 6 + x4 x5 y 1 − x4 y 1 y 5
+ x8 y4 − x9 y3 + x10 y2 .
It can be checked that L = (V, ◦) is a loop in which the Moufang identity (1)
holds. We omit the heavily technical veriﬁcation. The identity element of L is the
zero vector of V and, for x ∈ L, we have
x−1 = −x + h,

(3)

where h = (h1 , . . . , h11 ) and the polynomials hk are as follows:
h1 = h2 = h3 = h4 = h8 = h9 = h10 = 0,
h5 = −x2 x3 , h6 = −x2 x4 , h7 = −x3 x4 ,
h11 =

x2 x10 − x3 x9 + x4 x8 .

Let e1 , . . . , e11 be the standard basis of V , i. e., ei = (. . . , 0, 1, 0, . . .) with ‘1’ at the
ith place. Deﬁne
a = e1 , b = e2 , c = e3 , d = e4 .

ON THE COMMUTATIVE CENTER OF MOUFANG LOOPS

3

Lemma 2. a ∈ C(L).
Proof. This is true because using the multiplication formula (2) one can check that,
for an arbitrary x ∈ L, both a ◦ x and x ◦ a equal
x + a + (0, 0, 0, 0, 0, 0, 0, −x2x3 , −x2 x4 , −x3 x4 , x2 x7 − x3 x6 + x4 x5 ).

Lemma 3. [(a, b, c), d] 6= 1 in L.
Proof. Using (2) it can be shown that the following equalities hold in L:
e5 = [b, c], e6 = [b, d], e7 = [c, d],
e8 = (a, b, c), e9 = (a, b, d), e10 = (a, c, d),
e11 = [(a, b, c), d].
Since e11 6= 0 in V , it follows that [(a, b, c), d] is not the identity element of L.



Therefore, Lemma 1 implies
Corollary 1. C(L) is not a normal subloop of L.
The following properties of L are worth mentioning. L has exponent 3 and satisﬁes
the identity [[x, y], z] = 1. C(L) = he1 , e5 , e6 , e7 , e11 iF is associative of order 35 ,
where h iF denotes the F -linear span in V . C(L) is close to being normal — it has a
subloop N of index 3 which is normal in L. We have N = he5 , e6 , e7 , e11 iF and L/N is
commutative. Also, Nuc(L) = he8 , e9 , e10 , e11 iF and Z(L) = Nuc(L) ∩ C(L) = he11 iF .
In particular, if F = F3 , we obtain an example of order 311 .
4. Triplication of Moufang loops of exponent 3
A group G possessing automorphisms ρ and σ that satisfy ρ3 = σ 2 = (ρσ)2 = 1 is
called a group with triality S = hρ, σi if
2

(x−1 xσ )(x−1 xσ )ρ (x−1 xσ )ρ = 1
for every x in G. In such a group, the set
M(G) = {x−1 xσ | x ∈ G}
is a Moufang loop with respect to the multiplication
2

m.n = m−ρ nm−ρ

(4)

for all n, m ∈ M(G). Conversely, every Moufang loops arises so from a suitable group
with triality. Observe that taking powers or inverses of elements of M(G) is the
same, whether considered under the loop or group operation. We will also require the
following properties.
Lemma 4 ([6, Lemma 2.1]). Let G be a group with triality S = hρ, σi. Then, for all
m, n ∈ M(G), we have
(i) mσ = m−1 ;
2
(ii) m, mρ , mρ pairwise commute;
2
2
(iii) m−ρ nm−ρ = n−ρ mn−ρ ;
(iv) m.n.m = mnm.

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ALEXANDER N. GRISHKOV AND ANDREI V. ZAVARNITSINE

For more details on groups with triality, see [3, 4].
In the loop (M(G), .), we write Jx, yK = x−1 .y −1 .x.y instead of [x, y] = x−1 y −1 xy
to make a distinction between the loop and group commutator.
Let G be a group with triality S = hρ, σi and let M = M(G). It was observed
e = G ⋋ hρi is also a group
in [5, Lemma 3] that the natural semidirect product G
with triality S if and only if M has exponent 3. In this case, we will denote the
e by M(M, 3) and call it the triplication of M . It
corresponding Moufang loop M(G)
e with
contains M as a normal subloop of index 3 and coincides as a subset of G
M ∪ ρM ρ ∪ ρ2 M ρ2 .
Lemma 5. Let M be a Moufang loop of exponent 3 and let L = M(M, 3). Then
(i) L has exponent 3;
(ii) The multiplication rule of L is as given in Table 1;
(iii) ρ ∈ C(L);
(iv) (ρ, m, n) = Jn−1 , m−1 K for all m, n ∈ M .
(v) L is associative iff M is an abelian group.
Proof. We will use Lemma 4 and the fact that M has exponent 3 implicitly throughout
the proof.
(i) Let m ∈ L. If m ∈ M then m3 = 1. If m = ρnρ then m3 = ρnρ2 nρ2 nρ =
2
2
nρ nnρ = 1. If n = ρ2 nρ2 then m3 = ρ2 nρnρnρ2 = nρ nnρ = 1. Hence, L has
exponent 3.
(ii) Let m, n ∈ M . Then the product rules are
2

2

m.ρnρ = m−ρ ρnρm−ρ = ρm−ρ nm−ρ ρ = ρ(n.m)ρ,
2

ρmρ.n = n−ρ ρmρn−ρ = ρn−1 mn−1 ρ = ρ(n−1 .m.n−1 )ρ = ρ(m−1 .n.m−1 )ρ,
2

2

ρmρ.ρnρ = (ρmρ)−ρ ρnρ(ρmρ)−ρ = ρm−1 ρnρm−1 ρ = ρ2 m−ρ nm−ρ ρ2
= ρ2 (m.n)ρ2 ,
2

2

ρ2 mρ2 .ρnρ = (ρ2 mρ2 )−ρ ρnρ(ρ2mρ2 )−ρ = ρm−ρ ρ2 nρ2 m−ρ ρ = m−1 nm−1
= m−1 .n.m−1 ,
and similarly for other choices of the factors.
(iii) Let m ∈ M . Using (ii), we have
ρ.n = ρ2 ρ2 .n = ρ2 nρ2 ,
ρ.ρnρ = ρ2 ρ2 .ρnρ = n,
ρ.ρ2 nρ2 = ρ2 ρ2 .ρ2 nρ2 = ρnρ,

n.ρ = n.ρ2 ρ2 = ρ2 n−2 ρ2 = ρ2 nρ2 ,
ρnρ.ρ = ρnρ.ρ2 ρ2 = n,
ρ2 nρ2 .ρ = ρ2 nρ2 .ρ2 ρ2 = ρnρ.

Therefore, ρ ∈ C(L).
(iv) By (ii), we have
(ρ, m, n) = (ρ.(m.n))−1 .((ρ.m).n) = (ρ2 (m.n)ρ2 )−1 .(ρ2 mρ2 .n)
= (ρ(m.n)−1 ρ).(ρ2 (n.m)ρ2 ) = n.m.(m.n)−1 = Jn−1 , m−1 K.
(v) If M is not associative then neither is L, since M is a subloop of L. By (iv), L
is not associative if M is not commutative. Conversely, if M is a commutative group
then Table 1 turns into the multiplication rule of the direct product M × hρi, hence
L is associative in this case (an elementary abelian 3-group, in fact).


ON THE COMMUTATIVE CENTER OF MOUFANG LOOPS

5

Table 1. The product x.y in M(M, 3)
x \ y
m
ρmρ
ρ2 mρ2

n
ρnρ
m.n
ρ(n.m)ρ
−1
−1
ρ(m .n.m )ρ ρ2 (m.n)ρ2
ρ2 (n.m)ρ2
m−1 .n.m−1

ρ2 nρ2
ρ2 (m−1 .n.m−1 )ρ2
n.m
ρ(m.n)ρ

Lemma 5(ii) shows that up to isomorphism M(M, 3) does not depend on the group
G as long as M(G) ∼
= M (such a group is not unique). In other words, we can
construct M(M, 3) by adjoining to M an outer element ρ = ρ2 ρ2 and use Table 1 to
deﬁne multiplication on formal elements of the shape m, ρmρ, ρ2 mρ2 with m ∈ M .
Proposition 1. Let M be a Moufang loop of exponent 3 that does not satisfy the
identity [[x, y], z] = 1. Then the commutative center of M(M, 3) is not a normal
subloop.
Proof. Denote L = M(M, 3). By Lemma 5(iii), ρ ∈ C(L). By assumption, there
are x, y ∈ M such that [y −1 , x−1 ] 6∈ C(L). We have (ρ, x, y) = [y −1 , x−1 ] by Lemma
5(iv). Therefore, C(L) is not normal by Lemma 1.

It is known [7, §18.2] that the free r-generator Burnside group B(3, r) of exponent
3 has nilpotency class 3 for r > 3. In particular, it does not satisfy the identity
[[x, y], z] = 1. Hence, we obtain
Corollary 2. The commutative center of the Moufang triplication M(B(3, r), 3) is
not a normal subloop if r > 3.
The minimal example in this series is the loop M(B(3, 3), 3) of order 38 .
Finally, observe that the loops from Section 3 cannot be obtained by an application
of Proposition 1, because they satisfy the identity [[x, y], z] = 1.
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7. Berlin: Heldermann Verlag. (1990), 147 pp.
[2] S. M. Gagola III, A Moufang loop’s commutant, Math. Proc. Camb. Phil. Soc., 152, N 2
(2012), 193–206.
[3] S. Doro, Simple Moufang loops, Math. Proc. Camb. Phil. Soc., 83, N 3 (1978), 377–392.
[4] A. N. Grishkov, A. V. Zavarnitsine, Groups with triality, J. Algebra Appl., 5, N 4 (2006),
441–463.
[5] A. N. Grishkov, A. V. Zavarnitsine, Sylow’s theorem for Moufang loops, J. Algebra, 321, N 7
(2009), 1813–1825.
[6] A. N. Grichkov, A. V. Zavarnitsine, Multiplication formulas in Moufang loops, Internat. J.
Algebra Comput. 26, N 4 (2016), 705–725.
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Instituto de Matemática e Estatı́stica, São Paulo, Brazil
E-mail address: shuragri@gmail.com
Sobolev Institute of Mathematics, Novosibirsk, Russia
E-mail address: zav@math.nsc.ru

