Brauer character degrees and Sylow normalizers

If p and q are primes, and G is a p-solvable finite group, it is possible to detect that a q-Sylow normalizer is contained in a p-Sylow normalizer using the character table of G. This is characterized in terms of the degrees of p-Brauer characters. Some consequences, which include yet another generalization of the Itô–Michler theorem, are also obtained.

1 3 recurrent problems in character theory. Recently, in [12], we have asked if it is possible to characterize group-theoretically when Irr (G) = Irr (G) for sets of primes and .
In this paper, we fix a prime p, and we turn our attention to p-Brauer characters, within the universe of finite p-solvable groups. (Degrees of modular representations in characteristic p outside p-solvable groups are usually deemed an intractable subject.) If G is a p-solvable finite group and IBr(G) is the set of the irreducible p-Brauer characters of G, we let IBr (G) be the subset of ∈ IBr(G) such that all the primes dividing (1) lie in . As usual, if r is a prime, r ′ denotes the set of primes different from r.
The following is our first main result.
Theorem A Let G be a finite p-solvable group, and let q be a prime. Then, IBr q � (G) ⊆ IBr p � (G) if and only if there are Q ∈ Syl q (G) and P ∈ Syl p (G) such that Theorem A was the main result of [1], assuming that G is both p-solvable and q-solvable. To better understand our this new situation, we invite the reader to consider, for instance, G = PSL 2 (3 5 ).5 , for p = 5 . If q = 2 , then G (P) = G (Q) for some p ∈ Syl p (G) and Q ∈ Syl q (G) . Of course, G is p-solvable, but not q-solvable. There are families of almost-simple groups like this, and we are able to deal with this new situation thanks to the main result of [6]. In particular, the proof of Theorem A depends on the Classification of Finite Simple Groups.
From Theorem A, and using McKay bijections for Brauer characters in p-solvable groups together with the recent proof of the divisibility of degrees between Glauberman correspondents by M. Geck in [2], we can prove our second main result.
Theorem B Let G be a finite p-solvable group, and let q be a prime different from p. Then IBr p � (G) = IBr q � (G) if and only if there are P ∈ Syl p (G) and Q ∈ Syl q (G) such that G (P) = P G (Q) and Q is abelian.
As the reader can easily check, in the trivial case where p does not divide |G|, Theorem B is yet another restatement of the Itô-Michler theorem.
As happens with complex irreducible characters (see [12]), it does not seem easy to group-theoretically characterize when IBr (G) = IBr (G) for arbitrary sets of primes and , even if G is p-solvable. In the case, for instance, where = ℙ is the set of all primes and = q � , where q is a prime, this constitutes Problem 3.2 of [9]. (This problem was studied long before in [8] and more recently in [5].) It has now been conjectured that IBr q � (G) = IBr ℙ (G) if and only if the number of p-regular classes of G is the number of p-regular classes of G (Q)∕Q � , where Q ∈ Syl q (G) . This is a consequence of the Inductive McKay conjecture, and it seems difficult to obtain a direct proof. (See Conjecture D in [7].)

Proof of theorem A
Our notation for Brauer characters follows [9]. The deepest part of the proof of Theorem A comes from the main result in [MN].
Proof This is Theorem A of [6]. ◻ A very useful result to deal with the hypotheses in this paper appears in Suzuki's book.
Theorem 2.2 Let G be a p-solvable group, and let q be a prime. Let = {p, q} . Then, G has a unique conjugacy class of Hall -subgroups, and every -subgroup of G is contained in one of them.
Proof This follows from 5.3.13 of [14]. ◻ For the reader's convenience, let us prove the following standard result.

Lemma 2.3
Let G be a finite group. Let Q be a Sylow q-subgroup of G and let N be a normal subgroup of G. If ∈ IBr q � (G) , then N has a Q-invariant irreducible constituent and any two of them are G (Q)-conjugate.
Proof Let ∈ IBr q � (G) and let ∈ IBr(N) be an irreducible constituent of N . Let G be the stabilizer of in G. By Clifford's theorem (see Corollary 8.9 of [9], for instance), we have that |G ∶ G | divides (1) , and hence, it is a q ′ -number. It follows that there is an element g ∈ G such that Q g is contained in G . Hence, Q ⊆ G g −1 = G g −1 and g −1 is a Q-invariant irreducible constituent of N . Now, suppose that and are two Q-invariant irreducible constituents of N . Again by Clifford's theorem we have that there is an element g ∈ G such that = g . It follows that Q and Q g are contained in G , and hence, there is an element x ∈ G such that Q = (Q g ) x . Therefore gx ∈ G (Q) and gx = ( g ) x = x = , as wanted. ◻ The following result implies Theorem A. In its proof, we shall use Fong characters of Brauer characters, and we refer the reader to Chapter 10 of [9] for their main properties. (The term Fong character was coined by I. M. Isaacs after some results of P. Fong.) We also use the fact that if G is p-solvable, H is a p-complement of G, and ∈ IBr(G) has degree not divisible by p, then H ∈ Irr(H) . (See Theorem 10.9 of [9].) A complication when dealing with Brauer characters is that we do not have any form of Frobenius reciprocity, even in favorable conditions. For instance, in the previous situation where H is a p-complement of a p-solvable group G, if ∈ Irr(H) , and is an irreducible constituent of p ′ -degree of the induced Brauer character G , then needs not be the irreducible character H ; a fact that would simplify our proof below. (If G = 4 , p = 2 , and H = 3 , then the Brauer character (1 H ) G = 21 G + 1 + 2 , where i are distinct linear Brauer characters. Now take = 1 H and = i .) Theorem 2.4 Let G be a p-solvable group, let IBr(G) be the set of irreducible Brauer characters of G, and let q ≠ p be a prime. Let P ∈ Syl p (G) and Q ∈ Syl q (G) such that U = PQ is a subgroup of G. Suppose that H is a p-complement of G containing Q. Then, the following are equivalent.

Proof Set
= {p, q} . By Theorem 2.2, notice that we can find P ∈ Syl p (G) and Q ∈ Syl q (G) such that U = PQ is a Hall -subgroup of G. (In fact, given a Hall -subgroup U of G, then U = PQ for every P ∈ Syl p (U) and Q ∈ Syl q (U).) Assume (a). We prove (b) by induction on |G|. If K is a minimal normal subgroup of G, then we have that G (Q) ⊆ G (P)K , by using induction in G/K. Since G is p-solvable, then K is either a p-group or a p ′ -group. In the first case, G (P)K = G (P) and we are done. So we assume that K is a p ′ -group.
Let V be any subgroup of G containing KQ and let ∈ IBr q � (V) . By Lemma 2.3, there exists a Q-invariant irreducible constituent ∈ IBr(K) of the restriction K (using that Q ∈ Syl q (V) ). We claim that is also P-invariant. By Corollary 8.7 of [9], we have that is an irreducible constituent of the induced Brauer character V . Now consider the Brauer character G , which has degree |G ∶ V| (1) , which is not divisible by q. Hence, there exists an irreducible constituent ∈ IBr(G) of G of degree not divisible by q. By hypothesis, we have that has degree not divisible by p. Now, is an irreducible constituent of G , and therefore, is an irreducible constituent of the restriction K , again by Corollary 8.7 of [9]. If I = G is the stabilizer of in G, and ∈ IBr(I) is the Clifford correspondent of over , we have that |G : I| is a ′ -number. Therefore, using Theorem 2.2, we have that U ⊆ I g for some g ∈ G . Then, and g are Q-invariant constituents of , and thus, by Lemma 2.3, we have that g = x , for some x ∈ G (Q) ⊆ K G (P) . Hence, we may assume that g = y for some y ∈ G (P) . Then, I gy −1 = I , U y −1 ⊆ I , and we conclude that P = P y −1 ⊆ I . In other words, is P-invariant, as claimed. Now, let V = G (P)K . We claim that IBr q � (V) ⊆ IBr p � (V) . Let ∈ IBr q � (V) . By the claim in the previous paragraph, let ∈ IBr(K) be PQ-invariant under . By Theorem 8.11 of [9], there is a unique ̂∈ IBr(KP) over , extending , and we conclude that lies over ̂ . Now, KP ⊲ V , V/KP is a p ′ -group, and we have that (1)∕̂(1) divides |V : KP| by Theorem 8.30 of [9]. Since (1) is not divisible by p, we conclude that has p ′ -degree, as claimed. By induction, we may assume that V = G . That is, KP ⊲ G . We want to use Theorem 2.1.
Next, we show that every ∈ Irr q � (H) extends to G. By Lemma 2.3, let ∈ Irr(K) be Q-invariant under . By the claim in the third paragraph, we have that is P-invariant too. By Corollary 6.28 of [3], we have that has a canonical extension ∈ Irr(KP) . Using the uniqueness of , we easily check that G ∩ H = H . By Isaacs restriction theorem (Lemma 6.8(d) of [11]), the Clifford correspondence and Mackey's theorem (Theorem 1.16 of [10]), we have that extends to G. If 0 = p � is the set of primes dividing |G| different from p, by Theorem 2.1 applied to 0 , we conclude that there is Q 1 ∈ Syl q (H) such that G (Q 1 ) ⊆ G (P) . In particular, P ⊲ PQ 1 . By Theorem 2.2, PQ 1 and PQ are G-conjugate. Hence P ⊲ PQ and therefore Q, Q 1 ∈ Syl q ( G (P)) . Hence, Q z 1 = Q for some z ∈ G (P) , and G (Q) = G (Q 1 ) z ⊆ G (P).

Proof of theorem B
If N is a normal subgroup of G and ∈ IBr(N) , we write IBr(G | ) to denote the set of irreducible Brauer characters of G such that is an irreducible constituent of N . We write IBr p � (G | ) = IBr p � (G) ∩ IBr(G | ).
Next is the version for Brauer characters and p-solvable groups of Theorem A of [13], which we shall need to prove Theorem B. It is worth mentioning that its proof uses the recent proof of the divisibility of the degrees of the Glauberman correspondence in [2].

Theorem 3.1 Let G be a p-solvable group, and P ∈ Syl p (G) , then there is a bijection
Proof We argue by induction on |G|. Since p (G) is in the kernel of every ∈ IBr(G) , by induction we may assume that p (G) = 1 and hence K = p � (G) > 1 . Let S∕K = p (G∕K) and notice that P 0 = P ∩ S is a Sylow p-subgroup of S. By the Frattini argument, we have that G = K G (P 0 ) . Notice also that G (P 0 ) < G since p (G) = 1.
Let 1 , … , s be a complete set of representatives of the orbits of the action of G (P) on the P-invariant irreducible characters of K. By Lemma 2.3, we have that is a disjoint union. Fix i ∈ Irr(K) , P-invariant, and observe that i is also P 0 -invariant. Let * i ∈ Irr( K (P 0 )) = IBr( K (P 0 )) be the Glauberman correspondent of i and let T i = G i be the stabilizer of i in G. Since the Glauberman correspondence and the action of G (P 0 ) commute (see Lemma 2.10 of [10]), it follows that T i (P 0 ) = T i ∩ G (P 0 ) is the stabilizer of * i in G (P 0 ) . By Dade's theorem (see Theorem (6.5) of [15]), we have that (T i , K, i ) and ( T i (P 0 ), K (P 0 ), * i ) are isomorphic character triples. In particular, there is a bijection Δ ∶ Irr( . Now using Δ and Lemma 3.12 of [4] we can construct a bijection * ∶ IBr( is a bijection. Let ∈ IBr p � (G) and let i ∈ IBr(K) = Irr(K) be such that ∈ IBr p � (G | i ) . By the Clifford correspondence (Theorem 8.9 of [9]), we have that there is ∈ IBr p � (T i | i ) such that G = . Since T i (P 0 ) is the stabilizer of * i in G (P 0 ) , again by the Clifford correspondence we have that ( * ) G (P 0 ) ∈ Irr( G (P 0 )) . Furthermore, since P ⊆ T i (P 0 ) , we have that is a p ′ -number. Hence, we define Since 1 , … , s is a complete set of representatives of the action of G (P) on the P-invariant characters of K and the Glauberman correspondence commutes with the action of G (P) ⊆ G (P 0 ) , we have that * 1 , … , * s is a complete set of representatives of the action of G (P) on the P-invariant irreducible characters of Irr( K (P 0 )) = IBr( K (P 0 )) . By Lemma 2.3, we have that is a disjoint union and it follows that g is a bijection.

Theorem 3.2 Suppose that G is a p-solvable finite group and let q be a prime different from p. Then
if and only if there is a Sylow p-subgroup P of G and a Sylow q-subgroup Q of G, such that G (P) = P G (Q) and Q is abelian.
By the Itô-Michler theorem, we know that G/P has a normal and abelian Sylow q-subgroup PQ/Q. Hence, Q is abelian and PQ ⊲ G . Then G = P G (Q) by the Frattini's argument. Conversely, suppose that there is P ∈ Syl p (G) and Q ∈ Syl q (G) such that G (P) = P G (Q) and Q is abelian. Notice that since G (Q) ⊆ G (P) by Theorem 2.4 we have that IBr q � (G) ⊆ IBr p � (G) . We only need to prove the reverse containment. If G (P) < G , arguing by induction, we have that IBr q � ( G (P)) = IBr p � ( G (P)) . Let ∈ IBr p � (G) and let f ∶ IBr p � (G) → IBr p � ( G (P)) be the bijection given in Theorem 3.1. Hence (1)∕f ( )(1) divides |G ∶ G (P)| which is not divisible by q. Since IBr q � ( G (P)) = IBr p � ( G (P)) , we have that f ( )(1) is not divisible by q and thus ∈ IBr q � (G). Hence, we may assume that P is a normal subgroup of G and then IBr(G) = Irr(G∕P) . Also G = G (Q)P and hence PQ is normal in G. Then, PQ/P is an abelian normal Sylow q-subgroup of G/P. By Itô's theorem, we have that Irr(G∕P) = Irr q � (G∕P) . It follows that and we are done. ◻ To prove the assertion in the Abstract, it is enough to notice that in p-solvable groups, the ordinary character table of G uniquely determines the Brauer characters of G, by using the Fong-Swan theorem. (See Theorem 10.1 and Corollary 10.4 of [9].) Finally, using Isaacs -characters, the Glauberman-Isaacs correspondence, and some ad hoc arguments, it is possible to replace in Theorems A and B of this paper p ′ by , p-solvable groups by -separable groups, and Sylow p-subgroups by Hall -complements.
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