Solvable groups satisfying the twoprime hypothesis II
Abstract
In this paper, we consider solvable groups that satisfy the twoprime hypothesis. We prove that if is such a group and has no nonabelian nilpotent quotients, then . Combining this result with the result from part I, we deduce that if is any such group, then the same bound holds.
Keywords: character degrees, Finite groups, Solvable groups.
MSC[2010]: 20C15
1 Introduction
One general category of problems in the character theory of finite groups involves investigating how the structure of the set of degrees of irreducible characters, denoted , influences the structure of the group.
A wellknown result states that if is solvable and the elements of are pairwise coprime, then has at most three distinct character degrees. There are many possible extensions to this problem. One was investigated by the second author in [7]. He assumed that the group satisfies the “oneprime hypothesis,” that is, if you choose distinct, then is either 1 or a prime.
If is a positive integer, we will write to denote the total number of prime divisors of , counting multiplicity. As an abbreviation, we will write to mean . Thus, the oneprime hypothesis can be stated as: if with , then . Under this hypothesis, the second author proved in [8] that for a solvable group .
We can state the “prime hypothesis” in a similar way: if with , then . More generally, we say that a set of positive integers satisfies the prime hypothesis if for all with , we have . One hopes that for solvable and satisfying the prime hypothesis, for some function . At this time, we are not able to prove the existence of such a function for all , but we can prove that a bound exists when .
The main theorem of this paper is the following:
Theorem A.
Let be a solvable group satisfying the twoprime hypothesis. Assume does not have a nonabelian nilpotent factor group. Then .
This note contains work from the first author’s dissertation at the University of Wisconsin. The first part of that dissertation appeared in the published paper [2].
The main result of [2] was the following:
Theorem B.
Let be a solvable group satisfying the twoprime hypothesis. Assume has a nonabelian nilpotent factor group. Then .
Combining Theorem A with Theorem B, we obtain the main theorem of the first author’s dissertation.
Theorem C.
Let be a solvable group satisfying the twoprime hypothesis. Then .
We note that is the largest known value for where is a solvable group satisfying the twoprime hypothesis. Obviously, there is a wide gap between the upper bound we have found and the known groups. The goal in this paper is to prove that a bound exists, and we have made no attempt to find an optimal bound. However, it seems likely that even if one were to optimize the bound using the arguments in this paper that there would still be a large gap between the bound found here and the known groups, so there is still much room for the bound to be improved.
2 A Few Technical Results
We begin with a few technical lemmas which will be useful in the cases remaining to be considered. The following fact is quite useful and will be used several times. It is an easy application of the Glauberman correspondence (see Theorem (13.24) of [4]).
Lemma 2.1.
Let act on by automorphisms with solvable, abelian, and . Then the action of on and the induced action of on are permutation isomorphic.
We write for the intersection of all with nilpotent. Note that is a characteristic subgroup of and is the smallest normal subgroup of with a nilpotent factor group.
If is an integer, we use to denote the set of primes that divide . If is a subgroup of , we use .
Lemma 2.2.
Let be a solvable but not nilpotent group, and assume is the unique minimal characteristic subgroup of . Write .

Then is an elementary abelian Sylow subgroup of for some prime

If is abelian, then acts on with a regular orbit and

Let be an orbitsize of the action of on . Then .
Proof.
First note that since is solvable, must be an elementary abelian group for some prime . Let and observe that . We want to prove that . By the uniqueness of , we have . Hence, by the HallHigman theorem (“Lemma 1.2.3”) it follows that .
Since is not nilpotent, . Let be a complement for in . Because is nilpotent and is characteristic, it follows that is a characteristic subgroup of . Consider the action of on . As is abelian, we have that acts on , and this action is coprime. Hence, we may write .
Because , we see that centralizes . In particular, this implies that . Now since and are characteristic subgroups of , it follows that is also a characteristic subgroup of .
Assume . Then by the uniqueness of , we must have . Now and therefore . However, we now have , which is a contradiction.
Thus we may assume . Now . Therefore , as desired.
For (b), observe that , and so acts faithfully on . Now, Lemma (2.3) of [2] implies that has a regular orbit on , as desired. By Lemma 2.1, there must also be a regular orbit of on . Let be an element of this orbit. Then , and so by the Clifford correspondence (Theorem 6.11 of [4]), . We have , which gives (b).
To prove (c), let . Write and . Since divides , clearly . For the reverse containment, assume is a prime dividing but that does not divide . In particular, this implies that contains a full Sylow subgroup of .
Let . Note that since is nilpotent, it follows that is a characteristic subgroup of . Now since centralizes , we have . In particular, is a nontrivial characteristic subgroup of , and so by the uniqueness of . However, we now have . Since divides , it must be that , which is a contradiction. ∎
We now consider when is a normal subgroup of some overgroup . This lemma and its sequels should be compared with Lemmas 2.2 and 2.3 of [6].
Lemma 2.3.
Assume is solvable and is nonabelian with for all primes . Let be maximal with and nonabelian. Write . Then and is the unique minimal characteristic subgroup of .
Proof.
We first prove that is the unique minimal characteristic subgroup of . Let be a nontrivial characteristic subgroup of . Now and , so by the way was chosen we must have abelian. Then , as desired.
Since is nonabelian and for all primes , it follows that is not nilpotent. In particular, . By the uniqueness of , we have . Also, is abelian and in particular is nilpotent. Thus . Therefore , as desired. ∎
Lemma 2.4.
Assume is solvable and is nonabelian with for all primes . Let be maximal with and nonabelian. Write and . Assume further that is chosen so that is minimized.

is an elementary abelian Sylow subgroup of for some prime

Assume and is nonabelian. Suppose further that with and is maximal with nonabelian. Then .

Assume is cyclic. Suppose such that and is an abelian group. Suppose further that . Then the action of on is Frobenius.
Proof.
By Lemma 2.3, we have that is the unique minimal characteristic subgroup of . In particular, is not nilpotent and hence Lemma 2.2 applies to the group . Statement (a) clearly follows from Lemma 2.2(a).
We now work to prove (b). Let be as stated and write . Now, Lemma 2.3 implies that is the unique minimal characteristic subgroup of . Thus, is not nilpotent and we may apply Lemma 2.2 to the group .
Since is abelian, we have that is abelian. Hence . Now divides which is a power of . By Lemma 2.2(a) applied to , we have that is a Sylow subgroup of . Thus it cannot be that . Hence .
Now, , so we have , where the final inequality follows by the minimality of . Thus, and since , we have as desired.
Finally, we prove (c). Let and write . Note that since we have assumed , it follows that . We want to show that . Assume otherwise. Since is cyclic, note that .
Consider the action of on . Now is an abelian group and does not divide . Thus this action is coprime and we may apply Fitting’s Theorem. Write and so that . Note that since and are uniquely determined by subgroups normal in we have that .
Now acts trivially on , so . Since is cyclic it follows that is abelian. Let be a complement for . In particular, is a normal Hall subgroup of , and since it follows that .
Now , and so as modules. We next want to show that acts nontrivially on . It suffices to show that acts nontrivially on . Assume otherwise. Then we have and in particular,
This is a contradiction, and hence acts nontrivially on , as desired. Hence is nonabelian.
Let with , where is chosen maximal with nonabelian. Again by Lemma 2.3, we may apply Lemma 2.2 to the group . Write . Since is abelian, is abelian, so . Now , and in particular, divides , which is a power of . From Lemma 2.2 (a) applied to we conclude that , and thus . In particular, . Now, , where the last inequality follows by the minimality of . Hence , which contradicts the choice of . ∎
We obtain further information in this situation.
Lemma 2.5.
Assume is solvable and is not nilpotent. Let be maximal with and not nilpotent. Write . Then

is the unique subgroup of minimal with the property of being normal in

is an elementary abelian Sylow subgroup of for some prime

If is the size of a nontrivial orbit of the action of on , then .
Proof.
For (a), let with . Then by choice of , it must be that is nilpotent, and so . Hence , as desired.
It follows by (a) that is the unique minimal characteristic subgroup of . Hence (b) and (c) are immediate from Lemma 2.2 (a) and (c), respectively. ∎
We now consider a cyclic group acting semiregularly on the basis of a vector space.
Lemma 2.6.
Let act on , where is a cyclic group and is a vector space. Assume has a basis which is permuted semiregularly by . Write and let be a divisor of . Then there exists an orbit in of size .
Proof.
Since has a basis which is permuted semiregularly by , we can find a linearly independent set which is a complete orbit. Write . Choose notation so that for we have and .
Write . Since , we will be done if we can construct a vector with . Define
Note that since we have . Conversely, assume . Then since the are linearly independent, must be of the form for some with . In particular, we see that for some , and hence, , as desired. ∎
3 A unifying theorem
We now consider a useful generalization of the prime hypothesis, which we define here and will use several times.
First, we must define some notation. If is a positive integer and is a set of primes, write for the part of . Also, write for the complement of in the set of all primes. If is a set of positive integers, write .
Definition: Let be a set of positive integers, and let be a set of primes. Then satisfies the “” if whenever with , then . If satisfies this hypothesis, we will abbreviate by saying does. We will refer to the “” as the “”.
Note that the is equivalent to the prime hypothesis. Also, if a group satisfies the for some set of primes, then there is no information that can be obtained about the parts of elements of . The question then becomes: if a group satisfies the , can we bound the number of parts of character degrees in terms of ? Here are two results of this type that were proved in [2].
Theorem D.
Let be solvable and assume satisfies the . Then .
Theorem E.
Let be solvable and nonabelian, and assume satisfies the . Let be maximal with nonabelian. Write and assume is nilpotent. Then
The following theorem unifies many of the remaining cases, and allows us to apply the results of Section 6 of [2].
Theorem 3.1.
Let be a finite solvable group. Assume:

satisfies the twoprime hypothesis

are normal subgroups of , where is a Frobenius group with Frobenius kernel

is cyclic of order and is an elementary abelian group with


is abelian for all primes

For every with and , we have that the action of on is Frobenius

does not divide .
Then is nilpotent.
Proof.
Assume is not nilpotent. We will take a normal subgroup of , contained in , which is chosen maximal with a nonnilpotent factor. Using results from Section 2, we work towards a contradiction.
Let , where and is chosen maximal such that is not nilpotent. Write . Now, Lemma 2.5 applies to the group . From Lemma 2.5 (a), we have that is the unique subgroup of which is minimal with the property of being normal in . Also, by Lemma 2.5 (b), we conclude that is an elementary abelian Sylow subgroup of for some prime . Finally, by Lemma 2.5 (c), we have that if , then . Since is nilpotent, we may write , where and .
Step 1. is abelian.
Proof.
By assumption (5), it follows that is abelian for all primes . Since is nilpotent, we have
Now does not divide , and hence . ∎
Step 2. Assume with . Then .
Proof.
Assume otherwise. Write . Then . Since , it follows that . By the uniqueness of , we must have .
Now is a normal, central Sylow subgroup of . Hence, if we write , it follows that . On the other hand, , and since , we must have . However, , which contradicts the uniqueness of . ∎
Step 3. Assume with . Then does not divide .
Proof.
Since satisfies the twoprime hypothesis, it follows by Lemma (3.6) of [2] that satisfies the hypothesis. Since , the hypothesis implies that . Now, since , this implies , as desired. ∎
Step 4. Let . Then extends to .
Proof.
Write and . Now, since is a Sylow subgroup of , we have that is coprime to . In particular, if we write for the order of in the group , it must be that is coprime to . Then it follows by Corollary (6.28) of [4] that has a unique extension with .
Since uniquely determines , and certainly determines , we have that . Now is isomorphic to , which is a subgroup of . Hence is cyclic. Thus by Corollary (11.22) of [4], we have that extends to . Since is an extension of , we are done. ∎
Step 5. Let be nonprincipal. Then .
Proof.
Assume otherwise. Then in particular, .
We first work to show that is an irreducible character degree of . Note that this is trivial if . Hence, we may assume that .
We have , so . In particular, . Since is a Frobenius group with kernel , it follows that is a Frobenius group with kernel . Therefore, is an irreducible character degree of , as desired. Write .
Next, we prove that . By Step 4, we have that extends to . Now, Gallagher’s Theorem (Corollary 6.17 of [4]) implies that since we may choose with . By the Clifford correspondence (Theorem 6.11 of [4]), we have . Thus
as desired.
We now show that divides . Since is nilpotent and is a Sylow subgroup of , it suffices to show that . Assume otherwise. Then in the action of on there is a nontrivial fixed point. By Lemma 2.1, it follows that the action of on has a nontrivial fixed point. This contradicts Step 2 since .
Finally, we have . This character degree is divisible by both and , which contradicts Step 3. The contradiction arose from our assumption that , and this completes the proof of the step. ∎
Step 6. Assume and let be a Frobenius group. Suppose is the kernel of and assume . Assume . Then .
Proof.
Assume . Since , Step 2 implies that acts nontrivially on . Hence, Theorem 15.16 of [4] applies. We obtain a basis for which is permuted semiregularly by a Frobenius complement of . Viewing as a vector space over , we may appeal to Lemma 2.6.
Let , and write . By Lemma 2.6, there exists an orbit in of size . In view of Lemma 2.1, we must also have which lies in an orbit of size . Recall that is a group. Now, Lemma (2.5) of [2] implies that for some . By Step 2, we cannot have . Thus, we must have since .
By Step 4, we have that extends to . Using the Clifford Correspondence (Theorem 6.11 of [4]), we obtain . Now, consider . Since , we have divides . Now . But divides , which contradicts Step 3. ∎
Step 7. We cannot have both a Frobenius group (with kernel ) and with .
Proof.
First, we claim that . Assume this is not true, and let . By Lemma 2.5 (c), lies in a orbit of some size with . Now, we are assuming , and so . In particular, . By Step 3, does not divide . However, , so does divide . This is a contradiction. Hence, .
Assume is a Frobenius group with kernel , and assume there exists with . Write . Now, extends to by Step 4, so by Clifford Correspondence. Since , we have divides . It follows by Step 3 that .
Let . Then, by our arguments above. We obtain by assumption (4).
Note that , and so, is divisible by all the primes in . Let . Now, , and so, divides . However, is a group and by the way was defined. Therefore, .
We have , and hence, is a Frobenius group with kernel . In particular, , and so is a Frobenius group with kernel . We now have a contradiction with Step 6. Since is abelian, it follows that . Also, . Hence all the hypothesis of Step 6 are satisfied, but we have , which is the desired contradiction. ∎
Step 8. Assume . Then .
Proof.
Assume otherwise. By Step 4, . Hence, , so . Suppose . Recall that by Step 2 fixes no nontrivial element of . By Lemma 2.1, fixes no nonprincipal character of . However, fixes , which is a contradiction. We deduce that . In particular, and is a Frobenius group with kernel . We also see that . This contradicts Step 7. ∎
Step 9. All nontrivial orbitsizes of the action of on are divisible by . This action has at most two nontrivial orbitsizes. If there are two orbit sizes, then and one orbitsize must be .
Proof.
By Lemma 2.1, it will suffice to consider the action of on . Let . Applying Step 8, , so divides . In particular, divides , and this proves the first statement.
Since is a normal Sylow subgroup of , we have . Now, extends to by Step 4, and thus, is the part of some element of by the Clifford Correspondence (Theorem 6.11 of [4]).
Using Lemma (3.6) of [2], satisfies the hypothesis. Note that by assumption (7). Thus, Lemma (3.4) of [2] implies that satisfies the twoprime hypothesis. In particular, there are at most two elements of this set divisible by , as desired. Furthermore, only one element of can be divisible by . Finally, since divides each orbit size, the only way there can be two orbit sizes is if and one of the orbits has size . ∎
Step 10. The action of on has exactly two nontrivial orbitsizes.
Proof.
Assume this is false. The action of on is nontrivial by Step 2. By Step 9, the action has at most two orbit sizes. It must be that acts transitively on . Since and is a Frobenius group, clearly we have . Hence, is cyclic by Lemma 3 from [5].
Consider the action of on . This is a coprime action on an abelian group, so we may apply Fitting’s theorem. This gives a direct product decomposition of a cyclic group, so one of the factors must be trivial since there is a unique subgroup of order . This action is not trivial, hence, . Since is cyclic, we conclude that must be a Frobenius group with kernel .
Every subgroup of is characteristic in , hence normal in . In view of Step 2, if , then . This contradicts Step 7. ∎
Combining Steps 9 and 10, we obtain .
Step 11. and .
Proof.
By Step 10, there exist characters which lie in nontrivial orbits of different sizes. Write and , and Step 9 implies that one of and must equal 2. Without loss we may assume .
Assume , and let . Then divides for all by Lemma 2.5 (c). In particular, divides both and . Since and extend to their inertia groups in , we obtain with , , and both and are divisible by . Therefore, .
From hypothesis (1), we know that satisfies the twoprime hypothesis. Thus, satisfies the hypothesis by Lemma (3.6) of [2]. This implies . Since , we have , which is a contradiction. ∎
Step 12. has an orbit of size on .
Proof.
By Steps 9, 10, and 11, it follows that has an orbit of size on . From Lemma 2.1 we also have that has an orbit of size on . Thus, we may let with . By Step 8, . We obtain
where the second equality follows by the observation before Step 11. ∎
Step 13. is elementary abelian.
Proof.
Write . Since is a group, we have is elementary abelian, and it suffices to show that .
By Step 12 and Lemma 2.1, let such that has index in . In particular, is a maximal subgroup of , and thus .
Now is a characteristic subgroup of , and hence, is normal in . Also, since , we have by Step 2 that , as desired. ∎
Step 14. has rank 2.
Proof.
Since has an orbit on of size divisible by , it must be that . Hence, . Note that acts faithfully on by Step 2 since .
Assume the rank of is 3 or greater. Then by Lemma 2.6 from [11], we obtain at least three distinct nontrivial orbitsizes of the action of on , and we have three nontrivial orbitsizes of the action of on by Step 8. This contradicts Step 10. ∎
Step 15. is nonabelian.
Proof.
Assume otherwise. Apply Fitting’s lemma to the action of on to obtain where . Since “fixed points come from fixed points” in a coprime action, . By assumption (6), is a Frobenius group.
Now, is abelian, so let be a complement for in . Then is a Frobenius group with kernel . This contradicts Step 6. ∎
We now work for the final contradiction. By Steps 14 and 15, it follows that is extraspecial of order . Write . Observe that is a module and is a submodule. Also, does not divide by assumptions (2), (3), and (7). By Maschke’s Theorem, we may let be a complement for in .
First, assume is abelian. Let be a complement for in , and is a Frobenius group. Consider the action of on . Certainly this action is nontrivial since acts Frobeniusly on . Applying Fitting’s lemma and the fact that is the only nontrivial subgroup of which is normal in , we have . In particular, since is cyclic, for all subgroups with .
Consider , and write . By the previous paragraph, and thus, is a Frobenius group with kernel . This contradicts Step 6.
We now have is nonabelian. In particular, , and so, . By assumption (6), is a Frobenius group.
We can replace by . Let . Then Step 8 implies that . Hence, . Applying Step 2, we cannot have . Thus, , and hence, . We deduce that is a Frobenius group with kernel . The Frobenius complement of is an extraspecial group, which is impossible by 12.6.15 of [12]. This is our final contradiction. ∎
4 The Large Frobenius Case
Let be a group and let . We write and .
We can associate a graph with . We define the graph to be the graph whose vertex set is . There is an edge between and if . It has been proved that has at most two connected components when is solvable (see Theorem 30.2 of [3] or Theorem 18.4 of [9]).
Theorem 4.1.
Assume is solvable satisfying the twoprime hypothesis. Let and suppose that is a Frobenius group with kernel . Assume that is chosen among all such normal subgroups of so that is minimized. Suppose is cyclic and that . Assume further that either is elementary abelian of rank or is a direct product of two elementary abelian groups for different primes. Then .
Proof.
Write