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Abstract
Working over the field of complex numbers, the inequivalent irreducible projective representations of a finite group G with 2-cocycle α are considered. The greatest common divisor of the degrees of the α-characters of these representations is used to obtain information about the degrees of the irreducible αS-characters of a Sylow subgroup S of G. Suppose now that G has exactly three irreducible α-characters. Then the potential form that the three degrees can take is found if G has a nontrivial cyclic Sylow subgroup. However, the main result is to show that G is solvable when the degrees are of the form a, ra and ka with certain restrictions on r and k, most notably that r = 1 or k. This involves detailed analysis of the possible actions of G on the irreducible αN-characters of a normal subgroup N of G.
2020 MATHEMATICS SUBJECT CLASSIFICATION:
1 Introduction
Throughout this paper G will denote a finite group.
Definition 1.1.
A 2-cocycle of G over is a function
such that
and
for all x, y,
The set of all such 2-cocycles of G form a group under multiplication. Two 2-cocycles α and β are cohomologous if there exists a funtion
such that
where
for all x,
This defines an equivalence relation on
and the cohomology classes
form a finite abelian group, called the Schur multiplier
Definition 1.2.
Let Then
is α-regular if
for all
Obviously if is α-regular, then it is also αk-regular for any integer k. Now let
then x is α-regular if and only if it is β-regular and any conjugate of x is also α-regular (see [Citation23, Lemma 2.6.1]), so that we may refer to the α-regular conjugacy classes of G. Finally if o(x) and
are relatively prime, then x is α-regular from [Citation17, Lemma 1.2 (b)].
Definition 1.3.
Let Then an α-representation of G of dimension n is a function
such that
for all x,
An α-representation P is also called a projective representation of G with 2-cocycle its trace function ξ is its α-character and
which is the dimension of P, is called the degree of
To avoid repetition all α-representations of G in this paper are defined over We now let
denote the set of all irreducible α-characters of G (see [Citation23, p. 184] for details), but the relationship between
and α-representations is much the same as that between
and (ordinary) representations of G. Now
is α-regular if and only if
for some
(see [Citation24, Proposition 1.6.4]) and
is the number of α-regular conjugacy classes of G (see [Citation24, Theorem 1.3.6]).
Now from [Citation7, Lemma 4.5] there exists such that α has order equal to that of
and α is class-preserving, that is the elements of
are class functions. Throughout this paper we will assume without loss of generality that the initial choice of 2-cocycle α has these two properties. Under these assumptions the “standard” inner product
may be defined on α-characters of G and for which the “normal” orthogonality relations hold (see [Citation24, Section 1.11.D]).
Let α be a 2-cocycle of G and let P be an α-representation of G of dimension n. Then (see [Citation23, Proposition 3.1.1]). Also
where
is defined by
for all g,
It follows under the class-preserving assumption that x is α-regular if and only if
for all
In particular if
then
The absolute centralizer of a non-empty subset S of G with respect to α is defined by
Now for all x, y, we have that
where
On the other hand
so we conclude that
Thus is a one-dimensional
-representation of G.
Next let N be a normal subgroup of G. Then G acts on by
for
and all
Clifford’s theorem for projective characters applies to this action (see [Citation24, Theorem 2.2.1]). We need some terminology and results that arise from Clifford’s theorem as it applies to α-characters. The inertia subgroup
is just the stabilizer of ζ under the action of G. There exists an extension of ζ to
where
is the inflation of a 2-cocycle
of
the latter is called an obstruction 2-cocycle of
The cohomology class of
in
is independent of the choice of
Let
Then firstly the mapping from
into
defined by
is a bijection, where γ is the lift of
Secondly the mapping from
into
defined by
is another bijection. It is useful to observe using this notation that if
for i = 1, 2, then
so that degree ratios are preserved using the bijection from
onto
Finally by considering
restricted to N, we observe that
in
Lemma 1.4.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Then for each restricted to N is a linear representation of N. In particular if N is perfect,
Proof.
We observe that if then
is trivial. So if N is in addition perfect we conclude that
restricted to N is the trivial representation of N and so
□
Lemma 1.5.
Let α be a 2-cocycle of G. Suppose that N is a normal subgroup of G that is generated by α-regular elements of G. Then
Proof.
If N is generated by α-regular elements of G, then restricted to N is the trivial representation of N for all
and so
□
We also record [Citation15, Lemma 3.1].
Lemma 1.6.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Then the number of orbits of G on is equal to the number of α-regular conjugacy classes of G contained in N.
The set of degrees of the elements of is denoted by
although if
is trivial we use the standard notation
We now let
and
denote the greatest common divisor and minimum respectively of the elements of
Now
if and only if
is trivial, because of the trivial character. In general
is described in [Citation9, Proposition 1], where np denote the pth part of n for p a prime number:
Lemma 1.7.
Let P be a Sylow p-subgroup of G. Then
Let Then
(see [Citation24, Theorem 1.4.1]), but in general
However one consequence of Lemma 1.7 is that
For future use let
denote the set of prime divisors of n and
Then, as we have already seen,
and also
(see [Citation24, Theorem 1.4.8]).
Lemma 1.8.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose G acts transitively on and let
be an obstruction 2-cocycle of
Then
where
and
Proof.
Let Then
where
and ei is the degree of the element of
that corresponds to ξi under the bijections described earlier. Thus
and using the bijections in Clifford’s theorem for projective characters, □
2 Degree questions
We conjecture that if then G is solvable. This problem appears to be quite difficult in generality, but we will present some evidence to support the conjecture. First we consider a very easy case in which Cn denotes the cyclic group of order n and Sn the symmetric group on n letters.
Lemma 2.1.
if and only if
or
Lemma 2.2.
Let α be a 2-cocycle of G such that Suppose that N is a normal subgroup of G such that
is trivial. Then N is solvable.
Proof.
G has at most three orbits on and so
Thus N is solvable from [Citation21, Corollary 12.6 and Theorem 12.15]. □
Lemma 2.3.
Let α be a 2-cocycle of G such that Then 3 divides
Proof.
If for
then
So suppose that
for all
then
and hence
□
Consequently any p-group G with is a 3-group, but more is true. The next two results are essentially [Citation13, Lemma 1.2 and Corollary 1.3] in which a group G is of
-central type if
and
Lemma 2.4.
Let α be a 2-cocycle of a p-group G. Suppose that Then the number of elements of
of degree
is divisible by p and in particular G is of
-central type when
Lemma 2.5.
Let α be a 2-cocycle of a nilpotent group G and let Suppose that
Then
contains n elements of degree
where n is divisible by pi for some i. In particular G is of
-central type when
Proof.
For the part when k = pi we have from the proof of [Citation13, Corollary 1.3] that is the product of
for
, where Sj is a Sylow pj-subgroup of G. Consequently G is of
-central type if and only if Si is of
-central type and Sj is of
-central type for
and the result now follows from Lemma 2.4. □
Similarly we have
Lemma 2.6.
Let α be a 2-cocycle of G such that Let
but suppose that
Then either
for all
or
for precisely one
and p = 2 or
Proof.
Let We may assume without loss of generality that
Now
so that either two or three of the degrees must be coprime to p. In the case of two, we may assume that
Thus in
from which we obtain that
so that
□
Definition 2.7.
Suppose that there exists a 2-cocycle α of G such that and the elements of
have degrees
Then we say that G is of (a, b, c)-type with respect to
For convenience we will always assume that and in this paper we will be concentrating on the case when a divides b and c, so that
and
Lemma 2.8.
Let α be a 2-cocycle of G. Suppose but
Then p divides
where
The proof of this result follows from the observation that In the particular case when G is of (a, a, ka)-type in Lemma 2.8 we have that
and so in particular if
then
On the other hand if G is of (a, ka, ka)-type in Lemma 2.8 then
and
3 Sylow subgroups
Lemma 3.1.
Let α be a 2-cocycle of G, let P be a Sylow p-subgroup of G and set Then
P is of αP-central type if and only if
P contains a non-trivial α-regular element of G if and only if
Proof.
Let of smallest degree, so that
from Lemma 1.7. Firstly suppose
Then
and so P is of αP-central type. On the other hand suppose
and that P contains no nontrivial α-regular elements. Let
such that
Then
a contradiction. □
This result demonstrates that a Sylow p-subgroup of G is either of αP-central type or it contains a non-trivial α-regular element of G. The following corollaries are immediate from Lemma 3.1.
Corollary 3.2.
Let α be a 2-cocycle of G and let P be a Sylow p-subgroup of G. Then P contains a non-trivial α-regular element if and only if it contains a non-trivial αP-regular element.
Corollary 3.3.
Let α be a 2-cocycle of G and set Then
In particular equality holds if and only if for each
there is exactly one non-trivial α-regular conjugacy class containing elements of p-power order and there are no other non-trivial α-regular elements.
So in the case we have that
or 2. In this case we also note that
is the sum of three squares with at least one of these an odd number and hence
So if G contains an α-regular element of even order then
or
for p an odd prime number and
Corollary 3.4.
Let α be a 2-cocycle of G and set Suppose that
is square-free and
Then a Sylow p-subgroup P of G is of αP-central type and all αP-regular elements of P are α-regular.
Proof.
Using Lemma 1.7 we may let with
Now
and so from Lemma 2.3,
must consist of exactly p elements of degree
Now from Corollary 3.2 there exists a nontrivial
that is α-regular. Now
is a set of αP-regular conjugacy class representatives for P (see [Citation23, Lemma 2.6.1]) and by the same result each such element is α-regular in G. □
We note that if is a prime number then the Sylow structure of G described in Lemma 3.1 and Corollary 3.4 is sufficient to guarantee that G is of
-central type for
from [Citation16, Theorem 2.5]. However, even in the case that
and the Sylow structure of G is as described, then either G is of
-central type or G is of
-type with respect to
The theory of p-blocks may be easily applied to α-characters in the case that the particular application that we will use below is when G has a cyclic Sylow p-subgroup, details for which may be found in [Citation2, Section 68]. In the next result and subsequently we will consider the situation when
and G has a cyclic Sylow p-subgroup, we will eventually show that such a group is either of
-central type or of
-type with respect to
Proposition 3.5.
Let α be a 2-cocycle of G such that and let
Suppose that G has a cyclic Sylow p-subgroup P. Then either
and G is of
-central type or
is even and G is of
-type with respect to α for some a,
or
and the elements of
consist of two of the same odd degree and one of even degree.
Proof.
Every element of p-power order is α-regular in G, so that G contains one or two p-regular α-regular conjugacy classes of G. In the first case or p2 and in the second
Let
with
and let B be the p-block of G containing
so that it has defect group P.
Now if G has one p-regular α-regular conjugacy class then B contains elements of
and since it is the only p-block containing elements of
we conclude that
The decomposition numbers or Brauer graph yield that the elements of
all have the same degree.
On the other hand if G has two p-regular α-regular conjugacy classes then B contains either one or two irreducible projective Brauer characters. In the first case B contains elements of
and we conclude that
the two elements of
both have the same odd degree and the remaining element of
has even degree (it is of highest type). In the second case B contains
elements of
and we conclude that
. The Brauer graph yields that the degrees of the elements of
are a, b and a + b for some a, b. In particular at least one of these three numbers is even. □
4 Normal subgroups
One of the purposes of this section is to examine some properties of a normal subgroup N of a group G with Our results use Lemma 1.5 and the bijections in Section 1. We first consider in general what happens if
is of
-central type for
Lemma 4.1.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose G is not of α-central type, but there exists such that
is of
-central type. Let
and
Then there exists a non-trivial
such that
Proof.
We observe that ξ vanishes off N. Now
In either case there exists a nontrivial such that
□
Lemma 4.2.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose G has n obits on Then
if and only if
is of
-central type for all
Proof.
From the bijections in Section 1, we have that □
In the context of Lemma 4.2 if is of
-central type then the n elements of
all have the same degree.
Proposition 4.3.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose that G has two orbits on Then
if and only if
is of
-central type for k = 1, 2 with both values of k occurring as ζ ranges over
Proof.
Suppose and let
. Then
or 2 with each case occurring for a suitable choice of
Thus
or 2 respectively and in the latter case
is of
-central type from [Citation11, Theorem A]. The converse is proved as in Lemma 4.2. □
Let G be a group of -central type with
Let y be a nontrivial α-regular element of G. Then o(y) is a power of 2 and
from [Citation11].
Corollary 4.4.
Let α be a 2-cocycle of G and let N be a normal subgroup of G. Suppose and let x, y be non-trivial α-regular conjugacy class representatives of G. Suppose that
but
Then o(y) is even and for suitable labeling
and
Proof.
Let such that
is of
-central type. Then we may label the two elements of
as
Then ξ3 has the stated properties from Proposition 4.3 and Lemma 4.1. Now yN must be a representative of the nontrivial
-conjugacy class of
Thus o(y) is even and
moreover since y is α-regular
We also have that
from Proposition 4.3 and
from Lemma 4.1. □
We note in Corollary 4.4 that for all
but
for some
This observation yields the following result.
Corollary 4.5.
Let α be a 2-cocycle of G such that and let x and y be non-trivial α-regular conjugacy class representatives of G. Suppose there exists a normal subgroup N of G such that
but
Then there does not exist a normal subgroup M of G such that
but
The next results will become relevant when we consider an abelian minimal normal subgroup of an (a, b, c)-group G with respect to α for suitable values of b and c.
Corollary 4.6.
Let G be an (a, ra, ka)-group with respect to where
Suppose that N is a normal subgroup of G such that (i) G has two orbits on
(ii) the elements of
all have the same degree
and (iii)
is of the form
with q an odd prime number and b,
Let ξ,
with
and
Then r = 1 or k and the number of distinct irreducible constituents of ξN and
are respectively:
and
for r = 1 and k odd;
and
for r = 1 and k even;
and
for
Proof.
Let s1 denote the number of distinct irreducible constituents of ξN each occurring with multiplicity e1 and let s2, e2 denote the corresponding numbers for Then
so that
Suppose
Then
and
and hence
Thus,
and hence s1 and
divide
This implies that q = p and
Thus
or
for some
depending on whether k is odd or even respectively. The case r = k is similar, but easier using the fact that here
□
Corollary 4.7.
Let α be a 2-cocycle of G and set Suppose that N is a normal subgroup of G such that G has n orbits on
and the elements of
all have the same degree
Suppose that
such that
Let
Then the number of distinct irreducible constituents of
is
where
and t is the number for
Proof.
The elements of all vanish off N. Let
Then
where s is the number of distinct irreducible constituents of
Thus
□
We note that if we add the assumption that for p a prime number and
to the hypothesis in Corollary 4.7, then t is a power of p.
5 Solvable groups
Proposition 5.1.
Let α be a 2-cocycle of G, where G is solvable. Suppose and that K is a minimal normal subgroup of G containing more than one α-regular conjugacy class of G. Set
and
Then
is trivial,
and there exists a G-invariant
that is a constituent of some
with
Proof.
K is an elementary abelian p-group for some prime number p and G does not act transitively on By [Citation14, Lemma 1.1],
is trivial and
for all
So we can now assume
Set
and let R/C be a chief factor of G, also let
denote the dual group of K. Then R/C acts faithfully on
Now from the fixed-point theorem (see [Lemma 11.14, 22]) we obtain that R/C has order relatively prime to p. It follows from [Citation12, Proposition 1.5 and Corollary 2.4] that there exists an R-invariant
and so λg is R-invariant for all
Suppose for
that
for
Then
for all
Thus
for all
, so that λ is G-invariant.
So now suppose that K contains two α-regular conjugacy classes of G. Thus G has two orbits on by Lemma 1.6 and we let
be representatives for them. Then from Proposition 4.3 we can assume without loss of generality that
is of
-central type for the obstruction 2-cocycles
. Using the bijections detailed in Section 1 this means that there is one element of
vanishing off K and lying over the orbit containing ζ1 and the remaining two elements of
are of the same degree and lie over the other orbit. The squares of the degrees of the elements of
are thus
and
(twice), where ti is the length of the orbit containing
Now either t1 or t2 is 1 and
These two possibilities yield that
or
respectively.
Finally suppose that K contains all three α-regular conjugacy classes of G. Thus by Lemma 1.6, G has three orbits on of lengths
say. Then from Lemma 4.2 for each
we have that
is of
-central type for the obstruction 2-cocycle
It follows from the bijections in Section 1 that the three elements of
all vanish off K. So the squares of the degrees of the elements of
are
Now we know that at least one of
t2 and t3 is one, and so
□
We note for future reference that the proof only uses the solvability of G to give that both K and R are solvable.
Corollary 5.2.
Let α be a 2-cocycle of G and set and
Suppose that G is a solvable group with
Then
Proof.
Let K be a minimal normal subgroup of G. If K contains non-trivial α-regular elements, then the result is immediate from Proposition 5.1. So now suppose K contains no non-trivial α-regular elements. Then G acts transitively on by Lemma 1.6. Let
and let
be its obstruction 2-cocycle. Then the degrees of the elements of the elements of
are
for the three elements
. Thus
where e is the greatest common divisor of
Now
by Lemma 1.8. So by induction
for some
and hence
□
Corollary 5.3.
Let α be a 2-cocycle of G, where G is solvable and set Suppose that
and
Then at least two of the three elements of
have equal degrees and
Proof.
Let K be a minimal normal subgroup of G, so that K is an elementary abelian p-group for some prime number p.
Suppose K contains no nontrivial α-regular elements. Then using a similar argument to that in the proof of Corollary 5.2 we have that the two statements hold in G.
So now suppose that K contains nontrivial α-regular elements. Then and K does not contain all the α-regular elements of G from Lemma 3.1, so we obtain at least two elements of
of the same degree. Now as described in the proof of Proposition 5.1 there exists
such that
is of
-central type for the obstruction 2-cocycle
. Now
has one nontrivial
-regular conjugacy class which consists of elements of 2-power order. It follows that the α-regular class of G not contained in K consists of elements of even order and hence
□
We conclude this section by considering the situation when is odd.
Theorem 5.4.
Let α be a 2-cocycle of G, where G has odd order. Suppose that Then G is of type (a, ka, ka)-type with respect to α for some a and k, where
for some prime number
and r is odd.
Proof.
Choose if possible a normal subgroup N of G such that N contains no non-trivial α-regular elements of G. Then by induction, and using the degree ratio preserving bijection in Clifford’s theorem detailed in Section 1, is of type (b, kb, kb) with respect to the obstruction 2-cocycle
for some prime number
and r is odd. Set
then G is of type (tb, ktb, ktb) with respect to α as required. So now let K be a minimal normal subgroup of G. Then we may assume K contains a non-trivial α-regular element of G. From Proposition 5.1 there exists a G invariant
with
Let
with
Then
but v and
are in different orbits in the action of G on
So λv and
are in distinct G-orbits of length
Thus the degrees of the elements of
are
and two of degree ka, where
Now
and k is odd, so that
it follows easily that
and
where r is odd. □
One consequence of Theorem 5.4 is that we can now complete work on the situation when G has a cyclic Sylow 2-subgroup.
Corollary 5.5.
Let α be a 2-cocycle of G such that Suppose that G has a cyclic Sylow 2-subgroup P. Then
and G is of
type with respect to α, where
for some prime number
and r odd.
Proof.
From Proposition 3.5, and the elements of
are two
of the same odd degree and one ξ3 of even degree. Let H be a Hall
-subgroup of G. Then
is the sum of two elements of
of the same odd degree and ξ3 vanishes off H. Now every element of 2-power order is α-regular so we must have that
Thus
and the result is immediate from Theorem 5.4. □
6 Simple groups
Notation concerning simple groups is consistent with that used in [Citation1] and in which relevant projective character tables may be found.
Theorem 6.1.
Let α be a 2-cocycle of a non-abelian simple group G. Then
Proof.
Suppose then
Now by inspection of the projective character tables of G, we have that
with equality if and only if
So we may henceforward assume that
Now from [Citation11, p.10], G has a nontrivial cyclic Sylow p-subgroup P for some prime number p, so that
(see [Citation22, Propositions 2.1.1 and 2.1.2]), and there exists another prime number q such that
, but
Now all elements of p-power and q-power order are α-regular from Section 1.
Suppose that Then we conclude from Proposition 3.5 that P has order three. Thus the three α-regular conjugacy classes of G consist of the identity element, the elements of order 3 and the elements of order q.
Now let r be any other prime divisor of aside from 3 and q, then r must divide
So provided
is divisible by at least two prime numbers (
). The non-abelian simple groups G which have a Sylow 3-subgroup of order three are classified in Theorem A.1 and the Schur multipliers of groups divisible by at least two prime numbers other than three may be found in [Citation22, Table 8.5]. The intersection of these two lists is empty. Finally we have to consider the case when
and G has a Sylow 3-subgroup of order three. From [Citation8, Theorem 1] there are just two possibilities for G, either
or
both with Schur multiplier of order 2. However, using Lemma 2.1 it is sufficient to observe that
and
for
□
One consequence of the Classification Theorem is the following result, see [Citation5, Theorem 1.53]: Let G be a simple group and let p be a prime number. Suppose then
We have had reason to consider simple groups G whose order is divisible by 3 but not 9, so in this case
Lemma 6.2.
Let G be a simple group with a Sylow 3-subgroup of order three. Then is cyclic unless
where
p is odd and f is even, in which case
and
or
Proof.
The outer automorphism group of every simple group is known from [Citation1] and is trivial for and of order 2 for
For the classical groups we are considering we have the outer automorphism group is a semi-direct product of two groups: firstly the cyclic group of diagonal automorphisms (of order d) and secondly the direct product of the cyclic group of field automorphisms (of order f) and the cyclic group of graph automorphisms (of order g). For and
For
and
For
and
Now for or
we have that
and f is odd and so the outer automorphism group in this case is cyclic of order
For
or
we have that d = 1 and so the outer automorphism group is cyclic of order f in this case. Finally for
or
d is 2 if p is odd and is 1 for
also f even implies that
or
So the outer automorphism group is either cyclic or is
if p is odd and f is even. □
Lemma 6.3.
Let G be a group and suppose K is a non-solvable minimal normal subgroup of G. Then if K is the unique minimal normal subgroup of G, G/K is isomorphic to a subgroup of and G is isomorphic to a subgroup of
Proof.
We observe that is a normal subgroup of G and Z(K) is trivial. So
and hence G/K is isomorphic to a subgroup of
and G is isomorphic to a subgroup of
□
Proposition 6.4.
Let α be a 2-cocycle of G such that
Suppose G has a cyclic Sylow 3-subgroup. Then G is of
-central type or of
-type with respect to α for some a.
Suppose that K is a non-abelian simple normal subgroup of G. Then K is neither isomorphic to one of the groups listed in Theorem A.1 nor
Proof.
(a) From Proposition 3.5 a Sylow 3-subgroup of G has order three. Let x and y be non-trivial α-regular conjugacy class representatives. Let K be a minimal normal subgroup of G. Then either or
Also G has 1, 2 or 3 orbits on
from Lemma 1.6 according as to whether
or vice versa, or
respectively.
Firstly suppose Then
Let
Then
has a cyclic Sylow 3-subgroup and there exists an obstruction 2-cocycle
of
such that
By induction
is of
-central type or of
-type with respect to
for some b. Using the standard bijections in Clifford’s theorem detailed in Section 1 yields respectively that G is of
-central type or of
-type with respect to
where
Secondly assume and
or vice versa. Let
Then two of these α-characters restricted to K have the same irreducible constituents and using the bijections in Clifford’s theorem only the third vanishes off K. Suppose now that G is not of
-central type. Then from Proposition 3.5, we may assume
and
for some a, b, and from the proof of the same result G contains exactly one nontrivial α-regular conjugacy class consisting of 3-regular elements. Suppose
is both α-regular and 3-regular. Then
using the Brauer graph. Now the second orthogonality relation gives that
and so
We note that if and only if
and consequently
if and only if
So if
the only possibility is that
So suppose
and
Then as noted above there exist two elements of
that have the same restriction to K up to a scalar multiple. All three possibilities lead, using the equalities above, to the impossible equation
The only possibility left is that and we can now also assume K is the only minimal normal subgroup of G. Then
and hence K is simple. If
, then
consists of three elements of degree
from Clifford’s theorem. Otherwise G/K is isomorphic to a subgroup of
by Lemma 6.3 and
is of
-central type for
Now,
is either cyclic or
for f even as detailed in Lemma 6.2. Also
or 3 and hence
is non-trivial from Lemma 2.2.
If is cyclic then
and each of the three orbits has length
Now M(K) is trivial for
or for the values of q we are considering for
or
for
, or
for
from [Citation1].
For . For the nontrivial cohomology class
there are two irreducible α-characters of degree 2, one of degree 4 and one of degree 6, contradicting equality of orbit lengths. Similarly
For the nontrivial cohomology class
there are two irreducible α-characters of degree 4, one of degree 6 and one of degree 8, again contradicting equality of orbit lengths.
For and the values of q we are considering,
For the non-trivial cohomology class
and for
there are from [Citation24, Theorem 5.8.10]
irreducible α-characters of degree
the same number of degree q – 1 and finally 2 of degree
So
and hence
which isn’t one of the values being considered. Similarly for
there are from [Citation24, Theorem 5.8.12]
irreducible α-characters of degree
of degree q – 1 and 2 of degree
So
which is clearly impossible.
If is not cyclic, then either
or
and each of the three orbits either has length
or
As shown in the previous paragraph equality of orbit lengths is impossible, so either there are two orbits of length
and one of length
or vice versa.
Then using the same information as above we obtain for and q > 5 that
which is impossible. Similarly for
we obtain that
which is also impossible.
(b) We first note that must be nontrivial from Lemma 2.2, so in particular we need not consider those groups K with M(K) trivial. We also note that if K has three distinct irreducible projective character degrees, then G has three orbits on
and each orbit contains all the irreducible αK-characters of a specific degree. In this situation suppose that
consists of one element ζ1 of one degree, two elements
of a second degree and the rest of a third degree. Then
is of
-central type for i = 1, 2 with the only irreducible
-character of degree ei where
. However
for
The elements of
all vanish off K and so
which is impossible. This result applies for
or
as detailed in (a).
Suppose and
Suppose
Then using the detailed information in [Citation24, Theorem 5.8.10], the sum γ of the two irreducible elements of
of degree
is non-zero on
conjugacy classes of K. However, there is an element of
that is a scalar multiple of this character. So these conjugacy classes of K become the three α-regular conjugacy classes of G. Now either from the projective character values of γ or the orders of the elements in the conjugacy classes we conclude that this can only occur if
which we already know is impossible. Now suppose that
Then a similar result holds using [Citation24, Theorem 5.8.12], although this time it is the sum γ of two elements of degree
that is non-zero on
conjugacy classes of K, and we conclude that
which we again know is impossible. Finally we consider
and
or 6. In the case that
we have that
which is impossible. If
consists of two elements of degree 6 and two of degree 12. Now if G has three orbits on
then G must have one orbit of length 2 and two orbits of length one and we obtain a contradiction as in the first paragraph. If on the other hand G has two orbits on
then both orbits have length 2 and also Proposition 4.3 applies. So if ζ1 and ζ2 are orbit representatives then we can assume without loss of generality that there exist obstruction 2-cocycles
for i = 1, 2 such that
is of
-central type with irreducible
-character degrees
Thus
which cannot occur. □
Next we consider a simple counting result:
Lemma 6.5.
Let for
be non-empty sets of natural numbers with the elements in each set arranged in increasing order. Define
Then
If there exist at least two values of i with then equality holds if and only if
for all i with
and some fixed d.
Proof.
We prove the inequality by induction on n, with the result being trivially true for Assume the result is true for
then we may list the elements of
as
where and
Now
So as required.
Now suppose that Then there is nothing to prove if
so we may assume that
Now
Thus (by permuting the indices)
for any pair of distinct indices i, j from
Also if
then
. Consequently we can assume that
for all i.
We now consider the case when Then the
elements of
are
Now for
Thus
for
Now as noted above this result holds true for any pair of distinct indices i, j from
Conversely suppose
for all i. Then
of order □
We will apply Lemma 6.5 to when G is non-abelian and simple.
Example 6.6.
Suppose that in the above lemma and that
or 3. Then
or n + 2 respectively. Now the left hand side is greater or equal to 2n with equality if and only if all the
This yields that
or 2 respectively.
Now if and only if
Also
if and only if either
or
and the sets S1 and S2 are as described in Lemma 6.5.
Lemma 6.7.
Let be non-abelian simple groups and let
Now let α be a 2-cocycle of G. Then
Proof.
The result is true for since
in this case from [Citation11, Theorem 3.2]. Now
from [Citation23, Corollary 1.2.5] and so we may assume without loss of generality that
The result is now immediate from these two observations and Lemma 6.5. □
7 Minimal normal subgroups
This section will aim to prove the existence of a solvable minimal normal subgroup of a group G of (a, b, c)-type with respect to some 2-cocycle α under certain restrictions on the values of a, b and c.
In a similar vein to Lemma 6.3 we have
Lemma 7.1.
Let α be a 2-cocycle of G and set Suppose that K is a non-solvable minimal normal subgroup of G and that G has n orbits on
Now suppose
in
for some
Then
and G/K is isomorphic to a subgroup of
Proof.
By Lemma 1.4, Now
for all
also
must be trivial using the facts that K is non-abelian and minimal. Now suppose that
is trivial. Then
since
is of
-central type. We conclude that
is trivial. Thus G/K is isomorphic to a subgroup of
□
Corollary 7.2.
Let α be a 2-cocycle of G. Suppose that and
Suppose also that K is a non-solvable minimal normal subgroup of G and that G has three orbits on
with orbit representatives
Then
in
for some i.
Proof.
Suppose that is trivial for
Then from Lemma 7.1 the orbit lengths are all
Let the elements of
have degrees a, a and b. Then without loss of generality we have that
and
So
consists of 2t elements of degree
and t of degree
We conclude from Lemma 6.7 that K is simple and Now, from [Citation25, Theorem 3.1] (see also [Citation18, p.556]), there exists a
such that K has a cyclic Sylow p-subgroup and
Thus, using Lemma 7.1, G has a cyclic Sylow p-subgroup and we conclude from Proposition 3.5 and Corollary 6.4 that
and G is of
-type with respect to
However, M(K) is either trivial or of order 2 using Theorem A.1 and the known Schur multipliers of simple groups. On the other hand if then
It follows that all the elements of
have degree a power of 2, which would imply that K is solvable from [Citation10, Theorem B]. □
Theorem 7.3.
Let α be a 2-cocycle of G such that and set
Then G has a solvable minimal normal subgroup unless there exists an odd prime number p such that
and
Proof.
Let x and y be representatives of the nontrivial α-regular conjugacy classes of G. Suppose first that there exists a normal subgroup N of G such that and let K be a minimal normal subgroup of G with
. Then G acts transitively on
and so the elements of
all have the same degree. Thus K is abelian from Lemma 6.7.
Now suppose that there does not exist a normal subgroup N of G with . Let K be a minimal normal subgroup of G, then K is the unique minimal normal subgroup of G from Corollary 4.5. Now suppose
, where H is a non-abelian simple group. Thus Lemma 6.3 applies and G/K is isomorphic to a subgroup of
moreover as in the proof of Corollary 7.2 there exists a prime number p such that H has a nontrivial cyclic Sylow p-subgroup Q and p does not divide
.
Label the isomorphic copies of H from 1 to n so that Now G has two or three orbits on
and so
Thus
from Lemma 6.7.
If n = 1, we have that . Thus Q is isomorphic to a Sylow p-subgroup of G and hence
from Proposition 3.5, but Proposition 6.4 (b) then gives a contradiction.
Finally suppose n = 2. Now In particular
and therefore
. Hence
Moreover
as in the proof of Lemma 6.7, so
Let R be the Hall
-subgroup of
where
Then R is isomorphic to a subgroup of
see [Citation22, Corollary 2.1.1]. Thus
Consequently all elements of p-power order in G are α-regular. Let be nontrivial for
where Q(i) is a Sylow p-subgroup of
Then x1, x2 and
are α-regular in G, so
and nontrivial α-regular class representatives for G must be x1 and
This in turn implies that
otherwise there would be further nontrivial α-regular elements of order coprime to p from Lemma 3.1 (b). □
We note that if G is of (a, a, ka)-type with respect to then
for p a prime number and so G contains a solvable minimal normal subgroup from Theorem 7.3.
8 The solvability of groups of (a, ra, ka)-type
In this section it will be shown that a group G of (a, ra, ka)-type with respect to a 2-cocycle is solvable under certain restrictions on the values of r and k. We begin by recapping some information about simple groups that have a solvable subgroup of p-power index for p a prime number, these results are a summary of [Citation25, Theorem 3.2] (although this result omitted PSL(3, 3)), [Citation6, Theorem 1] and [Citation3, Theorems 2.1 and 2.2]. It is more convenient to use the notation PSL(n, q) rather than in this section.
Theorem 8.1.
Let S be a non-abelian simple group with a solvable subgroup T of p-power index for p a prime number. Then, up to isomorphism, firstly S and secondly T are as follows:
, where
is a prime number and
For
and all such subgroups are conjugate in S. For
there is again one conjugacy class of subgroups T of index 8 that are isomorphic to
but also two conjugacy classes of subgroups T of index 7 that are isomorphic to S4.
, where
is a prime number and
and all such subgroups are conjugate in S.
PSL(2, 8),
and all such subgroups are conjugate in S.
PSL(3, 3),
and there are two conjugacy classes of such subgroups in S.
Note that, up to isomorphism, PSL(2, 7) is the only non-abelian simple group in Theorem 8.1 that has a solvable subgroup of prime power index for more than one prime number. The following is also a direct result of [Citation6, Theorem 1].
Corollary 8.2.
Let S be a non-abelian simple group and let p be a fixed prime number. Suppose that there exist of p-power index in S. Then
, so in particular all such subgroups are maximal in S.
A consequence of [19, Satz II.1.5], which will be utilized at least twice in the rest of this section, is that if a group acts transitively on a set then every orbit of a subnormal subgroup acting on Ω has length dividing
Lemma 8.3.
Let G be a group of (a, ra, ka)-type with respect to the 2-cocycle α with r = 1 and k even or Suppose K is a solvable minimal subgroup of G and let H be a subnormal subgroup of G. Then there exists an H-orbit on
of p-power length, where p is the prime number dividing
Proof.
It is sufficient to show that G has an orbit on of p-power length.
Suppose that K contains no nontrivial α-regular elements of G. Then G acts transitively on and hence for any
Suppose G has two orbits on Then the orbit lengths are pd and
for r = 1 and k even; pd and
for r = k from Corollary 4.6.
Finally suppose G has three orbits on Then the orbit lengths are
, and
from Corollary 4.7 and the note following it. □
We will prove that a group G of (a, ra, ka)-type with respect to a 2-cocycle α is solvable under the following conditions:
(a) (1) k even and
is not of the form
where
is a prime number; or
(2) r = k and is not of the form
where
is a prime number; also
is not the square of a prime number p for
(b) and either
(i) or
(ii) and either
or k is odd.
We note that if G is of the type described then with equality certainly being obtained in cases (a) (1) and (b) with (i). All these conditions ensure that there exists a solvable minimal normal subgroup of G from Theorem 7.3. Also note that
and
are not congruent to 0 modulo 5, 7 or 13 for any value of k.
Now assume henceforward that G is a non-solvable group of minimal order of (a, ra, ka)-type with respect to the 2-cocycle α with r and k as specified. The forthcoming results follow the argument used in [Citation25, Section 2] and we give due acknowledgment to the authors of that paper.
Lemma 8.4.
There exists a solvable minimal normal subgroup K of G such that is the maximal solvable normal subgroup of G,
is trivial in
and
is solvable for all
Proof.
Suppose there exists a minimal normal subgroup K of G that contains no nontrivial α-regular elements of G. Then K is solvable and is of (b, rb, kb)-type for some b with respect to
using the notation of Lemma 1.8 and the degree ratio preserving bijections of Section 1. Thus
is solvable. If the supposition is false then G contains a unique minimal normal subgroup K using Corollary 4.5. In this case G has two or three orbits on
and hence
is solvable from Proposition 4.3 and Corollary 4.4.
Now is trivial from [Citation14, Lemma 1.1]. From the same result and its proof either
for all
and
acts transitively on
or
for all
In either case
is solvable. In the first case
and hence
contrary to the assumption that G is not solvable.
It remains to show that is maximal with respect to being solvable and normal. Suppose that
is a solvable chief factor of G. Then the argument given in the first paragraph of Proposition 5.1 yields that there is a G-invariant element of
contrary to the assumption that G is non-solvable. □
We conclude that if is a chief factor of G, then
is not solvable. Thus if
is a composition factor of
then
, where
is a non-abelian simple group. We now examine some properties of the group
which we know has a solvable subgroup of p-power index from Lemmas 8.3 and 8.4 with such groups being listed in Theorem 8.1.
Lemma 8.5.
There exists a solvable minimal normal subgroup K of G as in Lemma 8.4 such that all the H-orbits on have the same length,
where
Suppose also that G satisfies the conditions in (a), then every solvable minimal normal subgroup of G contains no non-trivial α-regular elements of G.
Proof.
From Lemma 8.4 we have that H has a solvable subgroup of p-power index. Suppose that G has two orbits on Then the orbit lengths are of the form pd and
for r = 1 and k even or pd and
for r = k as in Corollary 4.6. So
or
respectively. So
is of the form
in the first case and hence p is odd, whereas in the second case
is of the form
Now suppose that G has three orbits on
. Then similarly the orbit lengths are of the form
, and
and so
is a power of p and
Under the conditions specified in (a), this is impossible from Lemma 8.4 and Theorem 8.1. So in this case G must act transitively on On the other hand if G does act transitively on
, then for each
we have
is a solvable subgroup of H of p-power index, moreover this index is uniquely determined from Corollary 8.2.
So now suppose that every solvable minimal normal subgroup of G contains a non-trivial α-regular element of G. Then K is unique and the conditions in (b) give that Thus
and one orbit length of H acting on
is 9 from Theorem 8.1, moreover all subgroups I of H with
are solvable.
In the two orbit case let be in the G-orbit of length
for r = 1 and
or
for r = k and k odd. Then
is a solvable subgroup of H of index l dividing
or
Given the assumptions on k, we must have that
So
is a subgroup of the maximal subgroup of
of index 9 and the only possibilities for l are 9 and 18. Now a maximal subgroup of PSL(2, 8) of index 9 is a Frobenius group of order 56 with a Frobenius kernel of order 8, and so
The same argument applies in the three orbit case starting with any G-orbit of length
□
Using the first paragraph in the proof of Lemma 8.5 we note that G cannot have three orbits on for r = 1 and k even, whereas G can only have two (or three) orbits on
in the case r = k when
We continue with K as specified in Lemma 8.5.
Theorem 8.6.
or PSL(2, q) unless q = 7 and
.
Proof.
Let so that
is a power of p independent of the choice of
Suppose
is isomorphic to one of the groups listed in Theorem 8.1. To avoid repetition, we begin by building a relationship between the groups
and
, where
Now
is a maximal subgroup of
by Corollary 8.2, so
Let
then
and so
Case 1: Suppose , where q is as described in Theorem 8.1 but
.
Let Then
and
from Lemmas 8.4 and 8.5. Thus
for some
from Theorem 8.1. Suppose that
for some
. Then
and so
.
Now let F be the set of elements of fixed by
Then
is the number of H-orbits on
each of which has equal length. So
We obtain
So
Note that acts faithfully on
as in the proof of Lemma 5.1. In particular no nontrivial element
fixes every element of
and so
acts faithfully on
and on
. Thus there exists an injective mapping,
from
into
and
by [Citation26, Theorem 9.14], where
Suppose that , where
is a prime number and m > 3. Then
from Theorem 8.1 and
. Now the highest power of q dividing
is q itself, so that
is the normalizer of a cyclic Sylow q-subgroup
of
. Now [Citation4, Lemma 15.1.1] applies and
is a Frobenius group with a cyclic complement acting irreducibly on
. So
is a maximal abelian subgroup of
.
Suppose that fixes
. We know that
fixes exactly one element in each
-orbit of
, so
fixes λ for some
. Then
fixes λ, but since
and
are maximal in
we have that either
or
. Clearly
fixes λ for the first possibility. For the second possibility
is a normal Sylow q-subgroup of
since
. In particular we have
, and so we have equality. Hence any λ fixed by
is also fixed by
and therefore
. However
since
, so
.
Finally, consider . Now from above
and no non-trivial element of Z is fixed by
. So, since the size of any orbit of
on Z divides
the only option is that
has just two orbits on Z with one being the trivial one. In particular Z is a simple
module and [Citation20, Theorem 31.7] applies, so
divides m. However
for any
and so we rule out these groups.
Suppose . Then
from Theorem 8.1 and therefore
Moreover there exists an injective map
from
to GL(4, 3) and
. Now
is a normal subgroup of
. However
is a simple group and so
Here 7 divides
but not
so this is impossible.
Suppose , where
and
is a prime number. Then
and there exists an injective homomorphism
from
to GL(2, r). Now
and by a similar argument to that in the previous paragraph we obtain that
. However, this is not possible as it would imply that
.
Case 2: Suppose and
. In this case, there are exactly two
-orbits on the set of subgroups of
of index 13 and each
for
is contained in one of these orbits. So once ζ is fixed, then for any
either
is conjugate to
in
or it is contained in a second
-orbit. Choose δ such that
is conjugate to
, then by the same argument as in Case 1 there exists
in the H-orbit of δ which is fixed by
. Moreover we can choose
to maximize the number of H-orbits containing an element of
that is fixed by
. Then by a similar argument to that in Case 1 and using the same notation we obtain
Following the same argument we obtain
. Now since
and
is a power of 13, this simplifies to
which yields
Thus we can define W and as in Case 1. This time we have
and
is an injective map from
into GL(2, 13). Now
is simple so we obtain that
is isomorphic to a subgroup of SL(2, 13). However 9 divides
but does not divide
Case 3: Suppose and
. Similarly to Case 2,
has two orbits on the set of subgroups of
of index 7 and we can choose
such that
Following the same argument with this time
and
being powers of 7, we obtain
and we can define W and
as before. Now
and PSL(2, 7) is isomorphic to a subgroup of GL(2, 7). More specifically since PSL(2, 7) is simple, it is isomorphic to a subgroup of SL(2, 7) by a similar argument to that for PSL(2, 8). Now consider the natural homomorphism
. Then
is abelian. Since
is simple we obtain that
is trivial and
. Hence
. However SL(2, 7) and
are both perfect so we obtain
. This is not possible and so
. □
We have thus disposed of the case that G does not contain a minimal normal subgroup K containing no nontrivial α-regular elements of G. So we can now assume that G acts transitively on . From the previous result we need to consider
with
so that K is a 2-group. Let
denote the Fitting subgroup of G.
Proposition 8.7.
and
is a proper solvable subgroup of 2-power index in G for all
Proof.
Suppose is nontrivial for q an odd prime number and let
be a minimal normal subgroup of G. Then the previous results of this section show that L must contain two or three α-regular classes of G. Now Lemma 8.3 yields that there exists
such that
is a power of q. Then by Lemma 4.2, Propositions 4.3 and [Citation11, Theorem B] we also obtain that
is solvable. It follows from Lemmas 8.3 and 8.4 applied to L and K respectively that
has a solvable subgroup of q-power index and hence
By assumption G is of (a, ra, ka)-type with respect to
where either r = 1 or
however, as previously noted
and
So F contains no non-trivial α-regular elements of G by Lemma 3.1 and Corollary 4.4. Thus G acts transitively on and hence
is a power of 2. Let
then
for some
and so
□
Proposition 8.8.
Proof.
Suppose and let M/F be a chief factor of G with
. Then M/F is an elementary abelian q-group for q an odd prime number. Let Q be a Sylow q-subgroup of M. Then QF = M and
. Let
Then
for some positive integer d from Proposition 8.7, so that
contains a Sylow q-subgroup of G/F. Thus
and hence
Let Then
and hence
from [Citation22, Theorem 2.2.10] and [19, Hilfssatz V. 25.9]. It follows that
for all
and all
and hence Q fixes every element of
Now (G, F, Q) satisfies (*) of [Citation18, Definition 3.6], so by Theorem 3.7 of the same paper there exists a bijection from to
. Also if
is the image of τ under this bijection then
. Now by the Frattini argument,
and hence
. Moreover G acts transitively on
so
Thus acts transitively on
.
Now Clifford’s theorem for projective characters as described in Section 1 defines a bijection between and
so
Now let
. Then we similarly have a bijection between
and
Finally by [Citation18, Theorem 3.7] there exists a bijection
, where if
then
We conclude that
Moreover the degree ratios in
are preserved by these bijections, so
is of (b, rb, kb)-type with respect to
for some b and so
must be solvable. However,
and thus G/F is solvable giving a contradiction. □
Proposition 8.9.
.
Proof.
Suppose is a composition factor of the chief factor
of G. Then
and
is perfect, so
by [Citation23, Corollary 1.2.5]. Now
is a 2-group from Propositions 8.7 and 8.8, so
from [Citation22, Corollary 2.1.15].
Let Q be a Sylow 7-subgroup of G. Then there exists a Q-invariant with
by [Citation12, Lemma 1.6]. However, Q is of αQ-central type by Lemma 3.1, so Q also acts transitively on
. Therefore
and since
this gives
which is impossible since
is a Sylow 7-subgroup of
and
as PSL(2, 7) is a composition factor of
□
We have thus proved:
Theorem 8.10.
Let G be a group of (a, ra, ka)-type with respect to the 2-cocycle Then G is solvable under conditions (a) or (b).
Corollary 8.11.
Suppose that k satisfies the conditions in (a) (2) or (b) with (ii), but with in either case. Then there does not exist a group G of (a, ka, ka)-type with respect to any 2-cocycle α of G.
Proof.
Suppose such a group G exists. Then G is solvable from Theorem 8.10, but then Corollary 5.3 gives a contradiction. □
To summarize we have produced considerable evidence to support the conjecture that if G is a finite group with a 2-cocycle α such that then G is solvable. The techniques used rely on the existence of a solvable minimal normal subgroup K of G such that every G-orbit on
has the same length. It would be of interest to see if some of the conditions in (a) and (b) to ensure that this occurs can be removed. So for example is G solvable if the three irreducible α-characters have degrees of the form a, 2a and
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A.
Simple groups with order divisible by three but not nine
It is well-known that the only non-abelian simple groups that have order coprime to three are those of the form , where
. Therefore we list the non-abelian simple groups whose order is divisible by three but not nine. We leave the reader to check the veracity of this result using the notation and orders of non-abelian simple groups given for example in [Citation1, Tables 1 and 6].
Theorem A.1.
S is a non-abelian simple group with a Sylow 3-subgroup of order three if and only if it is isomorphic to exactly one of the following groups:
, where
, where
and
, where
Note that
and