JOURNAL OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 22, Number 3, July 2009, Pages 889–908
S 0894-0347(09)00631-6
Article electronically published on January 5, 2009
PROPER MOUFANG SETS WITH ABELIAN ROOT GROUPS
ARE SPECIAL
YOAV SEGEV
1. Introduction
Moufang sets are the Moufang buildings of rank one. They were introduced by
J. Tits [T1] as a tool for studying algebraic groups of relative rank one. They are
essentially equivalent to split BN -pairs of rank one, and as such they have been
studied extensively. In some sense they are the basic ‘building blocks’ of all split
BN -pairs. In the finite case, it had been a major project to classify split BN -pairs
of rank one. This project culminated in [HKSe]. Moufang sets are also essentially
equivalent to Timmesfeld’s ‘abstract rank one groups’ (see [Ti1]). In recent years
there has been a revived interest and significant progress in this area; for additional
information, see the bibliography at the end of the paper.
Let us recall that a Moufang set is essentially a doubly transitive permutation
group such that the point stabilizer contains a normal subgroup which is regular on
the remaining points. These regular normal subgroups are called the root groups,
and they are assumed to be conjugate and to generate the whole group. In [DW,
DS1, DS3] the notation M(U, τ ) is used for a Moufang set (and this notation is of
course explained there). The group U in this notation is isomorphic to any one of
the root groups of the Moufang set.
Recall that M(U, τ ) is special iff (−a)τ = −(aτ ), for all a ∈ U ∗ . We say that
M(U, τ ) is a sharply 2-transitive Moufang set if its little projective group is sharply
2-transitive. Recall that a Moufang set is proper if it is not sharply 2-transitive. In
this paper we prove:
Main Theorem. Let M(U, τ ) be a proper Moufang set such that U is abelian.
Then M(U, τ ) is special.
The Main Theorem is proved in section 11. It resolves a main conjecture in
the area of Moufang sets. We note that although the condition that M(U, τ ) is
‘special’ seems at first somewhat technical, this condition has many implications
on M(U, τ ) (see [DW, DS1, DST]). Indeed, the condition that M(U, τ ) is special
should eventually lead to the identification of M(U, τ ) as M(J) for some quadratic
Jordan division algebra J. We refer the reader to [Ti1, DW, DS1] and [Ti2] for
results on special Moufang sets with abelian root groups and their connection to
quadratic Jordan division algebras.
Received by the editors February 19, 2008.
2000 Mathematics Subject Classification. Primary 20E42; Secondary 17C60.
Key words and phrases. Moufang set, root group.
The author was partially supported by BSF grant no. 2004-083.
c
2009
American Mathematical Society
Reverts to public domain 28 years from publication
889
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890
YOAV SEGEV
For the sake of completeness we also recall that the converse of the Main Theorem
should also hold:
Conjecture 1. Let M(U, τ ) be a special Moufang set. Then U is abelian.
We refer the reader to [DST] and [S2] for partial results on this conjecture.
Recall from [S1, section 3] the notion of a ‘root subgroup’ (see also [DS3, section
6]). Here are the main steps in the proof of the Main Theorem. The first crucial
step is to show that, under the hypotheses of the Main Theorem, if a ∈ U ∗ is such
that µa is an involution and ∼a = −a (see the notation below for (the important)
∼a), then {b ∈ U ∗ | µb = µa } ∪ {0} =: Va is a root subgroup of U (Theorem 7.1).
Note that a priori there is no reason in the world why Va should be a subgroup (!).
The second crucial step is to show that if we assume in addition that all the µ-maps
are involutions, then the theorem holds (Theorem 8.1). The third important step is
to show that µa is an involution, for any a ∈ U ∗ such that ∼a = −a (Theorem 9.1).
Finally, letting I = {a ∈ U ∗ | µ2a = 1}, the final crucial step is to show that I ∪ {0}
is a root subgroup of U (Theorem 11.5). The Main Theorem follows quickly.
Notice that a Moufang set M(U, τ ) such that µx = µy , for all x, y ∈ U ∗ is
a sharply 2-transitive Moufang set, because by [DW, Theorem 3.1(ii)] the Hua
subgroup H, which is, by definition, the pointwise stabilizer of 0 and ∞, is generated
| x, y ∈ U ∗ }, so it is trivial. In particular, for the root subgroup Va
by {µx µ−1
y
above, the corresponding Moufang set M(Va , τ ) is sharply 2-transitive. Thus, along
our work on the Main Theorem we have encountered situations where, working
for a contradiction, we get nontrivial sharply 2-transitive root subgroups properly
contained in U (for example, the root subgroups Va or the root subgroup I ∪ {0}
above), where M(U, τ ) is a proper Moufang set. We conjecture that this never
happens:
Conjecture 2. Let M(U, τ ) be a proper Moufang set, and let V ≤ U be a root
subgroup with |V | ≥ 5. Then M(V, τ ) is proper.
The reason for the condition |V | ≥ 5 in Conjecture 2 is to exclude the cases
PSL2 (2) ∼
= S3 (the case |V | = 2), PSL2 (3) ∼
= A4 (the case |V | = 3) and the case
Sz(2) (|V | = 4)1 (Sz(2) is a Frobenius group of order 20). These cases occur as
(little projective groups of) root subgroups of proper Moufang sets.
Call a Moufang set M(U, τ ) Zassenhaus if the Hua subgroup H acts semiregularly on U {0} (i.e. the only permutation fixing 3 points is the identity).
It seems to us that it would be important to investigate Zassenhaus Moufang sets
as they are, in some sense, minimal Moufang sets. We cannot however, at this time,
formulate a conjecture on how these Moufang sets should look; so we ask:
Question 3. What is the structure of Zassenhaus Moufang sets?
A word on notation. We conclude the introduction with a word on notation.
The notation in this paper follows [DS1, section 2]; see also [DST, section 1]. The
notation M(U, τ ) is of course explained there. Recall that ∼a := (−aτ −1 )τ =
(−aµ−x )µx , for all x ∈ U ∗ , and that by [DS1, Prop. 3.10(3)], ∼a = −(−a)µa .
Notice that M(U, τ ) is special iff ∼a = −a, for all a ∈ U ∗ . Note also that ∼a =
∼b ⇐⇒ a = b, for all a, b ∈ U ∗ . Throughout this paper we let
(1.1)
1 We
Va := {b ∈ U ∗ | µa = µb } ∪ {0},
a ∈ U ∗.
thank Hendrik Van Maldeghem for pointing out this case to us.
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MOUFANG SETS WITH ABELIAN ROOT GROUPS
891
The following standard facts will be used without further reference:
(1.2)
µb
h
µ−a = µ−1
a = µ∼a , µaµb = µ−a , µa = µah ,
∀a, b ∈ U ∗ and h ∈ H.
Recall also that
h ∈ Aut(U ),
(1.3)
for all h ∈ H,
and that by [DS1, Proposition 3.10(5) and equation (3.3)]:
(1.4)
µc = µ−b µb−a µa
and
c = (a − b)µb + ∼b, where c = (aτ −1 − bτ −1 )τ.
Main Hypothesis. Throughout this paper M(U, τ ) is a Moufang set where τ = µe
for some e ∈ U ∗ . Further, except in section 2, we assume that U is abelian.
2. Lemmas for general Moufang sets
In this section M(U, τ ) is an arbitrary Moufang set.
Lemma 2.1. Let a ∈ U ∗ . Then
(1) if ∼a = a, then (aτ ) · 2 = 0;
(2) if ∼a = a and ∼−a = −a, then a · 2 = 0;
(3) if aµa = a and (−a)µa = −a, then a · 2 = 0;
(4) if (−a)µa ∈ {a, −a}, for all a ∈ U ∗ , then M(U, τ ) is special.
Proof. (1): We have aτ = (∼a)τ = −(aτ ), so (1) holds.
(2) and (3): We first note that the hypotheses in (2) and (3) are equivalent.
Indeed using [DS1, Prop. 3.10(3)] we see that ∼a = a iff (−a)µa = −a and that
∼−a = −a iff −aµ−a = −a iff aµa = a.
Assume that ∼a = a and ∼−a = −a. By (1), with τ = µa we get that (aµa )·2 =
0, so since aµa = a, we have a · 2 = 0.
(4): Assume that for some a ∈ U ∗ , (−a)µa = a. Then (−a)µa = −a and
a · 2 = 0. Thus, since µ2a ∈ Aut(U ), we have aµ2a = −[(−a)µ2a ] = a, and it follows
that aµa = aµ−a = a (because aµ−a = −a). By (2), a · 2 = 0, a contradiction.
Hence we see that
(−a)µa = a, for all a ∈ U ∗ .
Hence ∼a = −(−a)µa = −a, for all a ∈ U ∗ , so M(U, τ ) is special.
Corollary 2.2. Let M(U, τ ) be a Moufang set. Then M(U, τ ) is special iff the
implication
µa = µb =⇒ a = ±b
∗
holds for all a, b ∈ U .
Proof. If M(U, τ ) is special, then the implication holds by [DS1, Prop. 4.9(4)].
Suppose the implication holds. By equation (1.2), µ(−a)µa = µa , for all a ∈ U ∗ ,
and hence (−a)µa = ±a. Then, by Lemma 2.1(4), M(U, τ ) is special.
Lemma 2.3. Let a ∈ U ∗ . Then
(1) ∼a = −(−a)µa ;
(2) ∼−a = −aµ−a ;
(3) − ∼a = (−a)µa ;
(4) ∼− ∼a = aµa ;
(5) − ∼−a = aµ−a ;
(6) ∼− ∼−a = (−a)µ−a ;
(7) − ∼− ∼a = −aµa ;
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892
YOAV SEGEV
(8)
(9)
(10)
(11)
∼− ∼− ∼a = −aµ2a ;
− ∼− ∼−a = −(−a)µ−a ;
∼− ∼− ∼−a = aµ2−a ;
− ∼− ∼− ∼a = aµ2a .
Proof. First note that (−a)µx = ∼(aµx ) and (∼a)µx = −(aµx ), for all x ∈ U ∗ .
Part (1) is [DS1, Prop. 3.10(3)], and (2), (3) are immediate from (1). Part (5) is
immediate from (2). Next, by (3),
∼− ∼a = ∼(−a)µa = ∼(∼(aµa )) = aµa ,
so (4) holds. By (5),
∼− ∼−a = ∼aµ−a = (−a)µ−a ,
so (6) holds. Part (7) follows from (4). By (7) and (2) (with aµa in place of a),
∼− ∼− ∼a = ∼[−(aµa )] = −[aµa µ−aµa ] = −aµ2a ,
and (8) holds. Part (9) comes from (6), and (11) comes from (8). For (10) replace in
(11) a with −a to get ∼− ∼− ∼−a = −(−a)µ2−a = aµ2−a , since µ2−a ∈ Aut(U ). Lemma 2.4. Let a ∈ U ∗ . If Va ⊆ {0, a, −a}, then ∼a = −a.
Proof. Notice that − ∼a, ∼−a ∈ Va . Suppose that Va ⊆ {0, a, −a} and assume that
∼a = −a. Then − ∼a = −a, that is ∼a = a, and a · 2 = 0. Also ∼−a = −a, so by
Lemma 2.1(2), a · 2 = 0, a contradiction.
Lemma 2.5. Let a, b ∈ U ∗ and assume that a = b and µ := µa = µb . Then
(1) µa−b = µaµ−1 −bµ−1 ;
(2) µ−a+b = µ− ∼a+∼b .
Proof. (1): Using equations (1.2) and (1.4) we get
µ−1 µaµ−1 −bµ−1 µ = µ(bµ−1 −aµ−1 )µ
= µ−a µa−b µb
= µ−1 µa−b µ,
so (1) holds.
(2): Replacing a with −a and b with −b in (1) we must replace µ−1 with µ, so we
get
µ−a+b = µ(−a)µ−(−b)µ .
Notice however that (−a)µ = (−a)µa = − ∼a and similarly −(−b)µ = ∼b, so (2)
holds.
Lemma 2.6. Let a, b ∈ U ∗ and assume that a = b and µa = µ−b . Then µa−b =
2
2
−1 µ
µaµ−1
a = µa µaµa −bµa .
a −bµa
Proof. We have
−1 µa = µ
−1
µ−1
a µaµ−1
(bµ−1
a −bµa
a −aµa )µa
= µ−a µa−b µb
= µ−1
a µa−b µ−a ,
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MOUFANG SETS WITH ABELIAN ROOT GROUPS
893
so the first equality holds. The second equality holds because by equation (1.2) and
since µ2a ∈ Aut(U ) we get
2
2 2
2
2
−1 µ
−1 µ
µaµ−1
a = µa µ−a µaµ−1
a = µa µaµa −bµa .
a −bµa
a −bµa
3. Two basic lemmas
In this section we prove two lemmas that are some of the main tools used in this
paper. We remind the reader that from now on we assume that U is abelian.
Lemma 3.1. Let a, b ∈ U ∗ with a = b and µa = µb . Then
(1) (a − b)µb − (b − a)µ−b = ∼−b + a − ∼b;
(2) (aτ −1 − bτ −1 )τ − ((−a)τ −1 − (−b)τ −1 )τ = a;
(3) (a − b)τ − ((∼a) − (∼b))τ = aτ .
Proof. (1): We have
(3.1)
(aτ −1 − bτ −1 )τ = (−bτ −1 + aτ −1 )τ = ((∼b)τ −1 − (∼a)τ −1 )τ.
Now, by equation (1.4),
(aτ −1 − bτ −1 )τ = (a − b)µb + (∼b)
(3.2)
and
(3.3)
((∼b)τ −1 − (∼a)τ −1 )τ = (∼b − ∼a)µ−a + a.
Furthermore
(∼b − ∼a)µ−a = (−(−b)µb + (−a)µa )µ−a
= ((−a)µa − (−b)µa )µ−a
= (−a + b)µ−b + ∼−b.
Comparing (3.2) and (3.3) we get
(a − b)µb + ∼b = (−a + b)µ−b + ∼−b + a,
so (1) holds.
(2): By equation (3.2) we have
(a − b)µb = (aτ −1 − bτ −1 )τ − ∼b,
and replacing a with −a and b with −b we get
(b − a)µ−b = ((−a)τ −1 − (−b)τ −1 )τ − ∼−b.
Hence, by (1) we get
∼−b + a − ∼b = (a − b)µb − (b − a)µ−b
= [(aτ −1 − bτ −1 )τ − ∼b] − [((−a)τ −1 − (−b)τ −1 )τ − ∼−b]
= (aτ −1 − bτ −1 )τ − ((−a)τ −1 − (−b)τ −1 )τ − ∼b + (∼−b),
so (2) holds.
(3): This is obtained from (2) by replacing a with aτ and b with bτ .
Lemma 3.2. Let a, b ∈ U ∗ with a = b and µa = µ−b . Then
(1) (aτ −1 − bτ −1 )τ − ((−b)τ −1 − (−a)τ −1 )τ = ∼b − ∼−a;
(2) (a − b)τ − ((∼b) − (∼a))τ = ∼(bτ ) − ∼−(aτ ); in particular,
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894
YOAV SEGEV
(3) (a − b)µb − ((∼b) − (∼a))µb = a − ∼b;
(4) if ∼a = −a and ∼b = −b, then a = −b.
Proof. (1): By equation (1.4),
(aτ −1 − bτ −1 )τ = (a − b)µb + (∼b)
and
((−b)τ −1 − (−a)τ −1 )τ = (a − b)µ−a + (∼−a) = (a − b)µb + (∼−a);
subtracting we get (1).
(2): This part is obtained from (1) by replacing a with aτ and b with bτ .
(3): By (2), with τ = µb we have
(a − b)µb − ((∼b) − (∼a))µb = ∼(bµb ) − ∼−(aµb ) = ∼(bµb ) − ∼−(aµ−a ).
But by Lemma 2.3, bµb = ∼− ∼b and aµ−a = − ∼−a, so (3) holds.
(4): If ∼a = −a and ∼b = −b, then by (3), a − ∼b = 0, so a = −b.
4. Consequences of the equality µa = µb
Throughout this section a, b ∈ U ∗ are such that a = b and µa = µb . We denote
µ := µa = µb .
Lemma 4.1. We have
(1) µaµ−1 −bµ−1 = µa−b ;
(2) µa−b = µ∼a−∼b ;
(3) µaµ−bµ = µb−a ;
2
(4) µµa−b = µb−a .
Proof. (1): This is Lemma 2.5(1).
(2): This follows from Lemma 2.5(2), by interchanging a and b and using the fact
that U is commutative.
(3): By (2) with −a in place of a and −b in place of b we get
µb−a = µ−a−(−b) = µ∼−a−∼−b .
(∗)
Replacing b with ∼b and a with ∼a in (∗) and using (2) we get
(∗)
µb−a = µ∼b−∼a = µ∼− ∼a−∼− ∼b .
By Lemma 2.3, ∼− ∼a = aµ and ∼− ∼b = bµ, so (3) holds.
(4): Let x := aµ and y := bµ. Then x = y and µx = µ−1 = µy . We have
(1)
−1 = µx−y ,
µaµ2 −bµ2 = µxµ−yµ = µxµ−1
x −yµx
so (4) follows from (3).
Lemma 4.2. We have
(1) µµb−a = µ((∼a)−(∼b))−(a−b) ;
(2) µµa−b = µ(a−b)−((∼a)−(∼b)) ;
2
(3) µµa−b = µ−1 ;
(4) if aµ2 = a and bµ2 = b, then µ2 = 1.
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MOUFANG SETS WITH ABELIAN ROOT GROUPS
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Proof. (1): Using Lemma 3.1(3), equation (1.4) and Lemma 4.1(2), we get
µ = µ((a−b)τ −((∼a)−(∼b))τ )τ −1
= µ(∼b)−(∼a) µ((∼a)−(∼b))−(a−b) µa−b
= µb−a µ((∼a)−(∼b))−(a−b) µa−b
This shows (1).
(2): Set x := ∼a − ∼b and y := a − b. By Lemma 4.1(2), µx = µy , and by (1),
µ2
µ2
x
x
x
µ = µµx−y
. Further, by Lemma 4.1(4), µx−y
= µy−x . Hence, µµx = µx−y
= µy−x
as asserted.
(3): By (1) and (2),
−1
µµb−a = µ((∼a)−(∼b))−(a−b) = µ(a−b)−((∼a)−(∼b))
= (µ−1 )µa−b ,
and (3) follows.
(4): Assume that aµ2 = a and bµ2 = b. By Lemma 4.1(4) we get that
2
µa−b = µaµ2 −bµ2 = µµa−b = µb−a ,
so µ2a−b = 1. Then, by (3), µ2 = 1.
Lemma 4.3. Let ν := µa−b . Then
(1)
(2)
(3)
(4)
µν = µ−1 and ν µ = ν −1 ;
µ8 = ν 8 = 1;
µ is an involution iff ν is an involution;
aµ4 = a and bµ4 = b.
2
2
Proof. (1): By Lemma 4.1(4), ν µ = ν −1 , and by Lemma 4.2(3), µν = ν −1 .
(2): By (1), ν 4 = µ−2 ν −2 µ2 ν 2 = µ−4 and ν 4 = µ2 ν −2 µ−2 ν 2 = µ4 . It follows that
µ8 = 1 and then also ν 8 = 1.
(3): This is immediate from (1).
(4): Assume that aµ4 = a. Then we can take aµ4 in place of b in (1) to get that
µ2 inverts ρ := µaµ4 −a and hence µ4 centralizes ρ. It follows that
2
2
ρ = ρµ = µaµ8 −aµ4 = µa−aµ4 = ρ−1 ,
4
where we have used the fact that µ8 = 1. Thus ρ2 = 1, so by (3), µ2 = 1,
contradicting aµ4 = a. Hence aµ4 = a and by symmetry bµ4 = b.
Proposition 4.4. We have
(1) ∼−a + aµ2 − ∼a = −(∼−(bµ2 ) + b − ∼(bµ2 ));
(2) if ∼a = −a, then (∼−a + aµ2 − ∼a) · 2 = 0.
Proof. (1): By Lemma 4.3(4),
(a − b)µ2 = aµ2 − bµ2 = aµ−2 − bµ−2 = (a − b)µ−2 .
It follows that
[(a − b)µ − (b − a)µ−1 ]µ2 = (a − b)µ−1 − (b − a)µ
= −[(b − a)µ − (a − b)µ−1 ],
so
(4.1)
[(a − b)µ − (b − a)µ−1 ]µ2 = −[(b − a)µ − (a − b)µ−1 ].
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896
YOAV SEGEV
By equation (4.1) and Lemma 3.1(1) we get
(∼−b + a − ∼b)µ2 = − ∼−a − b + ∼a.
(4.2)
Since (−x)µ2 = −(xµ2 ) and (∼x)µ2 = ∼(xµ2 ), for all x ∈ U ∗ , and since µ2 ∈
Aut(U ), equation (4.2) implies (1).
(2): In equation (4.2), take the minus of both sides and then take b = − ∼a.
This gives
∼−a − (∼a) · 2 = (∼− ∼a − a · 2)µ2
= ∼− ∼aµ2 − (aµ2 ) · 2
(because µ2 ∈ Aut(U ))
= ∼− ∼aµ−2 − (aµ2 ) · 2
= − ∼−a − (aµ ) · 2
2
(by Lemma 4.3(4))
(by Lemma 2.3(10)).
We thus see that −(aµ2 ) · 2 = (∼−a − ∼a) · 2, so (2) holds.
Lemma 4.5. Assume that aµ = −a and bµ = −b. Then
(1) a = − ∼−b;
(2) if ∼a = −a and (− ∼a)µ = ∼a, then a · 2 = 0.
Proof. We have
(b − a)µ−1 = (aµ − bµ)µ−1 = (a − b)µ + ∼b.
Using Lemma 3.1(1) we get that
− ∼b = (a − b)µ − (b − a)µ−1 = ∼−b + a − ∼b.
This shows (1). Then under the hypotheses of (2) we may take b = − ∼a in (1) to
get a = −a, so (2) holds.
Lemma 4.6. Let a ∈ U ∗ . Then Va {0, a, −a} iff there exists c ∈ Va∗ such that
∼c = −c.
Proof. Suppose Va ⊆ {0, a, −a}. By Lemma 2.4, ∼a = −a, so also ∼(−a) = −(−a)
and hence there exists no c ∈ Va∗ such that ∼c = −c.
Conversely, assume that Va {0, a, −a}. We show that there exists c ∈ Va∗ such
that ∼c = −c. If ∼a = −a, take c = a. So assume that ∼a = −a and let c ∈ Va∗
such that c ∈
/ {a, −a}. Suppose that ∼c = −c and let x = −c. Then x = a and
µa = µ−x . By Lemma 3.2(4), x = −a, a contradiction.
5. Consequences of the equality µa = µ−b
In this section, a, b ∈ U ∗ are such that a = b and µa = µ−b .
Lemma 5.1. We have
2
2
−1 µ
(1) µa−b = µaµ−1
a = µa µaµa −bµa ;
a −bµa
(2) µa−b = µ∼b−∼a ;
(3) µa−b commutes with µ2a .
Proof. (1): This is Lemma 2.6.
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MOUFANG SETS WITH ABELIAN ROOT GROUPS
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(2): Replacing a with ∼a and b with ∼b in (1) we must replace µ−1
with
a
= µa , so we get
µ−1
∼a
µ∼a−∼b = µ(∼a)µa −(∼b)µa µ2−a
= µbµa −aµa µ2−a
(1)
= µ−1
a−b = µb−a .
Taking inverses we get (2).
(3): Suppose that aµ2a = a. Then b := aµ2a satisfies µa = µb , so we can apply
Lemma 4.3(4) to get aµ4a = a. Thus we always have
(5.1)
aµ4a = a,
for all a ∈ U ∗ .
We claim that
µa−b = µaµ2−a −bµ2a .
(5.2)
Indeed by (2) we have µa−b = µ−b−(−a) = µ(∼−a)−(∼−b) , so
(5.3)
µa−b = µ(∼−a)−(∼−b).
Applying equation (5.3) three times we get that µa−b = µ(∼− ∼− ∼−a)−(∼− ∼− ∼−b) .
By Lemma 2.3(10) we get
µa−b = µaµ2−a −bµ2−b = µaµ2−a −bµ2a ,
so equation (5.2) holds. But now we have
µa−b = µaµ2−a −bµ2a
= µaµ2a −bµ2a
(this is equation (5.2))
(by equation (5.1))
µ2a
= µa−b .
6. One more lemma for the equality µa = µb
Throughout this section a, b ∈ U ∗ are such that µa = µb . In this section we add
one more lemma to section 4. As in section 4 we let
µ := µa = µb .
Lemma 6.1. Assume that a = −b. Then
(1) µa+b = µ∼−a−∼b ;
(2) µa+b = µ(−a)µ−1 −bµ ;
(3) µa+b = µaµ−1 −(−b)µ−1 µ2 = µ2 µaµ−(−b)µ ;
2
2 −1
(4) µa+bµ2 = µ−1
a+b µ = µ µa+b .
Proof. (1): We have µa = µ−(−b) , so we can apply Lemma 5.1(2) with −b in place
of b to get
µa+b = µa−(−b) = µ∼−b−∼a .
Interchanging a and b we get (1).
(2): Replacing in (1) a with ∼−a and b with − ∼b (notice that ∼−a = −(− ∼b))
we get that µ∼−a−∼b = µ∼− ∼−a−∼− ∼b . But by Lemma 2.3 (6) and (7), ∼− ∼−a =
(−a)µ−1 and − ∼− ∼b = −(bµ). This shows (2).
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(3): As in the proof of (1) we can apply Lemma 5.1(1) with −b in place of b to
get
µa+b = µa−(−b) = µaµ−1 −(−b)µ−1 µ2 = µ2 µaµ−(−b)µ .
(4): Note that if a = −bµ2 , then µa = µ−bµ2 = µ−b = µ−a , and then µ2 = 1,
so a = −b, which is contrary to our assumption. Hence in (3) we can take bµ2 in
place of b. Notice that −bµ2 = (−b)µ2 ; consequently we get
µa+bµ2 = µaµ−1 −(−b)µ2 µ−1 µ2 = µaµ−1 −(−b)µ µ2 .
Notice now that aµ−1 = − ∼−a and −(−b)µ = ∼b. Hence aµ−1 − (−b)µ =
−(∼−a − ∼b). Using (1) we get
2
µa+bµ2 = µ−(∼−a−∼b) µ2 = µ−1
a+b µ .
This shows the first equality in (4), and the second equality follows from Lemma
5.1(3).
7. Consequences of the hypothesis µ2a = 1
Throughout this section a ∈ U ∗ is such that
µ2a = 1.
We recall from equation (1.1) the notation Va .
Theorem 7.1. Assume that ∼a = −a, and set t := ∼−a + a − ∼a. Then
(1) t · 2 = 0;
(2) ∼−v + v − ∼v = t, for all v ∈ Va ;
(3) Va is a root subgroup of U .
Proof. By Proposition 4.4(2), t · 2 = 0, and then (2) follows from Proposition 4.4(1)
and from the hypothesis that µ2a = 1.
Let u, v ∈ Va with u = v. By hypothesis, −u ∈ Va . By (2),
∼−u + u − ∼u = t = ∼−v + v − ∼v,
so
(u − v) − (∼u − ∼v) = ∼−v − ∼−u.
Using Lemma 4.1(2) with ∼−v in place of a and ∼−u in place of b we see that
µ(u−v)−(∼u−∼v) = µ∼−v−∼−u = µu−v .
But now by Lemma 4.2(2) with u, v in place of a, b we get that
µµuu−v = µ(u−v)−(∼u−∼v) = µu−v ,
so we see that µa = µu = µu−v and hence u − v ∈ Va . This shows that Va is a
subgroup of U . Also
µuµv = µ−v µ−u µv = µ−a = µa ,
so uµv ∈ Va∗ . By definition (see [S1], section 3) Va is a root subgroup of U .
Corollary 7.2. Assume that Va {0, a, −a}. Then Va is a root subgroup of U .
Proof. By Lemma 4.6 there exists b ∈ Va∗ such that ∼b = −b, so the corollary
follows from Theorem 7.1.
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Here is another very useful result:
Lemma 7.3. Let a, b ∈ U ∗ with a = b and let c := (aτ −1 − bτ −1 )τ . If µa , µb , µc
are involutions, then µa−b = µ∼a−∼b .
Proof. Since (∼a)τ −1 = −(aτ −1 ) and (∼b)τ −1 = −(bτ −1 ), we have
(aτ −1 − bτ −1 )τ = ((∼b)τ −1 − (∼a)τ −1 )τ
= (−((∼a)τ −1 − (∼b)τ −1 ))τ = ∼[((∼a)τ −1 − (∼b)τ −1 )τ ].
Since µc is an involution we see that
µc = µ((∼a)τ −1 −(∼b)τ −1 )τ .
Using equation (1.4) we get
µ−b µb−a µa = µ(aτ −1 −bτ −1 )τ = µ((∼a)τ −1 −(∼b)τ −1 )τ = µ− ∼b µ∼b−∼a µ∼a .
It remains to observe that µ−b = µ− ∼b and µa = µ∼a , because µa and µb are
involutions.
8. The case where all µ-maps are involutions
In this section we assume that all µ-maps are involutions. Our aim in this section
is to prove the following theorem:
Theorem 8.1. Let M(U, τ ) be a Moufang set and assume that
(i) U is abelian.
(ii) µa is an involution, for all a ∈ U ∗ .
Then either M(U, τ ) is sharply 2-transitive, or M(U, τ ) is special.
We start with
Lemma 8.2. Let b ∈ U ∗ and assume that µ2b = 1 and that ∼b = −b. Suppose
b · 2 = 0 and that Vb·2 {0, b · 2, −b · 2}. Then µb = µb·2 .
Proof. By Lemma 3.1(3), with b in place of a, −b in place of b, and µb·2 in place of
τ we see that
((b · 2)µb·2 − (−b · 2)µb·2 )µ−b·2 = b.
But by Corollary 7.2, Vb·2 is a root subgroup of U and hence the LHS of the above
equality is in Vb·2 . It follows that b ∈ Vb·2 as asserted.
Proposition 8.3. Assume that there exists a ∈ U ∗ such that ∼a = −a. Then
M(U, τ ) is either special or sharply 2-transitive.
Proof. We claim that
(8.1)
if b ∈ U ∗ and ∼b = −b, then µb = µa .
Suppose that (8.1) holds. Suppose also that M(U, τ ) is not special. We show
that if b ∈ U ∗ is such that ∼b = −b, then µb = µa . Indeed, let c ∈ U ∗ with
∼c = −c. Then, by (8.1), µa = µc = µb .
Next we claim that µb = µa , for all b ∈ U ∗ . So let b ∈ U ∗ . If ∼b = −b, then, by
the previous paragraph of the proof, µb = µa , while if ∼b = −b, then µb = µa , by
(8.1). We may thus conclude that M(U, τ ) is sharply 2-transitive as asserted.
Hence it remains to prove (8.1). So let b ∈ U ∗ such that ∼b = −b. We want
to show that µa = µb . Suppose that ∼b = b. Notice that ∼(−b) = −(−b) and if
∼(−b) = −b, then by Lemma 2.1(2), ∼b = b = −b, a contradiction. Hence, after
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replacing b with −b if necessary, we may assume without loss that ∼b = b. By
Theorem 7.1,
Vb is a subgroup of U .
(∗)
By Lemma 7.3, µa−b = µ−a−∼b . Assume first that a − b = −a − ∼b. Then a · 2 =
b − ∼b ∈ Vb∗ (because ∼b = b), so Va·2 = Vb . Further, since b ∈ Va·2 and ∼b = −b,
Lemma 4.6 implies that Va·2 {0, a·2, −a·2}. We can thus apply Lemma 8.2 to get
µa = µa·2 = µb−∼b = µb . Assume next that a−b = −a−∼b. We show that Va−b is a
subgroup of U . For that it suffices to show that Va−b {0, a−b, b−a} (see Corollary
∗
, it suffices to show that a − b = −a − ∼b = b − a. But
7.2). Since −a − ∼b ∈ Va−b
we are assuming that a − b = −a − ∼b and since ∼b = −b, we have −a − ∼b = b − a,
so Va−b is a subgroup of U . It follows that
µa−b = µa−b+(−a−∼b) = µ−b−∼b = µb ,
where the last equality is since Vb is a subgroup of U . But since Vb is a subgroup
of U the equality µa−b = µb implies that µb = µa−b+b = µa , as asserted.
We are now in a position to prove Theorem 8.1.
Proof of Theorem 8.1. Assume toward a contradiction that M(U, τ ) is neither
special nor sharply 2-transitive. In view of Proposition 8.3 and Theorem 7.1 we
have
(8.2)
∼a = −a, and Va is a subgroup of U ,
∀a ∈ U ∗ .
Let a, b ∈ U ∗ with a = b. We show that equation (8.2) leads to the conclusion
that µa = µb , thereby obtaining our desired contradiction; indeed this implies that
µx = µy , for all x, y ∈ U ∗ , and hence M(U, τ ) is a sharply 2-transitive Moufang set.
Let
c := (aτ −1 − bτ −1 )τ and d := ((−a)τ −1 − (−b)τ −1 )τ.
Notice that d = (∼(aτ −1 )−∼(bτ −1 ))τ . By Lemma 7.3, µaτ −1 −bτ −1 = µ∼(aτ −1 )−∼(bτ −1 )
and it follows that
µd = µc .
(8.3)
Next we claim that
c = (a − b)µb + ∼b = (∼b − ∼a)µa + a.
(8.4)
The first equality in (8.4) is by equation (1.4). Also since (∼a)τ −1 = −(aτ −1 ) and
(∼b)τ −1 = −(bτ −1 ) we get, using again equation (1.4), that
c = (aτ −1 − bτ −1 )τ = ((∼b)τ −1 − (∼a)τ −1 )τ = (∼b − ∼a)µa + a,
so equation (8.4) holds. Note now that ∼c = (bτ −1 − aτ −1 )τ , so interchanging a
and b in (8.4) gives
(8.5)
∼c = (b − a)µa + ∼a = (∼a − ∼b)µb + b.
Also replacing a by −a and b by −b in equation (8.4) gives
(8.6)
d = (b − a)µb + ∼−b = (∼−b − ∼−a)µa − a.
We now divide the proof into two cases:
Case 1. c + d = 0.
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Since d = −c, equations (8.4) and (8.6) imply that
−(a − b)µb − ∼b = (b − a)µb + ∼−b
or
−(a − b)µb − (b − a)µb = ∼−b + ∼b.
(8.7)
Notice that ∼−b + ∼b = 0; otherwise ∼− ∼−b = b, and then ∼− ∼− ∼−b = ∼−b.
But by Lemma 2.3, ∼− ∼− ∼−b = b, so we get that ∼−b = b and then ∼b = −b,
contradicting (8.2).
Returning to equation (8.7), consider x := −(a−b)µb −(b−a)µb . Working in the
group V(a−b)µb (see equation (8.2)) we see that µx = µ(a−b)µb . However working in
the group Vb shows that µ∼−b+∼b = µb , and we conclude from equation (8.7) that
µ(a−b)µb = µb .
(8.8)
But by equation (1.2), µ(a−b)µb = µb µb−a µb , so equation (8.8) implies that
µb−a = µb .
(8.9)
Working in the group Vb using equation (8.9) we see now that µa = µb .
Case 2. c + d = 0.
By equations (8.4) and (8.6),
c + d = (∼b − ∼a)µa + (∼−b − ∼−a)µa .
Working in the group Vc using equation (8.3) we see that µc+d = µc . Working in
the group V(b−a)µa noticing that by Lemma 7.3,
µ(∼b−∼a)µa = µ(b−a)µa = µ(∼−b−∼−a)µa ,
we see that for x := (∼b − ∼a)µa + (∼−b − ∼−a)µa we have µx = µ(b−a)µa . We
can now conclude that
(8.10)
µc = µ(b−a)µa .
Arguing as above using equation (8.5) (which implies that µc = µ∼c = µ(b−a)µa +∼a )
and working in the group Vc we obtain that µ(b−a)µa = µa . It follows that µa−b = µa
and arguing in the group Va we get that µa = µb . This completes the proof of the
theorem.
9. The proof that if ∼ a = −a, then µ2a = 1
Our aim in this section is to show
Theorem 9.1. If a ∈ U ∗ is such that ∼a = −a, then µ2a = 1.
This will be done in a series of lemmas. Thus, throughout this section,
a ∈ U ∗ is an element such that ∼a = −a.
We start with
Lemma 9.2. We may (and we will) assume that µ4a = 1.
Proof. Consider the root subgroup V ⊆ U of all elements fixed by µ4a . By Lemma
4.3(4), a ∈ V ∗ and hence also ∼a = −(−a)µa ∈ V ∗ . Hence the hypothesis ∼a = −a
holds in M(V, τ ) as well. Suppose the theorem holds for M(V, τ ); then µ2a is the
identity on V , so, in particular, aµ2a = a and (− ∼a)µ2a = − ∼a. But then by
Lemma 4.2(4) (with b = − ∼a), µ2a = 1 and we are done.
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We now assume that Theorem 9.1 is false; that is, we assume that µ2a = 1, and
our aim is to obtain a contradiction. Notice that a · 2 = 0, else µa = µ−a and then
µ2a = 1.
Lemma 9.3. We have
(1) aµ2a = a;
(2) µa·2 = µ∼−a−∼a ;
2
(3) µa+aµ2a = µ−1
a·2 µa ;
2
(4) aµa = −a.
Proof. (1): Suppose aµ2a = a. Then also (− ∼a)µ2a = − ∼a. But then by Lemma
4.2(4) with b = − ∼a we get that µ2a = 1, a contradiction.
(2) and (3): Just take a = b in Lemma 6.1, (1) and (4), respectively.
(4): If aµ2a = −a, then µa = µaµ2a = µ−a , so µ2a = 1, a contradiction.
Lemma 9.4. Set x = a + aµ2a . Then
(1) −aµ2a = ∼−a − ∼a;
(2) µa·2 = µ−1
a ;
(3) x = 0;
(4) µx = µ−a ;
(5) ∼x = −x;
(6) for y = ∼a + (∼a)µ2a we have x = −y.
Proof. (1): In Lemma 3.1(1) we take aµ2a in place of a and a in place of b. We get
(9.1)
(aµ2a − a)µa − (a − aµ2a )µ−a = ∼−a + aµ2a − ∼a.
Notice now that (aµ2a − a)µa = (a − aµ2a )µ−a , because µ4a = 1, and since µ2a ∈
Aut(U ). Hence the LHS of equation (9.1) is zero, so the RHS is also zero.
(2), (3) and (4): Part (2) follows from (1) and Lemma 9.3(2), and (3) comes
from Lemma 9.3(4). By Lemma 9.3(3) and by (2),
µa+aµ2a = µ−a·2 µ2a = µ3a = µ−a ,
so (4) holds.
(5): Assume that ∼x = −x. We claim that µ2x = 1. For this notice that by
(4), xµ2x = xµ2a = x. Also, (∼−x)µ2x = ∼−x. Thus applying Lemma 4.2(4) (with
a = x and b = ∼−x), we see that µ2x = 1. But then by (4), µ2a = µ2−a = µ2x = 1, a
contradiction.
(6): Replacing a with ∼a and arguing as in (5) we see that ∼y = −y and by
(4), µy = µ− ∼a = µa . Notice that if x = y, then µ−a = µx = µy = µa , so µ2a = 1,
a contradiction. Now taking x in place of a and y in place of b in Lemma 3.2(4),
we see that (6) holds.
We are now ready to obtain a contradiction and thereby to prove Theorem 9.1.
Proof of Theorem 9.1. First notice that by Lemma 9.4(1),
(9.2)
aµ2a = − ∼−a + ∼a.
By Lemma 9.4(6) we have a + aµ2a = −(∼a + (∼a)µ2a ) and hence
(9.3)
(a + ∼a)µ2a = −(a + ∼a).
Next, applying Lemma 3.1(1) with b = − ∼a gives
(9.4)
(a + ∼a)µa − (−(a + ∼a))µ−a = a · 2 − ∼− ∼a.
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But by equation (9.3), (a + ∼a)µa = (−(a + ∼a))µ−a , so from equation (9.4) we
see that
a · 2 = ∼− ∼a.
(9.5)
Replacing a with − ∼−a in equation (9.5) (note that this is possible because ∼b =
−b, for b = − ∼−a), and using Lemma 2.3(10) we get that
aµ2a = (− ∼−a) · 2.
But now equation (9.2) shows that − ∼−a = ∼a, or ∼− ∼−a = a. By Lemma
2.3(6) this shows that (−a)µ−a = a, so aµa = −a. Replacing a with − ∼a we get
that (− ∼a)µa = ∼a. Since we are assuming that ∼a = −a, we get from Lemma
4.5(2) that a · 2 = 0, a contradiction.
Corollary 9.5. Let b ∈ U ∗ such that Vb {0, b, −b} and set tb := ∼−b + b − ∼b.
Then
(1) there exists c ∈ Vb such that ∼c = −c;
(2) µ2b = 1;
(3) Vb is a root subgroup of U ;
(4) tb · 2 = 0 and tb = ∼−v + v − ∼v, for all v ∈ Vb∗ .
Proof. Part (1) holds by Lemma 4.6. By (1) and Theorem 9.1, (2) holds. By (2)
and Corollary 7.2, (3) holds. Finally, by Theorem 7.1(1 and 2), tb = ∼−c + c − ∼c
satisfies tb · 2 = 0, and tb = ∼−v + v − ∼v, for all v ∈ Vb∗ .
10. Some auxiliary lemmas
In this section we prove a few lemmas that will help us with the proof of the
Main Theorem. The purpose of the first three lemmas is to deal with elements of
order 3.
Lemma 10.1. Let a ∈ U ∗ such that µ2a = 1. Assume that a · 2 = 0 and ∼−a + a −
∼a = 0. Then (a · 2)τ = aτ · 2.
Proof. By Lemma 3.1(3) with b = −a we get
(a · 2)τ − (∼a − ∼−a)τ = aτ.
Now
∼a − ∼−a = −(∼−a − ∼a) = −(−a) = a,
so we get
(a · 2)τ − aτ = aτ.
∗
Lemma 10.2. Let a ∈ U such that a · 3 = 0 and such that ∼−a + a − ∼a = 0.
Then ∼a = −a.
Proof. If Va ⊆ {0, a, −a}, then by Lemma 2.4, ∼a = −a. Hence we may assume that
Va {0, a, −a}. Then, by Corollary 9.5, µ2a = 1. By Lemma 10.1, (a · 2)τ = (aτ ) · 2.
Similarly (−a · 2)τ = (−a)τ · 2. We thus have
aτ = (−a · 2)τ = (−a)τ · 2 = (a · 2)τ · 2 = aτ · 4.
Hence aτ ·3 = 0. But now (−a)τ = (a·2)τ = (aτ )·2 = −(aτ ). Hence (−(aτ ))τ −1 =
−a, so ∼a = −a.
Corollary 10.3. Assume that U is 2-torsion free and let a ∈ U ∗ such that Va {0, a, −a}. Then a · 3 = 0 if and only if ∼a = −a.
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Proof. First, by Corollary 9.5(2), µ2a = 1 and Va is a subgroup of U . Let t =
∼−a + a − ∼a. By Corollary 9.5(4), t · 2 = 0, so since we are assuming that U is
2-torsion free, t = 0.
Assume first that ∼a = −a. Then 0 = t = ∼−a + a − ∼a = a · 3. Conversely,
assume that a · 3 = 0. Then, by Lemma 10.2, ∼a = −a.
Lemma 10.4. Let a ∈ U ∗ such that ∼a = −a. Let ta = ∼−a + a − ∼a. Then
(1) either ta is the unique involution of Va , or ta = 0 and Va has no involutions;
(2) if ta = 0, or ta = 0 and ∼ta = ta , then ∼b = b for all b ∈ Va∗ {ta }, and
a · 2 = 0;
(3) there exists b ∈ Va∗ such that b · 2 = 0.
Proof. By Lemma 2.4, Va {0, a, −a}, so by Corollary 9.5, µ2a = 1 and Va is a root
subgroup of U .
(1): Let t ∈ Va∗ such that t · 2 = 0. By Corollary 9.5(4), t = ∼−t + t − ∼t = ta .
(2): Let x ∈ Va∗ such that ∼x = x. Suppose first that ta = 0. Then xµa =
(∼x)µa = −(xµa ). Thus t := xµa ∈ Va∗ satisfies t · 2 = 0. This contradicts (1).
Assume next that ta = 0 and that ∼ta = ta . Then ∼−x = ∼−x + x − ∼x = ta .
Hence x = − ∼ta = ta . Finally if a · 2 = 0, then, by (1), a = ta = 0, and then
∼a = −a, a contradiction.
(3): Assume that b · 2 = 0, for all b ∈ Va . Then, by (1), Va = {0, ta }. But since
∼a, −a ∈ Va we get that ∼a = −a, a contradiction.
Lemma 10.5. Let b ∈ U ∗ such that Vb {0, b, −b}. Then there exists a ∈ Vb∗ such
that a = ∼a = −a and ∼−a = −a.
Proof. Let t = tb as in Corollary 9.5(4). Suppose first that t = 0 and that ∼t = t.
Then we can take a = t.
Assume next that t = 0 or that t = 0 and ∼t = t. Let a ∈ Vb∗ such that ∼a = −a
(see Corollary 9.5(1)). Then a = t, so by Lemma 10.4(2), a satisfies the required
properties.
11. The proof of the Main Theorem
In this section we prove the Main Theorem (Theorem 11.6). We assume that
M(U, τ ) is not special, and we show that M(U, τ ) is sharply 2-transitive.
We denote
NS := {a ∈ U ∗ | Va {0, a, −a}}
and
S := U ∗ NS.
Thus S stands for ‘special’ (this happens in a special Moufang set) and NS stands
for ‘not special’. We also let
I := {a ∈ U ∗ | µ2a = 1}.
Notice that by Corollary 9.5(2),
NS ⊆ I.
Lemma 11.1. Let a, b ∈ U ∗ such that µa = µb . Then
(1) if a ∈ S, then ∼a = −a and a = ±b;
(2) if ∼a = −a, then a ∈ NS;
(3) if a ∈ NS, then aτ ∈ NS;
(4) if a ∈
/ I, then a = b.
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Proof. (1): Since a ∈ S, we have Va ⊆ {0, a, −a}, so by Lemma 2.4, ∼a = −a.
(2): This follows from Lemma 2.4.
∗
= Va∗ τ . By Corollary 9.5(1),
(3): Let a ∈ NS. It is easy to check that Vaτ
there exists c ∈ Va with ∼c = −c. It follows that ∼(cτ ) = −(cτ ). Since cτ ∈ Vaτ ,
Lemma 4.6 implies that aτ ∈ NS.
(4): Since a ∈
/ I we have a ∈ S, so by (1), a = ±b.
/ I, so a = b.
Now a = −b; otherwise µa = µb = µ−a , contradicting a ∈
Proposition 11.2. Let a ∈ NS. Then
(1) there exists b ∈ Va such that ∼b = −b;
(2) µ2a = 1;
(3) Va is a root subgroup of U ;
(4) Va∗ ⊆ NS.
Proof. Parts (1), (2) and (3) are Corollary 9.5. Part (4) follows from (1) and Lemma
4.6.
Proposition 11.3. Let a, b ∈ U ∗ and assume that ∼a = −a and that a = b. Then
(1) if a = ∼a, and b, (aτ − bτ )τ −1 ∈ I, then a − b ∈ NS;
/ I, then ∼a − ∼b ∈
/ I;
(2) if ∼−a = −a, b ∈ I and (aτ −1 − bτ −1 )τ ∈
/ I, then a − b ∈ NS.
(3) if ∼−a = −a, b ∈ I and (aτ −1 − bτ −1 )τ ∈
Proof. (1): First, by Lemma 7.3, µa−b = µ∼a−∼b . Suppose a − b ∈
/ NS. Then, by
Lemma 11.1(1), a − b = ±(∼a − ∼b), so
(11.1)
either a − ∼a = b − ∼b or a + ∼a = b + ∼b.
Suppose that b ∈ S. By Lemma 11.1(1), ∼b = −b. By equation (11.1), b · 2 =
b − ∼b = a − ∼a ∈ Va∗ . By Proposition 11.2(4), b · 2 ∈ NS. By Lemma 8.2,
∗
= Va∗ ⊆ NS, a contradiction.
µb = µb·2 , so b ∈ Vb·2
Thus b ∈ NS. Using equation (11.1) and the fact that Va and Vb are subgroups
we get
µa = µa±∼a = µb±∼b = µb .
Working in Va we see that a − b ∈ Va∗ ⊆ NS, a contradiction.
(2): Suppose that ∼a − ∼b ∈ I. Then
((−aτ −1 )τ − (−bτ −1 )τ )τ −1 = (∼a − ∼b)τ −1 ∈ I.
Let x := −aτ −1 and y := −bτ −1 . Suppose ∼x = −x. Then (−a)τ −1 = ∼(aτ −1 ) =
−(aτ −1 ) = (∼a)τ −1 , and then ∼a = −a, a contradiction. Suppose ∼x = x. Then
∼−(aτ −1 ) = −aτ −1 , and then
(− ∼a)τ −1 = ∼−(aτ −1 ) = −(aτ −1 ) = (∼a)τ −1 .
It follows that − ∼a = ∼a and then, using Lemma 2.3(11) (and the fact that
µ2a = 1), we get a = ∼− ∼a = − ∼−a, so ∼−a = −a, a contradiction. Also
y, (xτ − yτ )τ −1 ∈ I and x = y. We can thus apply (1) to get x − y ∈ I, that is,
bτ −1 − aτ −1 ∈ I. But this implies that (aτ −1 − bτ −1 )τ ∈ I, a contradiction.
/ I, we have c ∈ S, so
(3): Set c := (aτ −1 − bτ −1 )τ . Notice that since c ∈
∼c = −c. By equation (1.4) we have that
(aτ −1 − bτ −1 )τ = ((∼b)τ −1 − (∼a)τ −1 )τ = (∼b − ∼a)µa + a,
and since (bτ −1 − aτ −1 )τ = ∼c = −c we get
(aτ −1 − bτ −1 )τ = −(bτ −1 − aτ −1 )τ = −[(b − a)µa + ∼a].
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906
YOAV SEGEV
It follows that
(11.2)
(∼b − ∼a)µa + (b − a)µa = −(a + ∼a).
Interchanging a and b in equation (11.2) (notice that this is possible) we get that
(11.3)
(∼a − ∼b)µb + (a − b)µb = −(b + ∼b).
Notice that by (2), ∼a − ∼b ∈ S. Assume that a − b ∈ S. Taking the minus of both
sides of the equality in equation (11.3) and using Lemma 11.1(1), we get that
(11.4)
(∼b − ∼a)µb + (b − a)µb = b + ∼b.
Applying µ−a to both sides of equation (11.2) we get
(11.5)
((∼b − ∼a)µa + (b − a)µa )µ−a = (−(a + ∼a))µ−a .
Also, since the LHS of equation (11.2) is nonzero, the LHS of equation (11.4) is
nonzero (because µ−a µb ∈ Aut(U )), so we can apply µ−b to both sides of equation
(11.4) to get
(11.6)
((∼b − ∼a)µb + (b − a)µb )µ−b = (b + ∼b)µ−b .
Note next that since b + ∼b = 0, Lemma 11.1(2) implies that b ∈ NS. Since the
LHS of equation (11.5) equals the LHS of equation (11.6), the RHS of the equations
are equal. But (−(a + ∼a))µ−a ∈ Va and (b + ∼b)µ−b ∈ Vb , so we get that µa = µb ,
and now we can conclude that a − b ∈ Va∗ ⊆ NS. This contradicts the hypothesis
that a − b ∈ S.
As a corollary we get the following crucial result.
Corollary 11.4. Let a ∈ U ∗ such that ∼a = −a, and let b ∈ I such that a = b.
Then
(1) if ∼a = a and ∼−a = −a, then a − b ∈ NS;
(2) if U is 2-torsion free, then a − b ∈ NS;
(3) NS = I.
Proof. Part (1) is immediate from Proposition 11.3, (1) and (3). For (2) notice that
since in this part we are assuming that U is 2-torsion free, ∼x = x, for all x ∈ U ∗ ,
by Lemma 2.1(1). Hence (2) follows from (1).
Since we know that NS ⊆ I, to show (3) it suffices to show that I ⊆ NS. By
Lemma 10.5 there exists c ∈ Va∗ such that c = ∼c = −c and ∼−c = −c. Let b ∈ I.
If b = c, then b ∈ NS, so assume b = c. Using (1) twice we see that c − b ∈ NS and
then also b = c − (c − b) ∈ NS.
Theorem 11.5. I ∪ {0} is a sharply 2-transitive root subgroup of U .
Proof. Since aµx ∈ I, for every a ∈ I and x ∈ U ∗ , to show that I ∪ {0} is a root
subgroup of U , it suffices to show that I ∪ {0} is a subgroup of U . We work with
NS ∪ {0} (see Corollary 11.4(3)).
Suppose first that there exists an involution a ∈ U ∗ . Notice that a ∈ I and
hence, by Corollary 11.4(3), a ∈ NS. Let b ∈ NS; we show that Va = Vb . Suppose
first that ∼a = a. Pick c ∈ Vb∗ with c · 2 = 0 (see Lemma 10.4(3)). By Corollary
11.4(1), a + c ∈ NS. Then c · 2 = (c + a) · 2 ∈ Vc+a and then Vc = Vc·2 = Vc+a .
Hence a, c ∈ Vc+a , so Va = Vc+a = Vc = Vb .
We may thus assume that ∼t = t, for all involutions t ∈ U . Hence, either tb = 0
or tb = 0 and ∼tb = tb (see Corollary 9.5 for the notation tb ). In both cases pick
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MOUFANG SETS WITH ABELIAN ROOT GROUPS
907
any c ∈ Vb∗ with ∼c = −c. Then, by Lemma 10.4, (1) and (2), c · 2 = 0, and by
Lemma 10.4(2), ∼c = c and ∼−c = −c. By Corollary 11.4(1), c + a ∈ NS and
again c · 2 = (c + a) · 2 ∈ Vc+a . Again we see that Va = Va+c = Vc = Vb .
Hence we may assume that U is 2-torsion free. Assume first that there exists
a ∈ NS such that a · 3 = 0. We show that Va = Vb , for all b ∈ NS. So let b ∈ NS
and let c ∈ Vb with ∼c = −c. By Corollary 10.3, c · 3 = 0. By Corollary 11.4(2),
c + a ∈ NS. Hence c · 3 = (c + a) · 3 ∈ Vc+a . Thus Vc = Vc·3 = Vc+a . It follows that
Vb = Vc = Va , and we are done.
Hence we may assume that a · 3 = 0, for all a ∈ NS. By Corollary 10.3 we get
that ∼a = −a, for all a ∈ NS. By Corollary 11.4(2), a − b ∈ NS, for all a, b ∈ NS
with a = b. This completes the proof that NS ∪ {0} is a subgroup of U .
Since all µ-maps of M(I ∪ {0}, τ ) are involutions, Theorem 8.1 shows that it is
sharply 2-transitive (because there exists a ∈ I with ∼a = −a) and completes the
proof of the theorem.
Theorem 11.6. M(U, τ ) is either special or sharply 2-transitive.
Proof. Suppose that M(U, τ ) is not special. Then there exists a ∈ U ∗ such that
∼a = −a, and by Theorem 9.1, a ∈ I, so I = ∅. By Theorem 11.5, I ∪ {0} is a
sharply 2-transitive root subgroup of U . Also, if U ∗ = I, then M(U, τ ) is sharply
2-transitive, and we are done. Let a ∈ I; we claim that
(11.7)
xµa = −x,
for all x ∈ U ∗ I.
Indeed let x ∈ U ∗ I. Then aµx ∈ I, so since I ∪ {0} is a sharply 2-transitive root
x
subgroup, µa = µaµx = µµ−a
= µµa x . Hence µx = µµx a = µ−xµa . By Lemma 11.1(4),
xµa = −x, so equation (11.7) holds.
Now let b ∈ U ∗ I. Then a + b ∈ U ∗ I. Using equation (11.7) and equation
(1.4) we get
(−a)µa = (b − (a + b))µa = ((a + b)µa − bµa )µa = aµb − b,
where we have used the fact that ∼b = −b since b ∈
/ I. Hence we get that b =
aµb − (−a)µa . But aµb , (−a)µa ∈ I, which by Theorem 11.5 implies that b ∈ I, a
contradiction.
Acknowledgment
I thank Tom De Medts for numerous remarks that improved the exposition of
this paper, and for a very thorough reading of the paper.
Added after posting
One additional case should be excluded from Conjecture 2 of the Introduction,
the case of PSU3 (2) (|V | = 8). Recall that PSU3 (2) ∼
= 32 : Q8 is an elementary
abelian group of order 9 extended by the quaternion group Q8 . I thank Matthias
Grueninger from the Free University of Berlin for pointing out this omission to me.
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E-mail address: [email protected]
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