Table Of ContentRandom Generators of Given Orders and the Smallest
Simple Moufang Loop
7
0
0 Petr Vojtˇechovsky´
2
n
a Abstract. Theprobabilitythatmrandomlychosenelementsofafinitepowerassociative
J loopC haveprescribedordersandgenerateC iscalculatedintermsofcertainconstants Γ
relatedtotheactionofAut(C)onthesublooplatticeofC. Asanillustration,allmeaningful
4
probabilities of random generation by elements of given orders are found for the smallest
2
nonassociative simpleMoufangloop.
]
R
G
. 1. Random generators of given orders
h
t
a Let C be a power associative loop, i.e., a loop where every element generates a
m
group. Givenanm-tuple a=(a0, ..., am−1) ofelements ofC, letA={Ai}mi=0 be
[ the sequence of nested subloops A ≤C such that A is the smallest subloop of C,
i 0
and A =hA , a i. Note that A is independent of the orderof the elements a ,
1 i+1 i i m 0
v ..., am−1 in a.
3 Denote by Gen (C) the set of all m-tuples a ∈ Cm with A = C. Then the
m m
0 probability that m randomly chosen elements of C generate C is
7
1 pm(C)=|C|−m·|Genm(C)|. (1)
0
This notion can be refined in a natural way. For 1≤i≤n, let D be the set of
7 i
0 all elements of C of order i. Two m-tuples of integers r=(r0, ..., rm−1), s=(s0,
h/ ..., sm−1) are said to be of the same type if r0, ..., rm−1 is a permutation of s0,
t ..., sm−1. We say that a = (a0, ..., am−1) ∈ Cm is of type r if there is s = (s0,
ma ..., sm−1) of the same type as r satisfying ai ∈Dsi, for 0≤i≤m−1.
Let Genr(C)⊆Genm(C) be the set of generating m-tuples of type r. Then
v: pr(C)=|C|−m·|Genr(C)| (2)
i
X is the probability that m randomly chosen elements a0, ..., am−1 ∈C generate C
r and (a0, ..., am−1) is of type r.
a
For A, B ≤ C and an integer i, let Γ (A, B) be the cardinality of the set of all
i
elements x ∈ D such that hA, xi ∈ O , where O is the orbit of B under the
i B B
natural action of Aut(C) on the subloop lattice of C. Also, let Γ(A, B) be the
cardinality of the set of all elements x∈C such that hA, xi∈O .
B
1991 Mathematics Subject Classification: Primary: 20N05,Secondary: 20F05,06B99.
Key words and phrases: randomgeneratorsofgivenorders,Moufangloops,Paigeloops.
WhileworkingonthispapertheauthorhasbeenpartiallysupportedbyGrantAgencyofCharles
University,grantnumber269/2001/B-MAT/MFF.
1
2 PETRVOJTEˇCHOVSKY´
We are going to divide Cm into certain equivalence classes. Two m-tuples a,
b ∈ Cm with associated nested subloops {A }m , {B }m will be called orbit-
i i=0 i i=0
equivalent if A ∈O , for 0≤i≤m. We write a∼ b.
i Bi
Thesizeoftheequivalenceclassa∼ iseasytocalculatewithhelpoftheconstants
Γ(A, B). ThereareΓ(A , A )elementsxsuchthathA , xi∈O . Oncewearein
0 1 0 A1
theorbitO ,wecancontinueonthewaytoO byaddingoneoftheΓ(A , A )
Ai Ai+1 i i+1
elements x to hx , ..., x i. Thus,
i+1 0 i
m−1
|a∼|= Y Γ(Ai, Ai+1). (3)
i=0
Since
|Genm(C)|= X |a∼|, (4)
a∼∈Genm(C)/∼
we can combine (1), (3) and (4) to obtain
m−1
pm(C)=|C|−m X Y Γ(Ai, Ai+1). (5)
a∼∈Genm(C)/∼ i=0
Example 1.1. Let us illustrate (5) by calculating the probability that two ran-
domlychosenelementsofS generatetheentiresymmetricgroup. Letebetheneu-
3
tralelementofS . Thereare3subgroupsisomorphictoC (allinoneorbitoftran-
3 2
sitivity) and a unique subgroup isomorphic to C . As Γ(e, C ) = 3, Γ(e, C )= 2,
3 2 3
Γ(C , S ) = 4 and Γ(C , S ) = 3, we have p (S ) = (3·4+2·3)/36 = 1/2, as
2 3 3 3 2 3
expected.
Write ∼r for the restriction of ∼ onto Genr(C), and observe that
m−1
|a∼r|= X Y Γsi(Ai, Ai+1), (6)
s=(s0,...,sm−1) i=0
wherethesummationrunsoverallm-tuplessofthe sametypeasr. Consequently,
m−1
pr(C)=|C|−m X X Y Γsi(Ai, Ai+1). (7)
a∼r∈Genr(C)/∼r s=(s0,...,sm−1) i=0
Remark 1.2. All concepts of this section can be generalized to any finite universal
algebra C with subsets D closed under the action of Aut(C).
i
2. Random generators of given orders for M∗(2)
We assume from now on that the reader is familiar with the notation and termi-
nology of [2].
The value of Γ (A, B) in (7) can be calculated with help of Hasse constants,
i
provided A is maximal in B. Namely,
∗
Γ (A, B)=H (A|B)·|D ∩(B\A)|. (8)
i C i
RANDOM GENERATORS OF GIVEN ORDERS 3
This is obvious because H∗(A|B) counts the number of subloops of C containing
C
A and in the same orbit as B.
By (7) and (8), we should be able to calculate pm(C), pr(C) when all Hasse
constants for C are known and when the lattice of subloops of C is not too high.
These probabilities can be used alongside order statistics to recognize black box
loops, for instance (cf. [1]).
Buildingontheresultsof[2]substantially,weproceedtocalculateallmeaningful
probabilities pm(C), pr(C) for C = M∗(2)—the smallest nonassociative simple
Moufang loop.
AllHasseconstantsforC aresummarizedin[2,Fig.1],sowecaneasilyevaluate
allconstantsΓ (A, B)withAmaximalinB. Forexample,sinceH∗(E |M(A ))=
i C 8 4
3 and |D ∩(M(A )\E )|=8, we have Γ (E , M(A ))=24.
2 4 8 2 8 4
Apart from trivialities, the remaining constants to be calculated are
−
Γ (C , A ), Γ (S , C), Γ (A , C), Γ (E , C),
i 2 4 i 3 i 4 i 4
Γ (E−, M(A )), Γ (E+, M(A )), Γ (E+, C), Γ (E , C),
i 4 4 i 4 4 i 4 i 8
for i=2, 3. As some invention is needed here, we show how to obtain all of them.
WebeginwithΓ (S , C). LetGbeacopyofS inC. Foranyelementx6∈G,we
i 3 3
musthavehG, xi∼=M(S ),M(A ),orC. Therefore,Γ (S , C)=(i−1)·H (C )−
3 4 i 3 C i
Γ (S , M(S ))−Γ (S , M(A ))−(i−1)·H (C ), for i=2, 3. Consequently,
i 3 3 i 3 4 S3 i
Γ (S , C)=63−6−36−3=18, Γ (S , C)=56−0−18−2=36.
2 3 3 3
Similarly,
Γ (C , A ) = 0, Γ (C , A ) = 24,
2 2 4 3 2 4
Γ (A , C) = 48, Γ (A , C) = 48,
2 4 3 4
Γ (E , C) = 32, Γ (E , C) = 32.
2 8 3 8
A more detailed analysis of the subloop lattice of C allows us to calculate the
remaining eight constants.
Lemma 2.1. Let G ∈ O−, and let M , M , M be the three copies of M(A )
1 2 3 4
containing G. Then M ∩ M contains no element of order 3, for i 6= j, and
i j
M ∩M ∩M is the unique copy of E containing G. In particular,
1 2 3 8
− − − −
Γ (E , M(A ))=24, Γ (E , C)=24, Γ (E , M(A ))=24, Γ (E , C)=8.
3 4 4 3 4 2 4 4 2 4
Proof. Assume there is x∈M ∩M , |x|=3, for some i6=j. Then M =hG, xi=
i j i
−
M , because H (E |A ) = 0, a contradiction. Thus M ∪ M ∪ M contains
j C 4 4 1 2 3
3·8=24 elements x of order 3 such that hG, xi∈O .
M(A4)
LetH betheuniquecopyofE containingG. WemusthaveH =M ∩M ∩M ,
8 1 2 3
since H (E |M(A )) = 3. Therefore M ∪M ∪M contains 3·(12−4) = 24
C 8 4 1 2 3
−
involutions x such that hG, xi ∈ O . The constants Γ (E , C) are then easy
M(A4) i 4
to calculate with help of Figure 1. (cid:3)
Itis conceivablethatthereis G∈O+ andx∈C suchthathG, xi=C. Itisnot
so, though.
4 PETRVOJTEˇCHOVSKY´
Lemma 2.2. In C, we have
Γ (E+, M(A ))=48, Γ (E+, M(A ))=48, Γ (E+, C)=0, Γ (E+, C)=0.
3 4 4 2 4 4 3 4 2 4
Proof. PickG∈O+, andlet M , ..., M be the sevencopies ofM(A )containing
1 7 4
G. We claimthat (M ∩M )2 =e, fori6=j. Assume it is not true,andlet x be an
i j
elementoforder3containedin M ∩M . ThenA ∼=hG, xi≤M ∩M showsthat
i j 4 i j
H (A ) ≥ 2, a contradiction. Thus 7 M contains all 8·7 = 56 elements
M(A4) 4 Si=1 i
of order 3. In particular, we have hG, xi =6 C for any element x of order 3. This
translates into
Γ (E+, C)=0, Γ (E+, M(A ))=56−Γ (E+, A )=48.
3 4 3 4 4 3 4 4
We proceedcarefully to show that Γ (E+, C)=0. The groupG is containedin
2 4
a single copy A of A , that is in turn contained in a single copyM of M(A ). Let
4 1 4
H , H , H ≤M be the three copies of E containing G (see the proof of Lemma
1 2 3 1 8
13.1[2]). Observethat H ∪H ∪H =G∪Au, where Au is the second cosetof A
1 2 3
in M . Pick M , M , with 2 ≤ i < j ≤ 7. We want to show that M ∩M ⊆ M .
1 i j i j 1
Thanks to the first partof this Lemma, we know that M ∩M ∼=E or E . When
i j 4 8
M ∩M ∼= E then, trivially, M ∩M = G ≤ M . When M ∩M ∼= E then
i j 4 i j 1 i j 8
M ∩M =H for some k ∈{1, 2, 3}, else H (G|E )≥4, a contradiction.
i j k C 8
Consequently, 7 M contains at least 15+6·8 = 63 involutions; 15 in M ,
Si=1 i 1
and additional 8 in each M , i > 1. In particular, hG, xi =6 C for every involution
i
x. We get
Γ (E+, C)=0, Γ (E+, M(A ))=60−Γ (E+, E )=48.
2 4 2 4 4 2 4 8
This finishes the proof. (cid:3)
All constants Γ (A, B) have now been calculated. They are collected in Figure
i
1.
2.1. Random generators of arbitrary orders. According to Figure 1, there
are only five orbit-nonequivalent ways to get from e to C in 3 steps. Namely,
A = {{e}, C , A , C},
0 2 4
A = {{e}, C , E−, C},
1 2 4
A = {{e}, C , S , C},
2 2 3
A = {{e}, C , S , C},
3 3 3
A = {{e}, C , A , C}.
4 3 4
ThesesequencesandtherelatedconstantsΓ (A, B)arevisualizedinFigure2. Full
i
lines correspond to involutions (i=2), dotted lines to elements of order 3 (i=3).
Proposition 2.3. Let C = M∗(2). Then the probability that 3 randomly chosen
elements of C generate C is p (C)=955,584·120−3 =0.553.
3
Proof. By (4),
4
|Gen3(C)|=X|Ai∼|.
i=0
RANDOM GENERATORS OF GIVEN ORDERS 5
M∗(2)
48:48 54:54
8:24
M(A4) 48:48 M(S3)
32:32 18:36
24:8
12:0
24:24
6:0
48:48 24:24
E A
8 4 36:18
4:0 0:8
12:0 S
3
E+ E−
4 4
0:24
36:0
24:0 32:32
27:54
6:0
C C
2 3
63:0 0:56
{e}
Figure 1. The constants Γ (A, B) for M∗(2). If A is maximalin
i
a copy of B, then A and B are connectedby a thick, straightline;
else by a thin, curved line. The constants Γ (A, B), Γ (A, B) are
2 3
located in a box on the edge connecting A and B, separated by
colon.
By our previous calculation summarized in Figure 2, |A0∼| = 63·24·(48+48),
|A1∼|=63·24·(8+24),|A2∼|=63·(32+32)·(18+36),|A3∼|=56·36·(18+36),
and |A4∼|=56·(54+27)·(48+48). Thus |Gen3(C)|=955,584. We are done by
(1). (cid:3)
2.2. Randomgeneratorsofgivenorders. Theonlypossibletypesofordersfor
three generatorsin C are (2, 2, 2), (2, 2, 3), (2, 3, 3), and (3, 3, 3). The sequences
ofsubloopscorrespondingtoeachofthesetypesaredepictedinFigure3. Wemust
be careful, though, since not all combinations of lines in Figure 3 correspond to
6 PETRVOJTEˇCHOVSKY´
M∗(2)
...........8...2.4..............1.8.. .................36 .............4.8.4..8.............
E− S A
4 3 4
24 ....3...2..............3....2..2..4.....3.6.............2..7....................5..4.
C2 C3
63 .............5..6.
{e}
Figure 2. Sequences of subloops in M∗(2)
M∗(2) M∗(2) M∗(2) M∗(2)
8 18 ...2..4.....3...6........................ 1848 36 ........................4...8......4...8 ..............4...8...
E− S A E− S A E− S A E− S A
244 323 4 244....3....2........2...4....3...3.2....3..6.......247 4.....3....2.....2...4...........3...3.6...2.7............4..5..4 4 3 .............454
C C C C C C C C
263 3 263 ...........5..6 3 263 ...........5..6 3 2 ...........5..6 3
{e} {e} {e} {e}
(2, 2, 2) (2, 2, 3) (2, 3, 3) (3, 3, 3)
Figure 3. The shortest sequences of subloops in M∗(2)
sequences with correct types of orders. The possible continuations are emphasized
in Figure 3.
RANDOM GENERATORS OF GIVEN ORDERS 7
Proposition 2.4. Let s = (s , s , s ) be a 3-tuple of integers, s ≤ s ≤ s ,
1 2 3 1 2 3
C = M∗(2), and let ps(C) be the probability that 3 randomly chosen elements
a , a , a of C generate C and (|a |, |a |, |a |) is of type s. Then p (C) =
1 2 3 1 2 3 (2,2,2)
48,384 · 120−3 = 0.028, p (C) = 326,592· 120−3 = 0.189, p (C) =
(2,2,3) (2,3,3)
435,456·120−3=0.252, and p (C)=145,152·120−3 =0.084.
(3,3,3)
Proof. Use Figure 3 and (7). (cid:3)
References
[1] Groups and computation III, proceedings of the 3rd International Conference
held at The Ohio State University, Columbus, OH, June 15–19, 1999. Edited
by William M. Kantor and A´kos Seress. Ohio State University Mathematical
Research Institute Publications, 8, Walter de Gruyter, Berlin, 2001.
[2] P. Vojtˇechovsky´, Investigation of subalgebra lattices by means of Hasse con-
stants, to appear in Algebra Universalis.
Departmentof Mathematics,Iowa StateUniversity,Ames,IA50011,U.S.A.
E-mail address: petr@math.du.edu
