Regular orbits of coprime linear groups in large characteristic
Benjamin Sambale

TL;DR
This paper proves that finite coprime linear groups in large characteristic have regular orbits and demonstrates the optimality of the characteristic bound, with applications to blocks with abelian defect groups.
Contribution
It establishes the existence of regular orbits for coprime linear groups in large characteristic and shows this bound is tight, also applying results to block theory.
Findings
Finite coprime linear groups in characteristic p >= (|G|-1)/2 have regular orbits.
The bound on p is proven to be optimal.
Application to blocks with abelian defect groups.
Abstract
We prove that a finite coprime linear group G in characteristic p>=(|G|-1)/2 has a regular orbit. This bound on p is best possible. We also give an application to blocks with abelian defect groups.
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Taxonomy
TopicsFinite Group Theory Research · Coding theory and cryptography · graph theory and CDMA systems
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LemThm \aliascntresettheLem \newaliascntPropThm \aliascntresettheProp
Regular orbits of coprime linear groups
in large characteristic
Benjamin Sambale111Fachbereich Mathematik, TU Kaiserslautern, 67653 Kaiserslautern, Germany, [email protected]
Abstract
We prove that a finite coprime linear group in characteristic has a regular orbit. This bound on is best possible. We also give an application to blocks with abelian defect groups.
**Keywords: coprime linear groups, regular orbits, minimal subgroups
AMS classification: 20G40, 20C15**
1 Introduction
Linear groups, i. e. subgroups of where is a vector space, play an important role in many branches of group representation theory. In the present note we are interested in the situation where is an elementary abelian -group and is a -group. Then by Maschke’s Theorem, the action of on is semisimple. A crucial fact used in the solution of the so-called -problem [15] is that often has a regular orbit on , that is, an orbit of size . This is well-known for abelian groups and many other special cases have been handled in the literature (see for instance [3, 4, 6, 9, 16]). It is also known that regular orbits exist if is “large”, for example if (see [10, Lemma 2.2]). The main result of this paper establishes the existence of a regular orbit under the weaker condition . The proof relies on a classification of finite groups with “many” minimal subgroups given by Burness-Scott [2]. On the other hand, for every odd prime we construct linear groups without regular orbits such that .
These results are motivated by Brauer’s -Conjecture which is still open even for blocks with abelian defect groups. In this situation, is a defect group of a block of a finite group and is the corresponding inertial quotient. The number of irreducible characters in is denoted by . By a result of Robinson [10], Brauer’s -Conjecture, i. e. , holds provided has a regular orbit on . In previous papers [12, 13, 14] we have applied some of the techniques from the -problem to this more general situation. Now in the present paper we will give new application of our main theorem.
2 Main theorem
In the following we denote a cyclic group of order by . A dihedral group of order is denoted by and the symmetric and alternating groups of degree are and respectively. Moreover, for a finite group we set ( copies). The exponent of is defined by . Note that the minimal subgroups of a finite group are precisely the subgroups of prime order.
Proposition \theProp ([2, Theorem 1.1]).
If a non-trivial finite group has more than minimal subgroups, then one of the following holds:
- (i)
* where is abelian and acts as inversion.* 2. (ii)
* for some .* 3. (iii)
* for some (Frobenius group).* 4. (iv)
* for some (central product).* 5. (v)
. 6. (vi)
* for some .* 7. (vii)
* for some .* 8. (viii)
.
Note that (i) in section 2 includes all non-trivial elementary abelian -groups. Now in our main theorem we prove slightly more than what was promised in the introduction.
Theorem 1**.**
Let be a -automorphism group on a finite -group such that . Then one of the following holds:
- (i)
* has a regular orbit on .* 2. (ii)
. 3. (iii)
* and .*
Proof.
Let be a minimal counterexample. By a result of Hartley-Turull [7, Lemma 2.6.2], we may assume that is elementary abelian. Then is non-abelian (see for example [4, Corollary 4.I]). If is a proper quotient of , then either or and . In particular, every proper quotient of has a regular orbit. By [4, Lemmas 2.I and 3.I] it suffices to assume that is an absolutely irreducible -module over a finite field with elements. In particular, the center of is cyclic. Let be the set of minimal subgroups of . Then for every there exists such that . Hence,
[TABLE]
Since acts faithfully, for every . This shows and
[TABLE]
Therefore, . Let with (here may be negative). We discuss four cases.
Case 1: .
Here has odd order. Since two distinct minimal subgroups of intersect trivially, it follows that . Equality can only hold if is a -group. Then according to [4, Theorem 1.II] and we easily get a contradiction. Hence, . If we have equality this time, consists of elements of order and just one subgroup of order . But then is nilpotent and must contain elements of composite order as well. This shows . Since the number of subgroups of prime order is always congruent to modulo (Frobenius’ extension of Sylow’s Theorem), equality in this case leads to . Of course, this is impossible and we conclude that and . However, it is not hard to see that there is no group with the desired number of minimal subgroups.
Case 2: .
Now has even order and . Hence, is not divisible by . It is well-known that in this situation has a normal -complement . As usual, the number of minimal subgroups of inside is at most . Let be a Sylow -subgroup of . Then the number of minimal subgroups of outside is . If , then . Since , we derive that . These cases have been handled in [11, Proposition 14.9]. Thus, let . Then acts as inversion on and must be abelian. Ito’s Theorem implies that (see [8, Theorem 6.15]). Since , is cyclic and so . Both cases are excluded by [11, Proposition 14.8].
Case 3: .
As before, . Suppose first that . Let and with . Setting we refine (2.2) to
[TABLE]
This implies . In particular, every acts faithfully on and we deduce that . Hence, and is a -group. But then we may choose and obtain the contradiction , since is -invariant and is irreducible. This argument implies . If is still a -group, must be a Fermat prime or a Mersenne prime by [4, Theorem 2.II]. It follows easily that (a Fermat prime) and . Here one can show with GAP [5] that as given in the statement of our theorem. Now suppose that is not a -group. If , then, by (2.3), contains just one minimal subgroup of odd (prime) order. Since the number of involutions is always odd, we conclude that . But this gives the contradiction . Hence, we may assume that . At this point we refer to the next case.
Case 4: .
First observe that this case includes the remaining possibility , since . Moreover, is given as in section 2. Since is cyclic, some cases can be excluded immediately. If has odd order, then which is impossible. Therefore, is even and .
Suppose first that where is abelian and acts as inversion. Let be a minimal subgroup of lying inside . Then and is -invariant. It follows that . Hence, in (2.1) we only need to consider the minimal subgroups outside . This leads to in (2.2) after taking the last cases into account. By Ito’s Theorem we have and is cyclic. Since also is cyclic, contains at most one involution. This shows that is cyclic and . The first case corresponds to the exception given in the statement of the theorem and the second case is excluded by [11, Proposition 14.8].
Next let with . Then and [1, Lemma 1.1] gives . A theorem of Taussky leads to (see [1, Proposition 1.6]). But then and . Now let . Then Ito’s Theorem implies and the -rank of is at most (see for example [11, Proposition 7.13]). Consequently, , and . One can show that always has a regular orbit (see [11, Proposition 14.9]).
Now we discuss the extraspecial group with factors (note that there are no elementary abelian direct summands, because is cyclic). Then by the last cases. Let be the unique central involution. Then , since is -invariant. Hence, inverts the elements of . The non-central involutions in can be paired in the form . Then and . This yields and
[TABLE]
As in (2.2), we obtain
[TABLE]
From [2, Table 1] it follows that . In the case we derive the contradiction
[TABLE]
Therefore, and . Then again has -rank at most which contradicts .
In the case , has odd order which was already excluded. Finally, the cases give (see [2, Table 1]) and . Again one can show that these groups have regular orbits (see [11, Proposition 14.9]). ∎
Theorem 1 does not extend to . For example the semidihedral group of order acts faithfully on without having regular orbits, since . Similarly, is an automorphism group on without regular orbits. There is no doubt that one can classify these examples as well, but this becomes increasingly tedious and probably will not reveal new insights.
One can check that the exception in Theorem 1 actually occurs. In the following we show that the other exceptions occur for every odd prime .
Proposition \theProp.
Let be an odd prime. Then there exists an automorphism group on a finite -group such that and has no regular orbit on .
Proof.
Let be a -dimensional vector space over . We regard as the additive group of the field . Let be a generator of . Then has order and acts by multiplication on . Let be the Frobenius automorphism of . Then and acts faithfully on . It suffices to show that fixes every orbit of on . Any non-trivial element of has the form for some . Then . Hence, and lie in the same orbit of . ∎
For the smallest -automorphism group without regular orbits is the semilinear group acting on . It is obvious from the group order that in this situation there are no regular orbits.
3 Application
In the next theorem we consider blocks of finite groups with respect to an algebraically closed field of characteristic . We use the standard notation which can be found for example in [11].
Theorem 2**.**
Let be a -block of a finite group with abelian defect group and inertial index . Then Brauer’s -Conjecture holds for , i. e. .
Proof.
By [11, Theorem 14.13], we may assume that and therefore . The inertial quotient of is a -group of order and acts faithfully on . In order to find a large orbit of on , we may assume by [7, Lemma 2.6.2] that is elementary abelian of rank . Then . Let . We use the arguments from the proof of Theorem 1 to show that has a regular orbit on . Suppose by way of contradiction that has no regular orbit. Let be the set of minimal subgroups of . Then . Since , we have and for . Consequently,
[TABLE]
and . This contradicts .
Hence, has a regular orbit, i. e. there exists such that . It follows that . Now if is a prime or , then the claim follows from [13, Proposition 11]. In the remaining case there exists a normal subgroup such that is or an odd prime. We conclude that . Theorem 1 implies that has a regular orbit on . Now the claim follows from [11, Lemma 14.5] and [13, Corollary 12]. ∎
Acknowledgment
I thank Julian Brough for making useful comments on a draft of this paper. This work is supported by the German Research Foundation (project SA 2864/1-1) and the Daimler and Benz Foundation (project 32-08/13).
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