The p–gen nature of M₀(V ) (I)
Let V be a finite group (not elementary two) and p ≥ 5 a prime. The question as to when the nearring M₀(V) of all zero-fixing self-maps on V is generated by a unit of order p is difficult. In this paper we show M₀(V) is so generated if and only if V does not belong to one of three finite disjoint fa...
Збережено в:
| Дата: | 2013 |
|---|---|
| Автор: | |
| Формат: | Стаття |
| Мова: | English |
| Опубліковано: |
Інститут прикладної математики і механіки НАН України
2013
|
| Назва видання: | Algebra and Discrete Mathematics |
| Онлайн доступ: | http://dspace.nbuv.gov.ua/handle/123456789/152293 |
| Теги: |
Додати тег
Немає тегів, Будьте першим, хто поставить тег для цього запису!
|
| Назва журналу: | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| Цитувати: | The p–gen nature of M₀(V ) (I) / S.D. Scott // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 2. — С. 237–268. — Бібліогр.: 6 назв. — англ. |
Репозитарії
Digital Library of Periodicals of National Academy of Sciences of Ukraine| id |
irk-123456789-152293 |
|---|---|
| record_format |
dspace |
| spelling |
irk-123456789-1522932019-06-10T01:26:02Z The p–gen nature of M₀(V ) (I) Scott, S.D. Let V be a finite group (not elementary two) and p ≥ 5 a prime. The question as to when the nearring M₀(V) of all zero-fixing self-maps on V is generated by a unit of order p is difficult. In this paper we show M₀(V) is so generated if and only if V does not belong to one of three finite disjoint families D#(1, p) (=D(1, p) ∪ {{0}}), D(2, p) and D(3, p) of groups, where D(n, p) are those groups G (not elementary two) with |G| ≤ np and δ(G) > (n − 1)p (see [1] or §.1 for the definition of δ(G)). 2013 Article The p–gen nature of M₀(V ) (I) / S.D. Scott // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 2. — С. 237–268. — Бібліогр.: 6 назв. — англ. 1726-3255 2010 MSC:16Y30. http://dspace.nbuv.gov.ua/handle/123456789/152293 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
| institution |
Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| collection |
DSpace DC |
| language |
English |
| description |
Let V be a finite group (not elementary two) and p ≥ 5 a prime. The question as to when the nearring M₀(V) of all zero-fixing self-maps on V is generated by a unit of order p is difficult. In this paper we show M₀(V) is so generated if and only if V does not belong to one of three finite disjoint families D#(1, p) (=D(1, p) ∪ {{0}}), D(2, p) and D(3, p) of groups, where D(n, p) are those groups G (not elementary two) with |G| ≤ np and δ(G) > (n − 1)p (see [1] or §.1 for the definition of δ(G)). |
| format |
Article |
| author |
Scott, S.D. |
| spellingShingle |
Scott, S.D. The p–gen nature of M₀(V ) (I) Algebra and Discrete Mathematics |
| author_facet |
Scott, S.D. |
| author_sort |
Scott, S.D. |
| title |
The p–gen nature of M₀(V ) (I) |
| title_short |
The p–gen nature of M₀(V ) (I) |
| title_full |
The p–gen nature of M₀(V ) (I) |
| title_fullStr |
The p–gen nature of M₀(V ) (I) |
| title_full_unstemmed |
The p–gen nature of M₀(V ) (I) |
| title_sort |
p–gen nature of m₀(v ) (i) |
| publisher |
Інститут прикладної математики і механіки НАН України |
| publishDate |
2013 |
| url |
http://dspace.nbuv.gov.ua/handle/123456789/152293 |
| citation_txt |
The p–gen nature of M₀(V ) (I) / S.D. Scott // Algebra and Discrete Mathematics. — 2013. — Vol. 15, № 2. — С. 237–268. — Бібліогр.: 6 назв. — англ. |
| series |
Algebra and Discrete Mathematics |
| work_keys_str_mv |
AT scottsd thepgennatureofm0vi AT scottsd pgennatureofm0vi |
| first_indexed |
2025-07-13T02:45:46Z |
| last_indexed |
2025-07-13T02:45:46Z |
| _version_ |
1837498112503971840 |
| fulltext |
Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 15 (2013). Number 2. pp. 237 – 268
c© Journal “Algebra and Discrete Mathematics”
The p–gen nature of M0(V ) (I)
Stuart D. Scott
Communicated by G. Pilz
Abstract. Let V be a finite group (not elementary two)
and p ≥ 5 a prime. The question as to when the nearring M0(V )
of all zero–fixing self-maps on V is generated by a unit of order
p is difficult. In this paper we show M0(V ) is so generated if and
only if V does not belong to one of three finite disjoint families
D#(1, p) (=D(1, p) ∪ {{0}}), D(2, p) and D(3, p) of groups, where
D(n, p) are those groups G (not elementary two) with |G| ≤ np and
δ(G) > (n − 1)p (see [1] or §.1 for the definition of δ(G)).
1. Introduction
Throughout this paper nearrings will be taken as zero–symmetric and
left distributive. This will mean functions are written on the right. Unless
indicated otherwise all nearrings will have an identity. Also groups other
than automorphism groups and groups of units will be written additively.
This convention is not taken as implying the groups are abelian.
If V is a group then M0(V ) is the nearring of all zero–fixing self–maps
of V . These structures are called transformation nearrings. Associated
with an element α of M0(V ) is the subnearring N(α) (with or without
identity) of M0(V ) generated by α and it is a question of considerable
interest as to when α with special properties can generate M0(V ) (ie. is
such that N(α) = M0(V )).
Addressing the question as to when any α in M0(V ) generates M0(V )
when V is infinite, is not difficult. Here because N(α) is countable and
M0(V ) is not, there are no such elements. So straight away we may assume
2010 MSC: 16Y30.
Key words and phrases: nearring, unit, cycles (p-cycles), fixed–point–free, p–gen.
238 The p–gen nature of M0(V ) (I)
V is finite (an assumption adopted from now on). However, there are
other restrictions that the N(α) = M0(V ) requirement imposes. If for
example an element of M0(V ) is not bijective (ie. not a unit of M0(V )),
then in no way can it generate M0(V ). This is because if β in M0(V ) is
such that aβ = bβ (a and b distinct elements of V ), then all γ in M0(V )
with aγ = bγ is a proper subnearring (without identity) containing β.
The requirement that α in M0(V ) is such that N(α) = M0(V ), implies
V is finite and α belongs to the group (under composition) u(M0(V )) of
all units of M0(V ). So what further sensible restrictions can be placed on
α and it still generate M0(V )? In broad terms there is only one property
of elements of u(M0(V )) that springs to mind. As u(M0(V )) is the full
symmetric group on V ∗ (the non–zero elements of V ) the order (under
composition) of an α in u(M0(V )) is surely its dominant feature. Thus for
any given integer n ≥ 2, the question as to when the nearring involved is
generated by a unit α of order n is (in the case of M0(V )), at the heart of
generation investigations. An M0(V ) so generated will be called n–gen. It
is the purpose of this paper to give a complete answer as to when M0(V )
(V not elementary two) is p–gen (p a prime).
In [4] and [5] the problem of when M0(V ) is p–gen was completely
solved for p = 2 and 3. The main theorem of [4] was that M0(V ) is 2–gen
if and only if V is not a C3 (cyclic group of order three) or an elementary
two group, while the main theorem of [5] was that M0(V ) is 3–gen if
and only if V is not a C3, an S3 (symmetric group of order six) or an
elementary two group of order incongruent to 1 mod 3 (apart from {0}
and C2 ⊕ C2). This paper will deal with primes ≥ 5 in the case of V not
elementary two. The elementary two case presents special problems, and
is dealt with in a separate paper which is called ‘The p–gen Nature of
M0(V ) (II)’.
We now move towards the statement of the theorem this paper proves.
If V is a group then δ(V ) will be the number of minimal subgroups of V
(ie. the number of subgroups of V of prime order). This group invariant
was investigated in [1], where its importance for deciding which M0(V )
are p–gen was explained. The fact that we know what groups V have
δ(V ) > 1
2 |V | − 1 is significant. It means that given a prime p (≥ 5 say)
and group V there is theory available that tells us if M0(V ) is p–gen.
To state the theorem this paper proves, we define finite disjoint families
D(n, p), n = 1, 2, . . . of groups as follows. A group G is in D(n, p) if, it
is not elementary two, |G| ≤ np and δ(G) > (n − 1)p. With D#(1, p) as
D(1, p) ∪ {{0}} we have:-
S. D. Scott 239
Theorem 1.1. If V is a group (not elementary two) and p ≥ 5 a prime,
then M0(V ) is p–gen if and only if, V does not belong to any of the three
mutually exclusive classes D#(1, p), D(2, p) or D(3, p).
2. Some group theory
This section develops group notation and theory that will be required
later. Because most theory that we cover tends to be quite well known,
proofs are nearly always omitted (references are of course supplied). This
should make the paper more self contained.
As indicated in the previous section groups being dealt with are finite.
A feature of developments is that groups of odd order remain very much in
the background. Given this fact, it might be expected that the involutions
of a group V (non–zero elements v of V such that v + v = 0) play a
prominent role. In a real enough sense, this is true. It is mainly the subset
η(V ) of V of all such elements that figures. Indeed we shall often need to
look at |η(V )| (the order of a subset S of V is denoted by |S|). Also the
subset i(V ) = η(V ) ∪ {0} of V will on occasions, also enter. The subset
τ(V ) = V ∗\η(V ) (here if S is a subset of V, S∗ denotes S\{0}) consisting
of all elements of order > 2, will come into play. One useful fact about the
subset τ(V ) of V is that it can be partitioned into two element subsets
{h, −h} (h ∈ τ(V )) of V . This has the consequence that:-
Proposition 2.1. For a group V of even order |τ(V )| is even and |η(V )|
is odd.
Another more indirect consequence of the above partitioning is that:-
Proposition 2.2. If V is a group and S a subset of τ(V ) of non-zero
order 2n (or 2n + 1) where n ≥ 0 is an integer, then S contains a subset
K of order n (of order n + 1) not containing any H∗ with H a non–zero
subgroup of V .
Proof. Certainly S can be written as a union of disjoint two element
subsets and a subset S1 of elements of S without additive inverses in S.
A subset K1 of S containing one element of each two element subset and
S1 has no element with additive inverse in K1 and 2.2 readily follows.
Corollary 2.3. The K of 2.2 can be taken to have order ≤ n (when
|S| = 2n and order ≤ n + 1 (when |S| = 2n + 1), simply by taking a new
K contained in the K of 2.2.
Above information on τ(V ) (V a group) is basic, but there is one
result to do with η(V ) (thus τ(V ) also) which we need and has far more
240 The p–gen nature of M0(V ) (I)
substance. For our purposes it is of value to know what groups V have
|η(V )| > 1
2 |V |−1. There are four families of such groups (type (I) to (IV))
that are now specified. As stated below a theorem due to C.T.C.Wall
tells us that a group V with |η(V )| > 1
2 |V | − 1 is of type (I) to (IV).
Groups of type (I) are simply the generalised dihedral groups. Such a
V is an abelian group A extended semidirectly by an element of order
two which, under conjugation additively inverts elements of A. Here V
is denoted by D(A) and is only abelian (in fact elementary two) if A is
elementary two. These are also the groups V where the characteristic
subgroup λ(V ) of V generated by τ(V ) is < V . Indeed it is easy to show
that if such a V is not abelian (in fact not elementary two) then λ(V ) = A
(V being a D(A)). A special case of a D(A) is where A is a cyclic group Cn
of order n. These are the dihedral groups which we denote by Dn. They
have order 2n. A group of the form 〈a, b〉 with a and b both involutions
is one of them.
The order of η(V ) (V a D(A)) reduces to knowing what |η(A)| is.
Proposition 2.4. If V is of the form D(A) (A not elementary two) then
|η(V )| = 1
2 |V | + |η(A)|.
Groups of type (II) are those of the form D4 ⊕ D4 ⊕ E, where E is
elementary two. Here we have.
Proposition 2.5. A group V of type (II) has |η(V )| = 9
16 |V | − 1 (see
[4]).
Groups of type (III) are of the form H(r) ⊕ E (E elementary two
and r ≥ 1 an integer) where H(r) is the direct sum of r copies of D4
where the central subgroup consisting of a direct sum of r − 1 copies of
C2 that identifies the centers has been factored out so that Z(H(r)) is
a C2. If V = H(r) ⊕ E where E has order 2n, then |V | = 22r+n+1 and
|η(V )| = |V |/2 + 2n+r − 1 (see [6] or [1]). The only overlap between (I)
and (III) occurs when r = 1. Here |η(V )| = 3
4 |V | − 1 while for r = 2 or
≥ 3 the above value of |η(V )| gives.
Proposition 2.6. A group V = H(r) ⊕ E of type (III) has |η(V )| =
5
8 |V | − 1 when r = 2 and |η(V )| ≤ 9
16 |V | − 1 where r ≥ 3.
Groups of type (IV) are of the form S(r) ⊕ E (E elementary two of
order 2n and r ≥ 1 an integer) where S(r) is the direct sum of r copies
of C2 ⊕ C2 extended semidirectly by a C2 which is a wreath product
(ie. D4) on each C2 ⊕ C2 component. Again we have the only overlap
between (I) and (III) with (IV) is r = 1 where |η(V )| = 3
4 |V | − 1. In
general |V | = 22r+n+1 and |η(V )| = |V |/2 + 2n+r − 1 (see [6] or [1]) so
that as for type (III) we have:-
S. D. Scott 241
Proposition 2.7. A group V = S(r) ⊕ E of type (IV) has |η(V )| =
5
8 |V | − 1 when r = 2 and |η(V )| ≤ 9
16 |V | − 1 where r ≥ 3.
The theorem of C.T.C.Wall (see [6]) that was mentioned above states.
Theorem 2.8. (Wall) If V is a group with |η(V )| > 1
2 |V | − 1, then V
is of type (I) to (IV).
Before leaving this section we need some information about fixed–
point–free automorphisms. Such an automorphism µ (of a group V ) is
defined by the property that vµ 6= v for all non–zero v of V . When µ also
has prime order q certain things happen. One is that a subset S of V ∗
with Sµ = S has order divisible by q. Another is that when q = 2, V is
necessarily abelian (see [2, Ch.10]). Stated as a proposition:-
Proposition 2.9. If the group V has a fixed–point–free automorphism
µ of order two, then V is abelian and vµ = −v for all v in V .
In the case of q being a larger prime quite a lot can still be said about
the structure of V . A fundamental theorem due to Thompson (see [2,
Ch.10]) states:-
Theorem 2.10. (Thompson) If the group V has a fixed–point–free
automorphism of prime order, then V is nilpotent
3. Elements of M0(V )
In this section we cover notation and elementary theory about elements
of M0(V ) that will assist with developments of sections that follow.
If V is a group and α an element of M0(V ), then an element v of
V ∗ is said to be moved by α if vα 6= v, otherwise v in V ∗ is said to be
fixed by α. The set of all v in V ∗ moved (fixed) by α is denoted by M(α)
(by F (α)). Two elements α1 and α2 of M0(V ) are said to be disjoint if
M(α1) ∩ M(α2) = ∅.
There are certain elements of u(M0(V )) that will play a prominent
role in what follows. These are the cycles. If n ≥ 2 is an integer and
a1, . . . , an, n elements of V ∗ then [a1, . . . , an] will denote the element of
u(M0(V )) that fixes elements of V ∗\{a1, . . . , an} (of course 0 goes to 0),
takes ai to ai+1 for i = 1, . . . , n − 1, and takes an to a1. It is called a cycle
(more descriptively an n–cycle on V ). As a unit of M0(V ) it has order
n (ie. [a1, . . . , an]n = 1 and n is the smallest integer with this property).
It is clear that if α = [a1, . . . , an] then M(α) = {a1, . . . , an} and F (α) =
V ∗\{a1, . . . , an}. Disjoint cycles will be important. Here α1 = [b1, . . . , br]
and α2 = [c1, . . . , cs] are disjoint (see above) if {b1, . . . , br} ∩ {c1, . . . , cs}
242 The p–gen nature of M0(V ) (I)
is empty. Also k cycles β1, . . . , βk, of u(M0(V )) are mutually disjoint if
every two element subset of the set {β1, . . . , βk} is disjoint. The product
β1β2 . . . βk of mutually disjoint βi may be rearranged in any way as βi
and βj (i and j in {1, . . . , k}) commute. Also the order of this element of
u(M0(V )) is the l.c.m of the orders of the βi, i = 1, . . . , k.
The notion of disjoint cycles is important when dealing with a unit α of
M0(V ) of order p (p a prime). This is because α can be expressed uniquely
(apart from rearrangement) as a product α1α2 . . . αk of mutually disjoint
p–cycles αi, i = 1, . . . , k, on V . In this regard, there is a feature that
subgroups of V may have relative to α when slightly more information on
the αi is given. How things go here is now covered. We call a subgroup
H of V with Hα ⊆ H, α–invariant. In the situation where V has no non–
zero proper α–invariant subgroups we say α confuses proper subgroups
of V . The useful feature that certain α in u(M0(V )) of order p have is as
follows:-
Lemma 3.1. Suppose V is a group, p a prime and n ≥ 2 an integer. If
|V | > np and g1, . . . , gnp are np elements of V ∗, then
α = [g1, . . . , gp][gp+1, . . . , g2p] . . . [g(n−1)p+1, . . . , gnp]
is a unit of M0(V ) of order p, and if gsp = −gsp+1, for s = 1, . . . , n − 1,
then any α–invariant subgroup H of V such that H ∩ {g1, . . . , gnp} 6= ∅
necessarily contains {g1, . . . , gnp}.
Proof. The cycles used to define α are disjoint. Since M(α) in non–empty
(ie. α 6= 1) it follows from above that α has order p. If some gi is in H,
then gi is in {grp+1, . . . , g(r+1)p}, where r is one of {0, . . . , n − 1}. Now,
H[grp+1, . . . , g(r+1)p]k (k ≥ 1 an integer) is contained in H. Thus all of
{grp+1, . . . , g(r+1)p} is contained in H, and if r ≥ 1, then −grp+1 (ie. grp)
is in H. This in turn means, {g(r−1)p+1, . . . , grp} must be contained in H.
Continuing in this way we see {gkp+1, . . . , g(k+1)p} ⊆ H where 0 ≤ k ≤ r.
A reasonably similar argument shows {gkp+1, . . . , g(k+1)p} ⊆ H whenever
r ≤ k ≤ n − 1 so that 3.1 holds.
For a group V with α = α1α2 . . . αk (k ≥ 1 an integer) an element
of u(M0(V )) of order p, where the αi, i = 1, . . . , k, are mutually disjoint
p-cycles, we seek information on an automorphism µ of V such that
αµ = µα. One thing to note here is that F (α)µ = F (α). This is because
gµα = gαµ = gµ for all g in F (α). The other thing to note is that
αiµ = αj (i and j in {1, . . . , k}). This is not hard to prove. For example
if α1 = [g1, . . . , gp] we see g1µα = g1α1µ = g2µ etc., so that [g1µ, . . . , gpµ]
is a p–cycle under α and must be an αr with r in {1, . . . , k}. Thus we
S. D. Scott 243
certainly have, if i is in {1, . . . , k}, then M(αi)µ = M(αj) for some j in
{1, . . . , k}.
Automorphisms having the feature just displayed will, in a sense, be
quite important. Ensuring there are no such fixed–point–free automor-
phisms of prime order, allows under certain circumstances the conclusion
that N(α) = M0(V ). Indeed, when α confuses proper subgroups of V this
is the case.
Theorem 3.2. Let V be a group and α a unit of M0(V ) of order p
(p a prime) which confuses proper subgroups of V . If there does not
exist a fixed–point–free automorphism µ of V with prime order such that
µα = αµ, then N(α) = M0(V ) and M0(V ) is a p–gen nearring.
Proof. Set N = N(α). If it is shown that N = M0(V ), then the theorem
will follow. It should first be noted that, V is a minimal N–group (ie.
of type two). Here because α is a unit of finite order, V is readily seen
to be unitary. Also, if v 6= 0 is an element of V , then vN is a non–zero
subgroup of V with the property that vNα ⊆ vN . Because α confuses
proper subgroups of V, vN = V and V is minimal. The next step is to
show N is a non–ring. To obtain a contradiction assume N is a ring.
Since α 6= 0 and V 6= {0} there exists v 6= 0 in V such that vN = V .
This implies that, the N–group V is a ring module, and α is in Aut(V ).
Thus the subset of elements of V fixed by α (0 included), is a subgroup
H of V . Since Hα ⊆ H and, because α confuses proper subgroups of V ,
we see H = V or H = {0}. If H = V , then α = 1, contrary to α having
order p. This means α is a fixed–point–free automorphism of V of prime
order which commutes with α. This situation has been excluded and N
is necessarily a non–ring.
If AutN (V ) (the group of N–automorphisms of V ) is distinct from
{1} then, by [3, 4.52], AutN (V ) contains a fixed–point–free automorphism
(µ say) of V of prime order. Because µα = αµ, we have a contradiction.
Thus AutN (V ) = {1} and, by [3, 4.52], N must coincide with M0(V ).
4. Groups with few involutions
We now start deriving results telling us about the p–gen nature of
M0(V ) (V a group). To begin with it would be too much to take on
arbitrary V . Considerations given here will focus on the situation where
|η(V )| ≤ 1
3 |V |. In sections that follow we shall be allowing a higher
proportion of involutions to occur. The 1
3 |V | restriction will help us later
to deal with more involutions.
244 The p–gen nature of M0(V ) (I)
Lemma 4.1. Let p be a prime. If V is a group such that |η(V )| ≤ 1
3 |V |
and |V | > 2p, then M0(V ) is p–gen.
Proof. By [4] and [5], it can be assumed that p ≥ 5 (V is not elementary
two because |η(V )| ≤ 1
3 |V |). Let |V | − 1 = rp + s, where r ≥ 1 is an
integer and s an integer ≥ 0 and ≤ p − 1. Since |V | − 1 ≥ 2p, it follows
that r ≥ 2. We take
α = [g1, . . . , gp][gp+1, . . . , g2p] . . . [g(r−1)p+1, . . . , grp],
where the rp elements g1, . . . , grp of V ∗ are to be specified. First consid-
erations are to show how to take
gp, gp+1, g2p, g2p+1, . . . , g(r−1)p, g(r−1)p+1. (4.1)
The number of such elements is < 2r which is ≤ 2
5(|V | − 1) (from above
p ≥ 5 and rp ≤ |V |−1). Thus the number of such elements is for |V | > 2p
necessarily ≤ 2
3 |V | − 1 which is ≤ |τ(V )| because |τ(V )| = |V ∗| − |η(V )|
and |η(V )| ≤ 1
3 |V |. As τ(V ) can be partitioned into two element subsets
{h, −h} (h in τ(V )) it is possible to take the elements of (1) such that
gp = −gp+1, . . . , g(r−1)p = −g(r−1)p+1. Three steps specifying more of the
elements of {g1, . . . , grp} are now taken.
Step 1. When |η(V )| = 0 all elements of V ∗ are partitioned into two
element subsets {h, −h} (h in V ∗) so we can (and do) now take g1 =
−g2, . . . , gp−2 = −gp−1.
Step 2. When |η(V )| > 2 the elements of the sequence
g1, g2, . . . , gp−1, gp+2, . . . , g2p−1, g2p+2, . . . , g(r−1)p−1, g(r−1)p+2 (4.2)
are successively chosen as being from η(V ) (note that none of these
elements are in those of (1)). It will be shown all of η(V ) is used up. This
happens if
p + (r − 2)(p − 2) ≥
1
3
|V |, (4.3)
because 1
3 |V | ≥ |η(V )|. However, r is equal to (|V | − 1)/p − s/p and, since
s/p ≤ (p − 1)/p, it follows that r ≥ |V |/p − 1. Thus (3) holds if
p + (|V |/p − 3)(p − 2) ≥
1
3
|V |
or
p − 3p + 6 ≥ |V |(
1
3
− (p − 2)/p).
S. D. Scott 245
Because p ≥ 5, 1
3 − (p − 2)/p is negative, and dividing by 1
3 − (p − 2)/p,
we see (3) holds if
(−2p + 6)/(
1
3
− (p − 2)/p) ≤ |V |
or
|V | ≥ (6p2 − 18)/(2p − 6).
This last inequality holds if |V | ≥ 3p. It remains to show step 2. holds
when |V | < 3p. Here r = 2 and the number of elements of (2) is p so that
because |η(V )| ≤ 1
3 |V | < p all of η(V ) is used in successively defining
elements of (2).
Step 3. In step 1. and 2. the elements of the set {g(r−1)p+3, . . . , grp} did
not enter into consideration. Also they did not come into those that occur
in (1), so we are still free to adjust the elements of this set. This is done
in the following way. This set has p − 2 elements while F (α) has order
s where s ≤ p − 1. The elements of these two sets are in τ(V ) (all of
η(V ) was used up in step two) so by 2.3 we may adjust the g(r−1)p+3 to
grp in such a way that F (α) does not contain all non-zero elements of a
non-zero subgroup of V .
The α given by steps 1. to 3. will now be shown to confuse proper
subgroups of V . By step 3. a non-zero α–invariant subgroup K of V
cannot be such that K∗ ⊆ F (α). This means K ∩ M(α) 6= ∅ and from the
definition of α and 3.1, M(α) ⊆ K so that simple order arguments ensure
K = V . Thus α confuses proper subgroups of V . By 3.2 the lemma will
follow if it is shown there is no fixed–point–free automorphism µ of V of
prime order q say so that αµ = µα. There are two cases involved here.
Case 1. This is where |η(V )| = 0. Here we have from the defini-
tion of α and step 1. that either [g1, . . . , gp] is the only p–cycle α1 in
the definition of α with a single element (viz. gp) without an addi-
tive inverse in M(α1) or [g1, . . . , gp] and [g(r−1)p+1, . . . , grp] are the only
two such p–cycles. Because α1µ is another such p–cycle (see §.3) we
have either [g1µ, . . . , gpµ] = [g1, . . . , gp] and the uniqueness of gp means
gpµ = gp (a contradiction) or [g1, . . . , gp]µ = [g(r−1)p+1, . . . , grp] and
[g(r−1)p+1, . . . , grp]µ = [g1, . . . , gp]. This last situation (the first is now
excluded) gives [g1, . . . , gp]µ2 = [g1, . . . , gp] so that 2 divides q and q = 2,
meaning vµ = −v for all v in V (see 2.9). However g1µ = −g1 (in
M(α1). As g1µ would, in this situation, be in {g(r−1)p+1, . . . , grp} we have
a contradiction here also.
Case 2. This is where |η(V )| > 0. Here [g1, . . . , gp] (α1 say) is the only
p–cycle in the definition of α having elements of M(α1) in η(V ) or the only
246 The p–gen nature of M0(V ) (I)
one with M(α1) containing p−1 elements of η(V ). It is not difficult to see
this from the definition of α and step 2. This fact implies M(α1)µ = M(α1)
so that q divides p and q = p. Furthermore M(α1) is an orbit of µ so that
g1µk = gp (some integer k ≥ 1). A contradiction has been reached (g1 is
in η(V ) while gp is in τ(V )) and 4.1 is therefore proved.
5. Nilpotent with a quantity of involutions
In the last section the order of η(V ) was quite restricted. It was
assumed that |η(V )| ≤ 1
3 |V |. Here we raise this upper bound to 9
16 |V | − 1.
This figure is not arbitrarily chosen. It does in fact fit in quite well with
what is known about groups with |η(V )| > 1
2 |V | − 1 (see §.2 eg. 2.8).
Groups with |η(V )| > 1
2 |V | − 1 are necessarily often nilpotent. This is
another assumption that will apply to groups studied here. What we
prove is as follows.
Lemma 5.1. Let p be a prime. If V is a nilpotent group with |η(V )| ≤
9
16 |V | − 1 and |V | > 3p, then M0(V ) is p-gen.
Proof. By [4] and [5] it can be assumed p ≥ 5 (elementary two groups
are excluded because |η(V )| ≤ 9
16 |V | − 1). Now, if V is a group of odd
order, then the result follows by 4.1. It can be assumed V has even order.
If V is not a 2–group, then because V is nilpotent, all of η(V ) is contained
in a Sylow 2– subgroup V1 of V . This means
|η(V )| ≤ |V1| ≤ (|V1|/|V |)|V | ≤
1
3
|V |
and, by 4.1, M0(V ) is p–gen. Thus it can be assumed p ≥ 5 and V is a
2–group.
Let |V |−1 = rp+s, where r is an integer ≥ 1 and s is in {0, . . . , p−1}.
Because |V | − 1 ≥ 3p, we see r ≥ 3. Define
α = [g1, . . . , gp][gp+1, . . . , g2p] . . . [g(r−1)p+1, . . . , grp],
where the rp elements g1, . . . , grp of V ∗, are to be specified. First we
specify some of the elements of the sequence
g1, . . . , gp−1, gp+2, . . . , g2p−1, g2p+2, . . . , g(r−1)p−1, g(r−1)p+2, . . . , grp.
(5.4)
These are chosen successively as elements of η(V ) and all of η(V ) is used
up, if the number of elements of (4) is not less than |η(V )|. Thus all of
S. D. Scott 247
η(V ) is used up if
(p − 1)2 + (r − 2)(p − 2) ≥
9
16
|V | − 1.
Because
r = (|V | − 1)/p − s/p ≥ (|V | − 1)/p − (p − 1)/p = |V |/p − 1,
we see all of η(V ) is used up if
(p − 1)2 + (|V |/p − 3)(p − 2) ≥
9
16
|V | − 1
or
1 + (p − 1)2 − 3(p − 2) ≥ (
9
16
− (p − 2)/p)|V |.
However, for p ≥ 5, 9
16 − (p − 2)/p is negative and all of η(V ) is used up
in successively defining elements of the sequence (4) if
|V | ≥ (−p + 5)/(
9
16
− (p − 2)/p)
or
|V | ≥ (16p2 − 80p)/(7p − 32)
which is the case because
(16p2 − 80p)/(7p − 32) ≤ 3p
(clearly 16p − 80 ≤ 21p − 96 for p ≥ 5) while |V | > 3p. Thus, some
elements of (4) are defined and all the remaining g’s come from τ(V ).
From the nature of τ(V ) it is now possible to choose all the remaining
g’s such that
gp, gp+1, g2p, g2p+1, . . . , g(r−1)p, g(r−1)p+1,
are chosen with gp = −gp+1 up to g(r−1)p = −g(r−1)p+1. So in this manner
α is defined.
Clearly, V ∗\{g1, . . . , grp} has no elements of order two and, a proper
subgroup K of V , is such that M(α) ∩ K 6= ∅. Certainly if, Kα ⊆ K,
then 3.1 ensures M(α) ⊆ K and, simple order arguments mean K = V .
Thus α confuses proper subgroups of V . Also, from the manner in which
α is defined, it follows that either [g1, . . . , gp] is the unique p–cycle con-
taining p − 1 elements of order two or [g1, . . . , gp] and [g(r−1)p+1, . . . , grp]
are the unique pair of p–cycles (in the definition of α) containing p − 1
248 The p–gen nature of M0(V ) (I)
elements of order two. We conclude that a fixed–point–free automor-
phism µ of V of prime order q, such that αµ = µα, must satisfy either
{g1, . . . , gp}µ = {g1, . . . , gp} or {g1, . . . , gp}µ = {g(r−1)p+1, . . . , grp} and
{g(r−1)p+1, . . . , grp}µ = {g1, . . . , gp}. The second case cannot occur, oth-
erwise q = 2 and µ fixes all elements of order two (see 2.9). In the first
case, p = q, {g1, . . . , gp} is an orbit under µ and because g1 is in η(V ) and
gp in τ(V ), we have a contradiction, so that 3.2 applies and 5.1 is proved.
6. Dealing with V where |η(V )| ≤ 1
2
|V |
Although in this section we allow V to have more involutions than was
the case in four, our groups may be non–nilpotent and the 9
16 |V |−1 bound
of section five is, for our present purposes too high. A bound that helps
us go quite a long way to establishing 1.1 is 1
2 |V |. Thus the result that
we aim at here, is showing V with |V | > 3p and |η(V )| ≤ 1
2 |V | has p–gen
M0(V ) (p any prime). However, the ultimate goal of sections four to nine
(see 9.1) is in fact to show that the restriction on |η(V )| is unnecessary.
All that 9.1 requires is that V is not elementary two. Although this is
eventually arrived at, intermediate results must be covered. Sometimes,
for specific V and p, it is not that difficult to establish the p–gen nature
of M0(V ). For example, this can be done for V with |V | = 16 and p = 5
without too much trouble. Before stating and proving this sections main
theorem (announced above) this very special case is attended to.
Proposition 6.1. If V (not elementary two) is a group of order sixteen,
then M0(V ) is 5–gen.
Proof. Let α be the product
[g1, . . . , g5][g6, . . . , g10][g11, . . . , g15]
of three disjoint 5–cycles on V where g1, . . . , g15 are the 15 elements of
V ∗ chosen in such a way that g1, g2, . . . , g15 are successively taken from
η(V ).
Now V has no proper α–invariant subgroup H. This is because if H
existed, H∗ would have five or ten elements contrary to the fact that |H|
divides |V |. Also, out of the three 5–cycles defining α there is either a
unique one containing elements of η(V ) and τ(V ) or {g1, . . . , g5} is η(V )
and {g6, . . . , g15} is τ(V ). This is true since η(V ) 6= V ∗ and |η(V )| being
of odd order must be five. In the case of a unique 5–cycle [h1, . . . , h5] with
elements in η(V ) and τ(V ) it is clear that a fixed–point–free automorphism
µ of prime order q must be such that {h1, . . . , h5}µ = {h1, . . . , h5}, q = 5
and hiµ
r = hj (r ≥ 1 an integer — hi in η(V ) and hj in τ(V )). This
S. D. Scott 249
cannot happen so η(V ) = {g1, . . . , g5} = {g1, . . . , g5}µ and again q =
5. Clearly{g6, . . . , g10}µ 6= {g11, . . . , g15} (otherwise {g11, . . . , g15}µ =
{g6, . . . , g10} and q is 2). Thus {g6, . . . , g10}µ = {g6, . . . , g10}, q = 5 and if
g7 (in τ(V )) where to be taken as −g6 we would have g6µr = −g6 (r ≥ 1
coprime to 5) contrary to the fact two does not divide five. Thus no such
µ exists and α (of order five) is such that N(α) = M0(V ) (ie. 3.2 applies).
The proposition has been proved.
The main result of this section is the following.
Theorem 6.2. Let p be a prime. If V is a group with |η(V )| ≤ 1
2 |V | and
|V | > 3p, then M0(V ) is p–gen.
Proof. By [4] and [5] we may assume p ≥ 5. Also by 5.1, it can be assumed
V is non–nilpotent (clearly 1
2 |V | ≤ 9
16 |V | − 1 for |V | > 3p when p ≥ 5).
The situation where V has odd order can be excluded by 4.1. Furthermore,
it can be assumed that, all elements of η(V ) do not additively centralize
each other. This is because, if they did, then i(V ) is an elementary two
proper subgroup of V . Because V is non–nilpotent, this normal subgroup
cannot have index ≤ 2 (otherwise V is a 2–group). Hence, in this situation
|η(V )| ≤ 1
3 |V | and the result follows by 4.1. Thus the assumption that all
elements of η(V ) do not additively centralize each other can (and will)
be made.
Let |V | − 1 = rp + s, where r is an integer ≥ 0 and s in {0, . . . , p − 1}.
Because |V | − 1 ≥ 3p, we see r ≥ 3. Define
α = [g1, . . . , gp][gp+1, . . . , g2p] . . . [g(r−1)p+1, . . . , grp]
where the rp elements g1, . . . , grp of V ∗ are to be specified. Let gp−1
and gp be in η(V ) and not additively centralize each other. Now, by §.2,
〈gp, gp−1〉 = Dn where n is an integer ≥ 3. Let Cn be the normal cyclic
subgroup of Dn of order n. Let d be a generator of Cn. Clearly d 6= −d and
we may take g2p as d and g2p+1 as −d. Now Dn\Cn consists of involutions
(λ(Dn) = Cn) and therefore contains an element c of η(V ), distinct from
gp and gp−1. Certainly 〈d, c〉 = Dn. Take gp+1 as c. Out of the elements
of the sequence
g1, . . . , gp−1, gp+2, . . . , g2p−1, g2p+2, . . . , g(r−1)p−1, g(r−1)p+2, (6.5)
only gp−1 is accounted for. This sequence has p + (r − 2)(p − 2) elements,
and the number of unaccounted for ones is p−1+(r −1)(p−2). Let these
unaccounted for g’s be taken successively from η(V )\{a, b, c} (a = gp−1,
and b = gp). All the remainder of η(V ) has been used up in defining these
250 The p–gen nature of M0(V ) (I)
g’s if
p − 1 + (r − 2)(p − 2) ≥ |η(V )| − 3.
We want to show this holds when |V | > 3p. This is certainly the case for
r = 3 because here |V | ≤ 4p and 2p ≥ 1
2 |V | ≥ |η(V )|. Now for |V | ≥ 4p
all of η(V ) has been used up defining g’s if
p − 1 + (r − 2)(p − 2) ≥
1
2
|V | − 3.
Because
r = (|V | − 1)/p − s/p ≥ (|V | − 1)/p − (p − 1)/p = |V |/p − 1,
we see all of η(V ) has been used up in so far defining g’s if
p + 2 − 3(p − 2) ≥ |V |(
1
2
− (p − 2)/p).
However, 1
2 − (p − 2)/p is negative, since p ≥ 5. Thus all η(V ) has been
used up in, so far defining g’s if
|V | ≥ (−2p + 8)/(
1
2
− (p − 2)/p) = [2p(−2p + 8)]/(−p + 4) = 4p
which is certainly the case.
The remainder of the g’s must be taken from τ(V )\{d, −d}. these
elements of V ∗, can be partitioned into two element subsets consisting
of an element and its additive inverse. Thus the remaining g’s (in the
definition of α) can be chosen in such a way that g2p = −g2p+1, up to
g(r−1)p = −g(r−1)p+1 (vacuous if r = 3). All g’s have been chosen from
V ∗ and α is apart from the minor adjustment that follows, defined.
The essential features of the above preliminary definition of α are that
gp−1 and gp are in η(V ), 〈gp−1, gp〉 = Dn, 〈gp+1, g2p〉 = Dn, g2p = −g2p+1
is in τ(V ), the elements g3p, . . . , g(r−1)p are in τ(V ) and g3p = −g3p+1
up to g(r−1)p = −g(r−1)p+1, while all of η(V ) is in {g1, . . . , g(r−1)p+2}
(this last statement follows because all of η(V )\{a, b, c} was used up in
defining S\{gp−1} — here S is the elements of (5)). These features of α are
clearly unchanged if we rearrange the elements of {g(r−1)p+3, . . . , grp} ∪
(V ∗\{g1, . . . , grp}) (= S1 say). The first of these sets has p − 1 elements
and the second has ≤ p − 1 elements (viz. s elements). Clearly S1 is a
disjoint union of the subsets indicated and S1 ⊆ τ(V ). It is now possible,
by 2.3, to write S1 as a disjoint union of sets S2 and S3, where |S2| ≤ p−2
and |S3| = s in such a way that, S3 does not contain the non–zero elements
of a non–zero subgroup of V . Rename the g’s so that they remain the
S. D. Scott 251
same, apart from the fact that g(r−1)p+3, . . . , grp, are now taken as the
elements of S2. In this manner we obtain the final definition of α. All the
properties of α listed above hold true, but now we have the additional
fact that V ∗\M(α) (ie. F (α)) is S3 and does not contain the non–zero
elements of a non–zero subgroup of V .
It will now be shown that α confuses proper subgroups of V . Firstly, a
non–zero subgroup H of V cannot be such that M(α)∩H = ∅. If Hα ⊆ H
and an element of {gp+1, . . . , grp} is in H, then {gp+1, . . . , grp} ⊆ H. This
follows by 3.1, because g2p = −g2p+1 up to g(r−1)p = −g(r−1)p+1. However,
in this case, g2p and gp+1 are in H, implying 〈gp+1, g2p〉 ⊆ H and gp is in H.
This means M(α) ⊆ H. On the other hand, if an element of {g1, . . . , gp}
is in H, then 〈gp, gp−1〉 is contained in H and gp+1 is in H.. Thus, in
this case also, M(α) ⊆ H. Simple order arguments mean H = V and α
confuses proper subgroups of V . Now V is non–nilpotent and by 2.10 has
no fixed–point–free automorphisms of prime order. It follows from 3.2
that N(α) = M0(V ) and because α is a disjoint product of p–cycles on
V , M0(V ) is p–gen. 6.2 has been proved.
7. When V is not a D(A)
Now that theorem 6.2 has been proved it is time to move towards
looking at V with |η(V )| > 1
2 |V |. The idea is to extend 6.2 to any V
not elementary two. The situation here is in one regard more complex
and in another simpler. The complexity comes from the fact that, in
constructing an α of order p in u(M0(V )) with N(α) = M0(V ), there are
more elements of order two that must be lost within the disjoint p–cycles
into which α decomposes. The other side of the coin is that V has limited
structure (see 2.8). In some sense the difficult issue is what to do about
groups of type (II), (III) and (IV). All these are 2-groups and many have
|η(V )| coming within the 9
16 |V |−1 bound used in 5.1. Groups of type (II),
(III) and (IV) certainly have |η(V )| ≤ 5
8 |V | − 1 so a fairly real problem
is handling those η(V ) of order > 9
16 |V | − 1 and ≤ 5
8 |V | − 1. This is
accomplished in this section. What is proved here is the following.
Lemma 7.1. Let p ≥ 5 be a prime. If the group V is not a D(A) and
has order > 3p, then M0(V ) is p–gen.
Proof. By 6.2 we may assume |η(V )| > 1
2 |V |. Since V is not a D(A) it
follows from 2.8 that V is a 2-group and by 2.5, 2.6 and 2.7 |η(V )| ≤
5
8 |V | − 1.
252 The p–gen nature of M0(V ) (I)
Let |V |−1 = rp+s, where r is an integer ≥ 1 and s is in {0, . . . , p−1}.
Because |V | − 1 ≥ 3p, we see r ≥ 3. Define
α = [g1, . . . , gp][gp+1, . . . , g2p] . . . [g(r−1)p+1, . . . , grp],
where the rp elements of V ∗ are to be specified. First we specify some of
the elements of the sequence
g1, . . . , gp−1, gp+2, . . . , g2p−1, g2p+2, . . . , g(r−1)p−1, g(r−1)p+2, . . . , grp.
(7.6)
These are chosen successively as elements of η(V ) and all of η(V ) is used
up, if the number of elements of (6) is not less than |η(V )|. Thus all of
η(V ) is used up if
(p − 1)2 + (r − 2)(p − 2) ≥
5
8
|V | − 1.
However, p does not divide |V | (V is a 2-group) so s ≤ p − 2 and
r = (|V | − 1)/p − s/p ≥ (|V | − 1)/p − (p − 2)/p = |V |/p − (p − 1)/p
so that all of η(V ) is used up if
1 + 2p − 2 + (1 − 3p)(p − 2)/p ≥ (
5
8
− (p − 2)/p)|V |.
This holds if
8p + 16p2 − 16p + 8(1 − 3p)(p − 2) ≥ (−3p + 16)|V |
or
8p + 16p2 − 16p − 16 + 56p − 24p2 ≥ (−3p + 16)|V |.
Thus all of η(V ) is used up if
(−8p2 + 48p − 16) ≥ (−3p + 16)|V |.
For p = 5 this reduces to 24 ≥ |V | which is satisfied because |V | is 16,
32„ etc., and |V | = 16 can, by 6.1, be excluded. For p ≥ 7, we see this
inequality reduces to
(8p2 − 48p + 16)/(3p − 16) ≤ |V |
since −3p + 16 is negative and is satisfied if 8p2 − 48p + 16 ≤ 3p(3p − 16)
or 8p2 − 48p + 16 ≤ 9p2 − 48p (certainly the case since |V | ≥ 3p and
p ≥ 7). In this way some of the elements of (6) are defined (all of η(V )
S. D. Scott 253
has been used up) and the remaining g’s must come from τ(V ). From the
nature of τ(V ) it is now possible to choose the remaining g’s such that
gp = −gp+1 up to g(r−1)p = −g(r−1)p+1. So in this manner α is defined.
Clearly, V ∗\{g1, . . . , grp} has no elements of order two and, a proper
subgroup K of V , is such that M(α) ∩ K 6= ∅. Certainly if Kα ⊆ K,
then 3.1 ensures M(α) ⊆ K and simple order arguments mean K = V .
Thus α confuses proper subgroups of V . Also, from the manner in which
α is defined, it follows that either [g1, . . . , gp] is the unique p–cycle con-
taining p − 1 elements of order two or [g1, . . . , gp] and [g(r−1)p+1, . . . , grp]
are the unique pair of p–cycles (in the definition of α) containing p − 1
elements of order two. We conclude that, a fixed–point–free automor-
phism µ of V of prime order q, such that αµ = µα, must satisfy either
{g1, . . . , gp}µ = {g1, . . . , gp} or {g1, . . . , gp}µ = {g(r−1)p+1, . . . , grp} and
{g(r−1)p+1, . . . , grp}µ = {g1, . . . , gp}. The second case cannot occur, oth-
erwise q = 2 and µ fixes all elements of order two (see 2.9). In the first
case, it follows g1µk = gp for some integer k ≥ 1 (here q = p) and, because
g1 is in η(V ) and gp in τ(V ), we have a contradiction. By 3.2, 7.1 stands
proved.
8. V a D(A) with p dividing |V |
In this section we continue to deal with groups having |η(V )| > 1
2 |V |.
The situation where V is of type (II), (III) or (VI) has been dealt with in 7.1.
By 2.8 there remain those groups of the form D(A) (A abelian). Because
this paper is not concerned with elementary two groups the possibility of
such an A (D(A) is elementary two precisely when A is elementary two
and in all other cases D(A) is non-abelian and λ(D(A)) = A — see §.2)
is excluded. However this section makes another assumption also. Here
we assume that p divides |V | (ie. p divides |D(A)|). In the next the full
cycle of dealing with all V having |V | > 3p is completed by showing when
p does not divide the order of a D(A) having |D(A)| > 3p, M0(D(A)) is
p–gen.
Lemma 8.1. Let p be a prime and V a group of type (I) (not elementary
two) which is such that p divides |V |. If |V | > 3p, then M0(V ) is p–gen.
Proof. The case where p = 2 or 3 respectively follows from [4] or [5].
Thus it can be assumed that p ≥ 5. Also, because |V | is even and p divides
|V |, we have |V | ≥ 4p. Furthermore |λ(V )| can be assumed to be even.
This is because otherwise η(λ(V )) is empty and |η(V )| = 1
2 |V | (see §.2),
so that the result would follow by 6.2. Thus |λ(V )| can be assumed even
and λ(V ) (which contains τ(V )) contains an element of order two and
254 The p–gen nature of M0(V ) (I)
order p. Thus λ(V ) contains a subgroup of the form C2 ⊕ Cp and has at
least p − 1 elements of composite order.
Now λ(V ) is an abelian group containing τ(V ) (see §.2). Thus λ(V )
has a subgroup A of index p and all elements of λ(V )\A have order
divisible by p. Thus |i(λ(V ))| ≤ 1
p
|λ(V )| and, since |λ(V )| = 1
2 |V | and all
elements of V \λ(V ) are in η(V ) (see §.2), it follows that
|i(V )| ≤
1
2p
|V | +
1
2
|V | = [(p + 1)|V |]/2p,
which means
|τ(V )| ≥ [1 − (p + 1)/2p]|V | = [(p − 1)|V |]/2p.
From what has been proved we can therefore assume p ≥ 5, |V | ≥ 4p, V
has at least p−1 elements of composite order and |τ(V )| ≥ [(p−1)|V |]/2p.
Take |V | − 1 = rp + p − 1, where r is an integer ≥ 3 (remember
|V | ≥ 4p and p divides |V |). Define α as a product
[g1, . . . , gp][gp+1, . . . , g2p] . . . [g(r−1)p+1, . . . , grp],
where the rp elements g1, . . . , grp, of V ∗ are to be specified. We do this
by first specifying the elements of V ∗\{g1, . . . , grp} (= F (α)) and then
the elements of
{gp, gp+1, g2p, g2p+1, . . . , g(r−1)p, g(r−1)p+1} (8.7)
and finally allowing the remaining g’s to be taken arbitrarily as the
remaining elements of V ∗.
First τ(V )) is partitioned into two element subsets of the form {h, −h},
with h in τ(V ). This means the elements of composite order are so
partitioned. Take (p − 1)/2 such pairs (p − 1 elements forming a set S —
possible from above) and the elements of F (α) (order p − 1) are those
of S. Thus τ(V )\S is partitioned into two element subsets (as indicated)
and the elements of (7) can be chosen so that
gp = −gp+1, g2p = −g2p+1. . . . , g(r−1)p = −g(r−1)p+1,
provided |τ(V )\S| ≥ (r − 1)2 (ie. |τ(V )| ≥ (r − 1)2 + p − 1). Thus the
elements of (7) can be chosen as indicated (see above) if
[(p − 1)/2p]|V | ≥ (r − 1)2 + p − 1.
S. D. Scott 255
However,
r = (|V | − 1)/p − (p − 1)/p = |V |/p − 1,
and the elements of (7) can be chosen as indicated if
[(p − 1)/2p]|V | ≥ (|V |/p − 2).2 + p − 1
or, on multiplying by 2p
(p − 1)|V | ≥ 4|V | − 8p + 2p2 − 2p = 4|V | − 10p + 2p2.
Thus, the elements of (7) can be chosen as indicated provided, (p−5)|V | ≥
2p2 − 10p. For p = 5 both sides are zero and all is well. If p > 5, then
since |V | ≥ 3p this is satisfied if,
(p − 5)3p = 3p2 − 15p ≥ 2p2 − 10p
or p ≥ 5, which is certainly true. Thus the elements of (7) can be chosen
as indicated, while the remaining g’s are simply taken as the rest of the
elements of V ∗, namely those of the set
V ∗\(F (α) ∪ {gp, gp+1, g2p, g2p+1, . . . , g(r−1)p, g(r−1)p+1}).
Now suppose H is a non-zero α–invariant subgroup of V . If F (α) ⊇ H∗,
then F (α) would contain a non-zero element of prime order. Since this
cannot happen (the elements of F (α) have composite order), we see
M(α) ∩ H 6= ∅. By 3.1, M(α) ⊆ H and simple order arguments ensure
H = V .
Suppose V has a fixed–point–free automorphism µ of prime order (q
say), such that αµ = µα. Now λ(V ) is by §.2, characteristic and, it readily
follows that λ(V )\{0} and V \λ(V ) are partioned under µ, into q element
orbits. However, |V \λ(V )| = |λ(V )| (see §.2) and |λ(V )\{0}| = |λ(V )|−1.
Since q cannot divide both |λ(V )| and |λ(V )| − 1, no such µ can exist
and an application of 3.2 completes the proof.
9. A substantial Theorem
Although the goal of this paper is the proof of 1.1, virtually all previous
material has been developed with a view to proving an intermediate result.
This is that for a prime p and group V (not elementary two) with |V | > 3p,
M0(V ) is p–gen. The theorem alluded to is proved in this section. Allowing
as it does, considerations to focus on groups where |V | ≤ 3p it is essential
to the proof of 1.1. There is only one small detail of all that has preceeded
that is very slightly different. All material so far has been developed for
256 The p–gen nature of M0(V ) (I)
V having |V | > 3p with one small exception. This is 4.1 where the weaker
assumption of |V | being > 2p is made. It seems this slightly more inclusive
detail for groups with |η(V )| ≤ 1
3 |V | is also needed for proving 1.1. This
however does not concern us here. The theorem we prove is the following.
Theorem 9.1. Let p be a prime and V a group (not elementary two)
with |V | > 3p, then M0(V ) is p-gen.
Proof. By [4], [5] and 7.1 we may (and will) assume p ≥ 5 and V is a
D(A). Also by 8.1 it can (and will) be assumed that p does not divide
|V |. One further assumption is also made. |λ(V )| is taken as even since if
|λ(V )| is odd, V \λ(V ) = η(V ) and, by 2.4, |η(V )| = 1
2 |V |.
An Initial Step. It will be proved here that, the number of elements
of λ(V ) of composite order is ≥ 1
2 |λ(V )| − 1. Now λ(V ) is an abelian
group with |λ(V )| even. Also, it is easy to see that it is not elementary
two. If λ(V ) is a 2–group, then because i(λ(V )) is a subgroup of λ(V ) of
index at least two and elements of λ(V )\i(λ(V )) have composite order,
this step follows. Otherwise, λ(V ) has a maximal subgroup H1 of odd
order and a maximal non–zero 2–subgroup H2, and λ(V ) = H1 ⊕H2. Now
elements a + b, a in H∗
1 and b in H∗
2 , have composite order. The number
x of such elements is |H1||H2| − |H1| − |H2| + 1. So that if |H2| = 2 ,
x = |λ(V )| − |H1| − 1 = 1
2 |λ(V )| − 1 and the result follows. Thus it can
be assumed that |H2| ≥ 4 and if |H1| > 3,
x ≥ |λ(V )| −
1
4
|λ(V )| −
1
4
|λ(V )| + 1 ≥
1
2
|λ(V )| − 1.
Hence we may assume |H1| = 3. Now the subgroup i(λ(V )) of λ(V ) has
index ≥ 3 in λ(V ), and further |i(λ(V ))| ≤ 1
3 |λ(V )| and |τ(λ(V ))| ≥
2
3 |λ(V )| > 1
2 |λ(V )| so that |τ(λ(V ))| ≥ 1
2 |λ(V )| + 1. However, every non–
zero element of τ(λ(V )) has order 2k13k2 (k1, k2 integers ≥ 0), and only
two have order three (none have order two). Thus the number of elements
of λ(V ) of composite order is ≥ 1
2 |λ(V )| + 1 − 2 = 1
2 |λ(V )| − 1 and our
initial step is proved.
From what has been proved above it can (and will) be assumed the
p ≥ 5, V is of type (I) (not elementary two), p does not divide |V |,
|λ(V )| is even and the number of elements of λ(V ) of composite order is
≥ 1
2 |λ(V )| − 1.
Now |V \λ(V )| = 1
2 |V | (see §.2) and |V \λ(V )| > 3
2p. Let |V \λ(V )| =
r1p + s1, where r1 ≥ 1 is an integer and s1 is an integer in {0, . . . , p − 1}.
Define
α1 = [v1, . . . , vp][vp+1, . . . , v2p] . . . [v(r1−1)p+1, . . . , vr1p],
S. D. Scott 257
where v1, . . . , vr1p are any r1p elements of V \λ(V ). Certainly there is at
least one p–cycle in the definition of α1 since r1 ≥ 1. Also α1 is a product
of disjoint p–cycles and has order p. Now let a1, . . . , as1
, be the remaining
elements of (V \λ(V ))\{v1, . . . , vr1p}. Certainly s1 ≤ p − 1 and, because
p does not divide |V |, s1 ≥ 1 and a1 exists. Let b1, . . . , bp−s1
, be p − s1
elements of λ(V )∗ (the element b1 of {b1, . . . , bp−s1
} is to be chosen). Take
α2 = [a1, . . . , as1
, b1, . . . , bp−s1
]. Clearly α2 is disjoint from α1.
In order to define α3 it must be shown that
λ(V )∗\{b1, . . . , bp−s1
}, (9.8)
has at least p elements. If |V | ≥ 4p, then 1
2 |V | ≥ 2p and |λ(V )∗| ≥
2p − 1 so that, because {b1, . . . , bp−s1
} has ≤ p − 1 elements,
|λ(V )∗\{b1, . . . , bp−s1
}| ≥ p. Now |V | > 3p, so to prove the set (8) has
≥ p elements we can assume, 4p > |V | > 3p. Thus 2p > 1
2 |V | > 3
2p and
|V \λ(V )| (= 1
2 |V |) is < 2p and > 3
2p (thus ≥ p + (p + 1)/2), implying
r1 = 1 and s1 ≥ (p + 1)/2 so that p − s1 ≤ (p − 1)/2. Since |λ(V )∗|
coincides with 1
2 |V | − 1, and is < 2p − 1 and greater than p + (p + 1)/2 − 1
(= p + (p − 1)/2), we see that
|λ(V )∗\{b1, . . . , bp−s1
}| ≥ p + (p − 1)/2 − (p − 1)/2 = p,
and the required fact about (8) is established. Let
|λ(V )∗\{b1, . . . , bp−s1
}| = r2p + s2, where r2 is an integer ≥ 1
and s2 is in {0, . . . , p − 1}.
Now take g1, . . . , gr2p, as r2p elements of the set (8) and define
α3 = [g1, . . . , gp][gp+1, . . . , g2p] . . . [g(r2−1)p+1, . . . , gr2p], (9.9)
where some of the g’s, b1 and the s2 elements of
(λ(V )∗\{b1, . . . , bp−s1
})\{g1, . . . , gr2p} (= S2 say),
are to be specified (the remaining g’s can be taken arbitrarily as the
elements of {g1, . . . , gr2p} that remain). In this way α3 will be specified as
an element of u(M0(V )) of order p (r2 ≥ 1 and the r2 p–cycles of (9) are
disjoint) disjoint from α1 and α2. Concerning the g’s (and b1) we wish to
specify, these elements are gp, gp+1, g2p, g2p+1, . . . , gr2p, b1. Thus if r2 = 1,
this simply reduces to specifying gp and b1.
It follows that the set of elements to be specifed is the disjoint union
S2 ∪ {gp, gp+1, g2p, g2p+1, . . . , gr2p, b1} (= S3 say),
258 The p–gen nature of M0(V ) (I)
which has |S2| + r22 elements. The set of elements chosen arbitrarily as
the remainder of λ(V )∗\{b1, . . . , bp−s1
} is necessarily
{g1, . . . , gr2p}\{gp, gp+1, g2p, g2p+1, . . . , gr2p, b1}
and has (p−1)+(r2−1)(p−2) elements which together with {b2, . . . , bp−s1
}
constitute the remainder set S4 of the elements of λ(V )∗. This remainder
set of λ(V )∗ has (p − 1) + (r2 − 1)(p − 2) + (p − s1) − 1 (= |S4|), elements.
Clearly S3 and S4 are disjoint with union λ(V )∗ and if it is shown that
|S3| ≤ |S4|, then |S3| ≤ 1
2 |λ(V )∗| ≤ 1
2 |λ(V )|− 1
2 , showing |S3| ≤ 1
2 |λ(V )|−1
(|S3| is an integer) and meaning, by our assumptions, all elements of S3
can be chosen from the set C of elements of λ(V ) with composite order.
In order to prove the last statement of the above paragraph, it must
be shown that
|S2| + r22 ≤ (p − 1) + (r2 − 1)(p − 2) + p − s1 − 1. (9.10)
Now p − s1 ≥ 1 and s2 ≤ p − 1 so, if s2 ≤ p − 3, (10) will hold, in this
case, if it is shown that p − 3 + r22 ≤ p − 1 + (r2 − 1)(p − 2). This holds
if 0 ≤ 2 + (r2 − 1)p − r22 + 2 − r22, which holds if 0 ≤ 4 + r2p − r24 − p
or p − 4 ≤ r2(p − 4), which is certainly so since r2 ≥ 1 and p ≥ 5. In the
case where s2 ≤ p − 3, (10) must therefore hold. Let us assume s2 = p − 2.
In this situation |λ(V )∗| = p − 2 + r2p + p − s1 and if p − s1 = 1 (p − s1
is non–zero), then |λ(V )∗| = r2p + p − 1 and |λ(V )| − 1 = r2p + p − 1
showing |λ(V )| = (r2 + 1)p, which is a contradiction (p does not divide
|V |). It follows in the case s2 = p − 2, we have p − s1 ≥ 2. Thus (10)
holds, in this case, if p − 2 + r22 ≤ p − 1 + (r2 − 1)(p − 2) + 1 or
p − 3 + r22 ≤ p − 1 + (r2 − 1)(p − 2), which was seen to hold in arguments
above. In the case where s2 = p − 2, (10) must therefore hold. It can
therefore be assumed that s2 = p − 1. If p − s1 = 1, then |λ(V )| − 1 =
(r2 + 1)p and |V \λ(V )| = (r2 + 1)p + 1, showing s1 = 1 and p − s1 ≥ 4
(a contradiction). If p − s1 = 2, then |λ(V )| − 1 = p(r2 + 1) + 1 and
|V \λ(V )| = (r2 +1)p+2, showing s1 = 2 and p−s1 ≥ 3 (a contradiction).
Thus p−s1 ≥ 3 and (10) holds if p−1+r22 ≤ p−1+(r2 −1)(p−2)+2 or
p − 3 + r22 ≤ p − 1 + (r2 − 1)(p − 2), which was seen to hold in arguments
above. In all cases |S3| ≤ |S4|. It has been shown that S3 can be chosen
from the set C of elements of λ(V ) with composite order.
Now C can be partioned into two element subsets {h, −h} with h in C.
Since S3 is the disjoint union of S2 and {gp, gp+1, g2p, g2p+1, . . . , gr2p, b1},
we may choose gp = −gp+1, g2p = −g2p+1, . . . , gr2p = −b1, in C, with S2
contained in the remainder of C.
S. D. Scott 259
Now α1α2α3 is an element of u(M0(V )) of order p. If H is a non–zero
α1α2α3–invariant subgroup of V , then it contains a non–zero element d
of V . If d is in {v1, . . . , vr1p}, then because Hα1 ⊆ H, d, dα1, . . . , dαp−1
1
are p elements of H and the subgroup K1 they generate has non–zero
intersection with λ(V ) (index two in V ). Since K1 ≤ H, in this case,
λ(V ) ∩ H 6= {0}. If d is in M(α2), then since Hα2 ⊆ H, M(α2) ⊆ H and
H contains {b1, . . . , bp−s1
} and intersects λ(V ) non–trivially. Clearly if d
is in V ∗\({v1, . . . , vr1p}∪M(α2)), then λ(V )∩H 6= {0}. It has been shown
λ(V ) intersects H non–trivially and λ(V ) contains an element d1 of H of
prime order. Since S2 consists of elements of λ(V ) of composite order, d1
is not in S2. Thus d1 is in M(α2) or M(α3) and H ∩ M(α2α3) 6= ∅. Now,
from the manner in which gp, gp+1, g2p, g2p+1, . . . , gr2p, b1, are defined and
3.1 we conclude that M(α2α3) ⊆ H. However,
〈M(α2α3)〉 ≥ 〈{g1, . . . , gr2p} ∪ {b1, . . . , bp−s1
}〉,
where the last group is, by simple order arguments = λ(V ). Thus H ≥ λ(V )
and, since M(α1) ⊆ H, H > λ(V ) and H = V (see §.2). It has been
shown that α1α2α3 confuses proper subgroups of V .
Suppose µ is a fixed–point–free automorphism of V of prime order (q
say) such that µα1α2α3 = α1α2α3µ. Since λ(V ) is, by §.2 characteristic,
the elements of λ(V )\{0} and V \λ(V ) (respective orders 1
2 |V | − 1 and
1
2 |V |) are, under µ, partitioned into q element subsets. This is a contra-
diction (q cannot divide 1
2 |V | and 1
2 |V | − 1). No such µ exists and an
application of 3.2 completes the proof of the theorem.
10. Proving 1.1 simplified
Let p ≥ 5 be a prime. It is a very pleasing fact that, the question of
which M0(V ) (V not elementary two) are p–gen can be solved. The full
solution of this problem was announced in the statement of 1.1. Here
we move toward the proof of this theorem. The material of sections two
to nine is a lead up to 10 to 13. What they have done is establish an
essential ingredient (viz. 9.1) needed for the proof.
At this stage in developments the group invariant δ(V ) comes into
its own. Possibly it may seem strange that it has not appeared sooner.
However, there is reason for this. Only V with |V | ≤ 3p need involve
its use. Thus 9.1 has fully covered the p–gen nature of M0(V ) where
consideration of δ(V ) is irrelevant. Since that is what 9.1 achieves, we
must now use this important invariant. Things start with a proposition.
260 The p–gen nature of M0(V ) (I)
Proposition 10.1. If V is a group, then a subset S of V ∗ of order
> |V ∗| − δ(V ), must contain the non-zero elements H∗ of a non-zero
subgroup H of V .
Proof. Clearly |S| ≥ 1 and V is non–trivial. Let Ai, i = 1, . . . , δ(V ), be
the collection of all cyclic subgroups of V of prime order. If S does not
contain an H∗ (H as indicated), then for each Ai, i = 1, . . . , δ(V ), we
can find non–zero ai in Ai, not in S. Thus S ⊆ V ∗\{a1, . . . , aδ(V )} and
|S| ≤ |V ∗| − δ(V ), contrary to the nature of S. 10.1 has been proved.
In one direction the proof of 1.1 is not that difficult. It is easy enough
to show that V (not elementary two) with p–gen M0(V ) (p ≥ 5 a prime)
cannot be in D#(1, p), D(2, p) or D(3, p). This is the contents of the
proposition that follows.
Proposition 10.2. If p ≥ 5 is a prime, then no group V in D#(1, p),
D(2, p) or D(3, p) has p–gen M0(V ).
Proof. If V is in D#(1, p) and M0(V ) is p-gen, then we can find α in
u(M0(V )) of order p such that N(α) = M0(V ). By §.3, M(α) (a subset
of V ∗) has order ≡ 0 mod p and has ≥ p elements. This implies |V | > p
contrary to the nature of D#(1, p). Thus, no V in D#(1, p), is such that
M0(V ) is p-gen.
If V is in D(2, p) or D(3, p) and M0(V ) is p–gen, then there exists α in
u(M0(V )) of order p such that N(α) = M0(V ). Because |M(α)| ≡ 0 mod p
and |M(α)| = p, when |V ∗| < 2p and |M(α)| = p or 2p, when |V ∗| < 3p, it
must follow that δ(V ) > |M(α)|. Thus |F (α)| = |V ∗\M(α)| > |V ∗|−δ(V )
and, by 10.1, F (α) contains the non–zero elements H∗ of a non–zero
subgroup H of V . Because M(α) is non–empty, H is proper and a proper
non–zero α–invariant subgroup of V . This would imply H.N(α) ⊆ H,
contrary to the fact that N(α) = M0(V ). Thus 10.2 stands proved.
The proof of 1.1 can be reduced to considering groups with |V | > 2p,
|V | ≤ 3p and δ(V ) ≤ 2p. This is what this section is about. We prove:-
Lemma 10.3. To prove 1.1 it is only required that we show that if p ≥ 5
is a prime then a group V (not elementary two) with 2p < |V | ≤ 3p and
δ(V ) ≤ 2p has p–gen M0(V ).
Proof. By 10.2 it must be shown that a V (not elementary two) not in
D#(1, p) ∪ D(2, p) ∪ D(3, p) with M0(V ) not p–gen has 2p < |V | ≤ 3p and
δ(V ) ≤ 2p.
Firstly 9.1 implies |V | ≤ 3p. Secondly as V is not in D#(1, p) and
D#(1, p) is clearly all groups G (G not elementary two) with |G| ≤ p
S. D. Scott 261
we see |V | > p. This means the possibilities for |V | are p < |V | ≤ 2p or
2p < |V | ≤ 3p.
Case 1. This is where p < |V | ≤ 2p. Here we have δ(V ) ≤ p or δ(V ) > p
but since when δ(V ) > p, V is in D(2, p) this situation cannot occur.
Thus we may define α on V as [v1, . . . , vp] where the p elements vi,
i = 1, . . . , p, of V ∗ (of order ≥ p) contain a non-zero element of every
non-zero cyclic subgroup of V of prime order (possible because δ(V ) ≤ p).
Clearly α has order p and confuses proper subgroups of V (a non–zero
α–invariant subgroup of V has an element in M(α) so that M(α) is
contained in it — also |M(α) ∪ {0}| > 1
2 |V |). Clearly there is no fixed–
point–free automorphism µ of prime order q such that αµ = µα. This is
because M(α)µ = M(α), q divides |M(α)| and must be p, while q divides
|F (α)| < p. So 3.2 applies and M0(V ) is p–gen.
Case 2. This is where 2p < |V | ≤ 3p. Here we have δ(V ) ≤ 2p or
2p < δ(V ) but when 2p < δ(V ), V is in D(3, p) and as this situation
does not occur we are only left with the possibility of V being such that
2p < |V | ≤ 3p and δ(V ) ≤ 2p. Lemma 10.3 is completely proved.
By 10.3 we are now looking at groups V (not elementary two) with
2p < |V | ≤ 3p and δ(V ) ≤ 2p, with a view to proving M0(V ) is p–gen
(here p is a prime ≥ 5). A lemma that will be of assistance with this is
the following.
Lemma 10.4. Let p ≥ 5 be a prime and V a group of even order > 2p
and ≤ 3p. If α in u(M0(V )) has order p, then there is no fixed–point–free
automorphism µ of V of prime order (q say) such that αµ = µα.
Proof. Suppose such a µ (prime order q) exists. Since M(α) ⊆ V ∗ and
has order divisible by p (α is a product of mutually disjoint p–cycles) we
must have |M(α)| = p or 2p. From sections two and three M(α)µ = M(α)
and M(α) is partitioned into q element subsets. Thus q divides p or 2p.
In the second case, q = 2 or q = p. The q = 2 possibility cannot occur
since by 2.9, µ would fix elements of η(V ). Thus, in either case, q = p.
Now, V ∗ = V \{0} is partitioned into q (= p) element subsets under µ,
and because |V ∗| < 3p and ≥ 2p, it follows that |V ∗| = 2p. This is a
contradiction (V has even order) and 10.4 is proved.
11. Moving into proving 1.1
The three sections ten to twelve are involved in final considerations
that establish 1.1. The point of the last was to reduce the proof of 1.1
to looking at those V with 2p < |V | ≤ 3p and δ(V ) ≤ 2p (see 10.3). As
262 The p–gen nature of M0(V ) (I)
far as this situation goes there are three cases. This section deals with
the two cases that arise when V has involutions that do not additively
centralize each other. This means that section twelve is about those V
which have a largish elementary two subgroup and finalizing the proof
of 1.1.
The first lemma we prove is the following.
Lemma 11.1. Let p ≥ 5 be a prime and V a group of order > 2p and
≤ 3p such that δ(V ) ≤ 2p. If V has a subgroup of the form Dp1
(p1 an
odd prime), then M0(V ) is p–gen.
Proof. Since |V ∗| ≥ 2p and δ(V ) ≤ 2p, we can take a subset S of V ∗
consisting of 2p elements and such that, S contains a non–zero element
of every cyclic subgroup of V of prime order. Now the subgroup Dp1
of
V , contains a cyclic subgroup Cp1
of order p1 with an element a 6= 0 in S.
Also Cp1
= λ(Dp1
) (see §.2) so that Dp1
\Cp1
consists of p1 involutions.
Take b, c and d as three such involutions (possible because p1 ≥ 3). Clearly
{b, c, d} is contained in S. Also, because 〈c, d〉 > 〈c〉 and 〈c〉 is a maximal
subgroup of Dp1
(|Dp1
| = 2p1), it readily follows that
〈a, b〉 = 〈c, d〉 = Dp1
. (11.11)
Let the elements of S be g1, . . . , g2p, where g1 = a, g2 = b, gp+1 = c
and gp+2 = d. The other g’s come from the remainder S\{a, b, c, d} of S
which by definition contains a non–zero element of every cyclic subgroup
of V of prime order (elements of {a, b, c, d} are amongst these). Define
α = [g1, . . . , gp][gp+1, . . . , g2p].
Clearly α is in u(M0(V )) and has order p. By 3.2 and 10.4 (V has even
order as Dp1
is a subgroup), the lemma will follow if it is shown that α
confuses proper subgroups of V .
If H is a non–zero α–invariant subgroup of V , then H contains a
subgroup of prime order and, the manner in which S was chosen, ensures
{g1, . . . , g2p} contains an element of H. If this is in {g1, . . . , gp}, then
because H is [g1, . . . , gp]–invariant, it follows that {g1, . . . , gp} and {a, b}
are contained in H. By (11), 〈a, b〉 = 〈c, d〉 and c and d are in H, so
that being [gp+1, . . . , g2p]–invariant, H contains {g1, . . . , g2p}. Thus if
{g1, . . . , gp} contains an element of H, then S ⊆ H. The alternative is
that {gp+1, . . . , g2p} contains an element of H. A very similar argument
to that above (again using (11)), shows that in this case also, S ⊆ H.
Simple order arguments mean H = V and 11.1 is proved.
S. D. Scott 263
The next lemma is, in some regards, very similar to 11.1. It deals with
p and V as in 11.1, except that, rather than V having a subgroup of the
form Dp1
(p1 an odd prime), it contains a subgroup of the form D4.
Lemma 11.2. Let p ≥ 5 be a prime and V a group of order > 2p and
≤ 3p such that δ(V ) ≤ 2p. If V has a subgroup of the form D4, then
M0(V ) is p–gen.
Proof. Since |V ∗| ≥ 2p and δ(V ) ≤ 2p, we can take a subset S of V ∗
consisting of 2p elements and such that, S contains a non–zero element
of every cyclic subgroup of V of prime order. Now the subgroup D4 of V ,
contains a cyclic subgroup C4 with generator d1. Since d1 has order four,
it is in λ(D4), which coincides with C4 (see §.2). Let a be an element of
D4\C4. All elements of a + C4 are in D4\λ(D4) and thus in η(D4). Let
b = a + d1, c = a + d12 and d = a + d13. It follows that because 〈a, b〉
contains a and d1, it must coincide with D4. Similarly, 〈c, d〉 contains c
and d1 and coincides with D4. Clearly, {a, b, c, d} (a subset of η(V )) is
contained in S and, from what has just been shown,
〈a, b〉 = 〈c, d〉 = D4. (11.12)
Let the elements of S be g1, . . . , g2p, where g1 = a, g2 = b, gp+1 = c
and gp+2 = d. The other g’s come from the remainder S\{a, b, c, d} of S
which by definition contains a non–zero element of every cyclic subgroup
of V of prime order (elements of {a, b, c, d} are amongst these). Define
α = [g1, . . . , gp][gp+1, . . . , g2p].
Clearly α is in u(M0(V )) and has order p. By 3.2 and 10.4 (V has even
order as D4 is a subgroup), the lemma will follow if it is shown that
α confuses proper subgroups of V . Here the proof that a non–zero α–
invariant subgroup of V coincides with V , follows as in the last paragraph
of the proof of 11.1, with (12) and (11) playing exactly the same role.
11.2 is therefore proved.
12. Finalizing the proof of 1.1
Lemma 10.3 reduced the proof of 1.1 to considering V with 2p <
|V | ≤ 3p and δ(V ) ≤ 2p. Furthermore 11.1 and 11.2 have looked at such
groups in the case of them having certain non–trivial dihedral subgroups.
This in fact covers such groups where two involutions do not additively
commute. Essentially all that remains is to consider the situation where
there is a number of commuting involutions. So now only one preliminary
result needs proving before 1.1 is finalized.
264 The p–gen nature of M0(V ) (I)
Lemmas 11.1 and 11.2 dealt with a wide assortment of groups and
covering the commuting involution situation does also. Only the existence
of seven mutually commuting involutions is required.
Lemma 12.1. Let p ≥ 5 be a prime and V a group (not elementary two)
of order > 2p and ≤ 3p such that δ(V ) ≤ 2p. If V has a subgroup of the
form C2 ⊕ C2 ⊕ C2, then M0(V ) is p–gen.
Proof. Since |V ∗| ≥ 2p and δ(V ) ≤ 2p, we can take a subset S of V
consisting of 2p elements and such that S contains a non–zero element
of every cyclic subgroup of V of prime order. From the assumptions, V
has a subgroup which is a direct sum A1 + A2 + A3 of the subgroups
Ai, i = 1, 2, 3, of order two. Let ai, i = 1, 2, 3, be the non–zero elements
of Ai. Define a4 as a1 + a2, a5 as a2 + a3 and a6 as a1 + a2 + a3. It
is clear 〈a1, a2, a3〉 must coincide with A1 + A2 + A3, but also, since
a4 + a5 = a1 + a3 and is distinct from a6, we conclude that 〈a4, a5, a6〉
coincides with A1 + A2 + A3. It has been shown that
〈a1, a2, a3〉 = 〈a4, a5, a6〉 = A1 + A2 + A3. (12.13)
Let the elements of S be g1, . . . , g2p, where g1 = a1, g2 = a2, g3 = a3,
gp+1 = a4, gp+2 = a5 and gp+3 = a6 (the other g’s come from the
remainder S\{a1, . . . , a6} of S). This is possible not only because p ≥ 3,
but also because,{a1, . . . , a6} is contained in η(V ) and therefore in S.
Define
α = [g1, . . . , gp][gp+1, . . . , g2p].
Clearly α is in u(M0(V )) and has order p. By 3.2 and 10.4, the lemma
will follow if it is shown that α confuses proper subgroups of V .
If H is a non–zero α–invariant subgroup of V , then H contains a
subgroup of prime order and, the manner in which S was chosen, ensures
that {g1, . . . , g2p} contains an element of H. If this is in {g1, . . . , gp},
then because H is [g1, . . . , gp]–invariant, it follows that, {g1, . . . , gp} and
{a1, a2, a3} are contained in H. By (13), 〈a1, a2, a3〉 = 〈a4, a5, a6〉 and a4,
a5 and a6 are in H, so that being [gp+1, . . . , g2p]–invariant, H contains
{g1, . . . , g2p}. Thus if {g1, . . . , gp} contains an element of H, then S ⊆ H.
The alternative is that {gp+1, . . . , g2p} contains an element of H. A very
similar argument to that above (again using (13)), shows in this case also,
S ⊆ H. Simple order arguments mean H = V . 12.1 is therefore proved.
Everything is now in place for the final assault on 1.1.
Proof of 1.1. By 10.3, 1.1 will be proved if it is shown that, when p ≥ 5 is
a prime and V a group of order > 2p and ≤ 3p such that δ(V ) ≤ 2p, then
it follows that M0(V ) is p–gen. By 4.1 it can and will also be assumed
S. D. Scott 265
η(V ) 6= ∅. For such a group, it will be shown that we may assume any
two elements a and b of η(V ), additively centralize each other. If a and b
do not additively centralize each other, then from §.2 〈a, b〉 is a dihedral
group of the form Dn, where n is an integer ≥ 3 (the case of n = 2 is
where Dn = C2 ⊕ C2 and a and b commute). The cyclic subgroup Cn
(= λ(Dn)) of Dn, therefore contains an element of order p1 (p1 ≥ 3 a
prime) or order four. It readily follows that Dn contains a subgroup of the
form Dp1
or D4 (this can be taken as the normal Cp1
or C4 subgroup of
Dn extended by 〈a〉). From 11.1 and 11.2, M0(V ) is p–gen and thus it can
be assumed 〈a, b〉 is abelian. Hence i(V ) is an elementary 2-subgroup of
V . It follows readily, from 12.1, that it can be assumed |i(V )| ≤ 4. Thus
|η(V )| ≤ 3 ≤ 1
3 |V | (p ≥ 5 and |V | > 2p). From 4.1, M0(V ) is necessarily
p–gen and theorem 1.1 is established.
13. Discussion
Theorem 1.1, [4], [5] and [1] mean we can determine when a V (not
elementary two) has p–gen M0(V ) (p a prime). For p = 2 and 3 it is
[4] and [5] that respectively allow this. For p ≥ 5 there is now enough
information on δ(V ) at hand (see [1]) so that theorem 1.1 can be applied.
Understanding of when δ(V ) > 1
2 |V | − 1 means, given a prime p ≥ 5
it can quickly be decided if V is in D(2, p) or D(3, p). Since D#(1, p) is
simply all V (not elementary two) with |V | ≤ p the problem is, in a sense
solved. Certainly for p ≥ 5 there is no general presentation of what D(2, p)
and D(3, p) look like. However, given p ≥ 5 and a V , rapid determination
of whether V lies in D(2, p) or D(3, p) is possible. This is an agreeable
fact that rests on what T. Burness and myself have proved in [1]. Just
how this is the case will now be explained.
A group V in D(2, p) has |V | ≤ 2p and δ(V ) > p so that δ(V ) >
1
2 |V | > 1
2 |V | − 1. A group V in D(3, p) has |V | ≤ 3p and δ(V ) > 2p so
that δ(V ) > 2
3 |V | > 1
2 |V | − 1. This means knowing the value of δ(V ) and
groups V with δ(V ) > 1
2 |V | − 1 supplies precise information on whether
V is in D(2, p) or D(3, p). In fact what we need only know is the V and
δ(V ) when δ(V ) > 1
2 |V | to do this.
In reference to the above, covering what [1] has achieved will now
be undertaken. There those V having δ(V ) > 1
2 |V | − 1 were classified.
However, since determining when M0(V ) is p–gen only requires we know
when δ(V ) > 1
2 |V | the full force of the main theorem of [1] is not used
here. What [1] states about groups V with δ(V ) > 1
2 |V | is now uncovered.
There arise in [1] eight possibilities for V with δ(V ) > 1
2 |V |. Type
(I) is groups G of the form D(A) (A abelian) having order 2|A| and
266 The p–gen nature of M0(V ) (I)
δ(G) = |G|/2 + δ(A). Type (II) is groups G of the form D4 ⊕ D4 ⊕ E (E
elementary two of order 2n) having order 2n+6 and δ(G) being 9
16 |V | − 1.
Type (III) is groups of the form H(r) ⊕ E (r ≥ 1 an integer) having order
22r+n+1 with δ(G) = |G|/2+2n+r −1, while type (IV) is those of the form
S(r) ⊕ E having the same order and same δ(G). Type (V) is groups G of
the form T (r) (see [1]) having order 3.22r and δ(G) being 2
3 |G| − 1. The
remaining three cases (types (VIII), (IX) and (X) in accordance with [1])
consist of individual groups. (VIII) is simply S3 ⊕ S3 (S3 the symmetric
group on three letters). Here the order of G is 36 and δ(G) = 19. (IX) is
simply S4 (the symmetric group on four letters). Here the order of G is
24 and δ(G) = 13, while type (X) is A5 with |G| = 60 and δ(G) = 31.
The type of groups found in D(2, p) or D(3, p) (p a prime ≥ 5) can
be restricted still further. It is now shown how types (VIII), (IX) and
(X) (the cases where we have isolated groups) need not concern us. For
V = S3 ⊕ S3 to be in a D(2, p) it is necessary that |V | = 36 ≤ 2p so
that p ≥ 18 is ≥ 19 while δ(V ) = 19 > p. For S3 ⊕ S3 to be in D(3, p) it
is necessary that p ≥ 12 while δ(V ) = 19 > 2p. Thus S3 ⊕ S3 is never
in D(2, p) ∪ D(3, p). For V = S4 to be in D(2, p) it is necessary that
|V | = 24 ≤ 2p so that p ≥ 12 is ≥ 13 while δ(V ) = 13 > p. For S4
to be in D(3, p) it is necessary that |V | = 24 ≤ 3p and p ≥ 11 while
δ(V ) = 13 > 2p. Thus S4 is never in D(2, p) ∪ D(3, p). For V = A5 to
be in D(2, p) it is necessary that |V | = 60 ≤ 2p and p be ≥ 31 while
δ(V ) = 31 be > p. For A5 to be in D(3, p) it is necessary that 60 ≤ 3p or
p ≥ 20 while δ(V ) = 31 be > 2p. Thus A5 is never in D(2, p) ∪ D(3, p).
The last paragraph shows that in determining groups in D(2, p) ∪
D(3, p) (p ≥ 5 a prime) we need only look amongst groups of type (I) to
(V). The statement of 1.1 therefore becomes:- A group V (not elementary
two) is p-gen (p ≥ 5 a prime) if and only if none of the following occur
(i) |V | ≤ p,
(ii) |V | ≤ 2p and V is of type (I) to (V) with δ(V ) > p, and
(iii) |V | ≤ 3p and V is of type (I) to (V) with δ(V ) > 2p.
Moreover, since for V of order ≤ 2p (or 3p) we know when δ(V ) > p
(when δ(V ) > 2p) the groups of (ii) and (iii) are determined.
It would now seem appropriate to give an example based on what is
set out above. This consists of specifying all V (not elementary two) which
are 31–gen. The prime 31 has been chosen because it is fairly easy to
establish which M0(V ) are 31–gen. For 37 things are more difficult. There
groups of the form S(r) ⊕ E and H(r) ⊕ E (E elementary two) creep
into those belonging to D(2, p) (p = 37). Also, in spite of this for p = 37
there are many more groups in D(2, p). First considerations for p = 31
involve finding the groups of D(3, p). V is in D(3, 31) if |V | ≤ 3.31 = 93
and δ(V ) > 2.31 = 62. Here δ(V ) > 2
3 |V | − 1 and since 2
3 > 5
8 no groups
S. D. Scott 267
of type (II), (III), (IV) or (V) (other than those of type (I)) can arise (see
[1]). So we are looking at D(A)’s where δ(A) > 1
6 |V | − 1 (ie. A abelian
with δ(A) > 1
3 |A| − 1). It is easily checked that the only possibilities for
A are C4 ⊕ E (E elementary two) or A an elementary 3–group (here
δ(A) = 1
2 |A| − 1
2). These give rise to a V (ie. D(A)) with order 2m or
order 2.3m1 (m and m1 integers ≥ 1). The only V with such order in the
range 62 to 93 has |V | = 26 (A = C2 ⊕ C2 ⊕ C2 ⊕ C4) but as V (= D(A))
has δ(V ) = 47 such a V is not in D(3, 31).
Continuing with the example of p = 31, we seek to determine D(2, 31).
If V is in D(2, 31), then |V | ≤ 62 and δ(V ) > 31. As T (r)’s have order
22r.3 (r ≥ 1 an integer) with δ(T (r)) = 2
3 |T (r)| − 1 no T (r) occurs (T (2)
just misses out because δ(T (2)) = 31). For V of the form S(r) ⊕ E or
H(r) ⊕ E (E elementary two) the only possibility not of type (I) is r = 2
and E = {0}. Here δ(V ) = 5
8 |V | − 1 = 19 so such a V is not in D(2, 31).
Clearly no groups of type (II) are in D(2, 31) (here |V | ≥ 64). Thus all V
in D(2, 31) are D(A)’s.
Specifying the groups of D(2, 31) is the same as specifying the D(A)’s
in D(2, 31). Now a D(A) (= V ) has δ(V ) = |A| + δ(A) and for A not
elementary two δ(A) ≤ 1
2 |A| − 1
2 (equality occurring when A is an elemen-
tary abelian 3–group). Thus δ(V ) ≤ 3
4 |V | − 1
2 . However a D(A) (= V )
in D(2, 31) has order ≤ 62 with δ(V ) > 31 so that 32 ≤ 3
4 |V | − 1
2 and
|V | ≥ 44 so |A| ≥ 22 and ≤ 31. We now look at possible A (abelian) that
can occur. |A| = 22 gives A = C22, δ(A) = 2 and δ(V ) not > 31. |A| = 23
gives A = C23, δ(A) = 1 and δ(V ) not > 31. |A| = 24 gives δ(A) ≤ 7
unless A = C2 ⊕ C2 ⊕ C2 ⊕ C3 (here δ(A) = 8). Thus for |A| = 24 only
D(C2 ⊕ C2 ⊕ C2 ⊕ C3) is in D(2, 31). |A| = 25 gives δ(A) ≤ 6 and δ(V )
not > 31. |A| = 26 gives δ(A) = 2 and δ(V ) = 28 not > 31. |A| = 27
gives δ(A) = 1, 4 or 13 (the last occurring for A = C3 ⊕ C3 ⊕ C3) so that
for |A| = 27 only D(C3 ⊕ C3 ⊕ C3) is in D(2, 31). For |A| = 28, δ(A) is
2 or 4 (when A = C2 ⊕ C2 ⊕ C7) so that only D(C2 ⊕ C2 ⊕ C7) is, for
|A| = 28, in D(2, 31). For |A| = 29, δ(A) = 1 and δ(V ) = 30 not > 31. For
|A| = 30, A = C30, δ(A) = 3 and δ(V ) = 33 > 31 so D30 is in D(2, 31).
For |A| = 31, δ(A) = 1 and D31 is in D(2, 31).
Putting together all the material of the above four paragraphs we see
that a V (not elementary two) has 31–gen M0(V ) if and only if V is not
one of the following. A group of order ≤ 31. Also not one of the groups
D(C2 ⊕ C2 ⊕ C2 ⊕ C3), D(C3 ⊕ C3 ⊕ C3), D(C2 ⊕ C2 ⊕ C7), D30 or D31.
Before finishing this paper we address questions that 1.1 raise. In the
statement of 1.1 only the families D#(1, p), D(2, p) and D(3, p) arise, so it
can be asked why D(n, p) with n ≥ 4 do not occur? If we look at groups
V (not elementary two) with δ(V ) > 2
3 |V |, paragraphs four to six of this
268 The p–gen nature of M0(V ) (I)
section supply the fact that such a V must be of type (I) (ie. D(A)’s)
and have δ(V ) = |V |/2 + δ(A). This means we are looking at abelian
groups A with δ(A) > 1
3 |A|. It is an easy matter to check that the only
such A are of the form C4 ⊕ E (here δ(A) = 1
2 |A| − 1) or A an elementary
abelian 3–group (here δ(A) = 1
2 |A| − 1
2). The V (ie D(A)) in these cases
has δ(V ) which is respectively 3
4 |V | − 1 and 3
4 |V | − 1
2 . Thus all V (not
elementary two) have δ(V ) ≤ 3
4 |V |. Since this implies D(n, p), n ≥ 4, is
empty it becomes apparent why such a family is excluded.
Above analysis yields even more. We see that D(3, p) can contain at
most only two groups (D(A)’s with A one of the two groups given above).
Often D(3, p) will be empty (simple order arguments ensure this). From
time to time it will contain one of the two D(A)’s specified and quite
infrequently, when the order of both D(A)’s conform with being ≤ 3p
and having δ(V ) > 2p, it necessarily contains both.
The mystery of when M0(V ) (V not elementary two) is p–gen has
been resolved. Basic (in a sense complete) understanding of the families
D#(1, p), D(2, p) and D(3, p) has been supplied. Why it is that the
D(n, p) (n ≥ 4) do not occur has been covered. Also the somewhat
rudimentary nature of the family D(3, p) has been given coverage. What
more remains? There are still questions that arise. Reasonably tight
upper and lower bounds for |D(2, p)| (in terms of p) would be of quite
real interest. Questions as to what happens when p is of a special nature
surface and, of course there remains the deep and intriguing question of
when M0(V ) is n–gen (n any integer ≥ 2).
References
[1] T. C. Burness and S. D. Scott, On the Number of Prime Order Subgroups of Finite
Groups, J. Austr. Math. Soc 87 (2009), 329–357.
[2] D. Gorenstein, Finite Groups, Harper and Row, New York (1968).
[3] G. Pilz, Near–rings, North–Holland Pub. Amsterdam (1983).
[4] S. D. Scott, Involution Near–rings, Proc. Edin. Math. Soc 22 (1979), 241–245.
[5] S. D. Scott, Transformation Near–rings Generated by a Unit of Order Three, Alg.
Col. 4:4 (1997), 371–392.
[6] C. T. C. Wall, On Groups Consisting Mostly of Involutions, Proc. Camb. Phil. Soc
67 (1970), 251–262.
Contact information
S. D. Scott University of Auckland, New Zealand
Received by the editors: 24.04.2010
and in final form 08.09.2012.
|