Abstract
If G is a finite group, we have proposed three new conjectures on the interaction between different primes and their corresponding Brauer principal blocks. In this paper,we give strong support to the validity of Conjectures B and C.
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1 Introduction
Meaningful interaction between the representation theory of finite groups from the perspective of different primes is extremely rare. However, in [26], we proposed the following three plausible conjectures, which extended work of several authors (see [3, 16, 19, 29]).
If p is a prime and G is a finite group, we denote by B_{p}(G) the principal pblock of G. The main subject of our work is the set \(\text {Irr}_{p^{\prime }}(B_{p}(G))\) of the irreducible complex characters in the principal pblock of G whose degree is not divisible by p. This set seems to possess remarkable properties.
Conjecture A
Let G be a finite group and let p and q be different primes. If
then there are a Sylow psubgroup P of G and a Sylow qsubgroup Q of G such that xy = yx for all x ∈ P and y ∈ Q.
Conjecture B
Let G be a finite group and let p and q be primes dividing the order of G. If \(\text {Irr}_{p^{\prime }}(B_{p}(G))=\text {Irr}_{q^{\prime }}(B_{q}(G))\), then p = q.
Conjecture C
Let G be a finite group, and let p and q be different primes. Then q does not divide χ(1) for all \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(G))\) and p does not divide χ(1) for all \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(G))\) if and only if there are a Sylow psubgroup P of G and a Sylow qsubgroup Q of G such that xy = yx for all x ∈ P and y ∈ Q.
The main result of [26] was to reduce Conjecture A to a problem on almost simple groups and to prove it in the case that one of the primes is 2. To prove Conjecture A for almost simple groups in the case where p and q are both odd remains quite a challenge.
In the present paper, we focus on Conjectures B and C. Using the Classification of Finite Simple Groups, in our first main theorem we prove the following.
Theorem D
Conjecture C implies Conjecture B.
After proving Theorem D, therefore, we concentrate our efforts in the remainder of the paper towards Conjecture C. Since the “if” direction of Conjecture C follows from “if” direction of the main result [19], we shall only focus on the “only if” direction.
Theorem E
Conjecture C holds for finite simple groups.
Besides Conjecture C being true for simple groups, in Theorem 3.3 below, we shall also prove that Conjecture C is true for psolvable groups, assuming the inductive Alperin–McKay condition. This gives strong support to the validity of this conjecture.
Unfortunately, at the time of this writing, we still do not know how to reduce Conjecture C to a question on almost simple groups.
2 Theorem D
In this Section we prove that Conjecture C implies Conjecture B. This will require the following result on simple groups.
Theorem 2.1
Let p, q be different primes and let S be a nonabelian simple group with pq∣S. Assume that [P,Q] = 1 for some Sylow psubgroup P of S and a Sylow qsubgroup Q of S. Then one of the following holds:

(a)
There exists \(\alpha \in \text {Irr}_{p^{\prime }}(B_{p}(S)) \text {Irr}(B_{q}(S))\) which is Aut(S)_{p}invariant, where Aut(S)_{p} is some Sylow psubgroup of Aut(S).

(b)
There exists \(\alpha \in \text {Irr}_{q^{\prime }}(B_{q}(S)) \text {Irr}(B_{p}(S))\) which is Aut(S)_{q}invariant, where Aut(S)_{q} is some Sylow qsubgroup of Aut(S).
Proof
First, note that by [19, Lemma 3.1], the condition [P,Q] = 1 implies that \([\bar {P}, \bar {Q}]=1\) for some Sylow p and q subgroups \(\bar {P}, \bar {Q}\) of any covering group of S. If S is one of the sporadic groups J_{1} or J_{4}, then we may use GAP [8] and its Character Table Library to see that the statement holds. Then [19, Propositions 3.2–3.4] further imply that we may assume S to be a simple group of Lie type that is not isomorphic to a sporadic or alternating group and is defined in characteristic r_{0}∉{p,q}. (We remark that this information was first found in [2].)
So, let S be of the form S = G/Z(G) for G a group of Lie type of simply connected type defined in characteristic r_{0}≠p,q. In this case, we may further assume that p and q are both odd and that the Sylow p and qsubgroups of G are abelian, using [19, Proposition 3.5] (see also [2, Theorem 2.1]).
Since p and q are odd, note that we may therefore assume without loss that p ≥ 5. Furthermore, using [18, Lemma 2.1 and Proposition 2.2], we have p is good for G and \(p\nmid \mathbf {Z}(G)\). This implies that p does not divide the order of any diagonal or graph outer automorphism, so that Aut(S)_{p} may be taken as a subgroup of \(S\rtimes \langle F\rangle \), where F is a generating field automorphism.
Now, let t be a pelement in G^{∗} whose G^{∗}class is 〈F〉_{P}invariant. Then the semisimple character χ_{t} of G lies in B_{p}(G) by [12, Corollary 3.4] and has degree prime to p since \(\chi _{t}(1)=[G^{\ast }:\mathbf {C}_{G^{\ast }}(t)]_{r_{0}^{\prime }}\) (see e.g. [5, Theorem 8.4.8]) and Sylow psubgroups of G^{∗} are abelian. But since χ_{t} lies in the Lusztig series \(\mathcal {E}(G, t)\) and B_{q}(G) contains only characters lying in Lusztig series \(\mathcal {E}(G, s)\) with s a power of q (see [4, Theorem 9.12]), we see that χ_{t} does not lie in Irr(B_{q}(S)). Furthermore, as \(p\nmid \mathbf {Z}(G)\), we have \(t\in O^{p^{\prime }}(G^{\ast })\), and hence χ_{t} is trivial on Z(G). Finally, since \(p\nmid \mathbf {Z}(G)\), we also have \(\mathbf {C}_{\mathbf {G}^{\ast }}(t)\) is connected, where G^{∗} is the ambient reductive group whose fixed points under an appropriate Frobenius endomorphism yields G^{∗} (see [21, Exercise 20.16]). Then for φ ∈〈F〉, we have \(\chi _{t}^{\varphi }=\chi _{t^{\varphi ^{\ast }}}\), where φ^{∗} is an appropriate field automorphism of G^{∗} (see [28, Corollary 2.5]). Hence χ_{t} is Aut(G)_{P}invariant by our choice of t. □
Lemma 2.2
Suppose that N is a minimal normal subgroup of G, which is a direct product of the different Gconjugates of a nonabelian simple group S. Let P ∈Syl_{p}(G). Suppose that α ∈Irr(S) is Aut(S)_{p}invariant, where Aut(S)_{p} is some Sylow psubgroup of Aut(S). Then there are g_{i} ∈ G, h_{i} ∈ P and \(\sigma _{i} \in \text {Aut}(S^{g_{i}})\) such that \((\alpha ^{g_{1}})^{\sigma _{1}h_{1}} \times {\cdots } \times (\alpha ^{g_{m}})^{\sigma _{m} h_{m}} \in \text {Irr}(N)\) is Pinvariant.
Proof
Suppose that N is the direct product of Ω = {S^{g}  g ∈ G}. Now, write
where \(\mathcal {O}(S_{i})\) is the Porbit of some \(S_{i}=S^{x_{i}}\), for some x_{i} ∈ G. Then N = N_{1} ×⋯ × N_{t}, where N_{i} is the product of the elements in \(\mathcal {O}(S_{i})\). Let us fix an i until the end of the proof. Of course, N_{i} is Pinvariant. Let \(\alpha _{i}=\alpha ^{x_{i}} \in \text {Irr}(S_{i})\). Suppose that \(\{S_{i}^{y_{1}}, \ldots , S_{i}^{y_{r}}\}\) are the different Pconjugates of S_{i}, where y_{j} ∈ P. Hence \(N_{i}=S_{i}^{y_{1}}\times {\cdots } \times S_{i}^{y_{r}}\), and
is a disjoint union.
By hypothesis, α is invariant under X ∈Syl_{p}(Aut(S)). Therefore α_{i} is invariant under \(X_{i}=X^{x_{i}} \in \text {Syl}_{p}(\text {Aut}(S_{i}))\). Let M_{i} = N_{G}(S_{i}) and C_{i} = C_{G}(S_{i}). We have that M_{i}/C_{i} embeds into Aut(S_{i}). Then N_{P}(S_{i})C_{i}/C_{i} is a psubgroup of Aut(S_{i}), and therefore, there is σ_{i} ∈Aut(S_{i}) such that \((\mathbf {N}_{P}(S_{i})C_{i}/C_{i})\subseteq X_{i}^{\sigma _{i}}\). Since α_{i} is X_{i}invariant, it follows that \(\beta _{i}=(\alpha _{i})^{\sigma _{i}}\in \text {Irr}(S_{i})\) is \(X_{i}^{\sigma _{i}}\)invariant, and therefore N_{P}(S_{i})invariant. We claim that \(\gamma _{i}=\beta _{i}^{y_{1}} \times {\cdots } \times \beta _{i}^{y_{r}}\in \text {Irr}(N_{i})\) is Pinvariant. Indeed, if x ∈ P, then y_{k}x = w_{k}y_{σ(k)} for some w_{k} ∈N_{P}(S_{i}), 1 ≤ k ≤ r, and σ a permutation of S_{r}. Now, if u ∈ S_{i}, then
This proves that γ_{i} is Pinvariant. Hence γ = γ_{1} ×⋯ × γ_{t} ∈Irr(N) is Pinvariant. □
We will need the following wellknown result of J. Alperin and E. C. Dade.
Theorem 2.3
Suppose that N is a normal subgroup of G, with G/N a \(p^{\prime }\)group. Let P ∈Syl_{p}(G) and assume that G = NC_{G}(P). Then restriction of characters defines a natural bijection between the irreducible characters of the principals pblocks of G and N.
Proof
The case where G/N is solvable was proved in [1] and the general case in [7]. □
Theorem 2.4
Assume that the “only if” direction of Conjecture C is true for all finite groups. Let p and q be primes. Assume that G is a finite group of order divisible by p and q. If \(\text {Irr}_{p^{\prime }}(B_{p}(G))=\text {Irr}_{q^{\prime }}(B_{q}(G))\), then p = q.
Proof
We argue by induction on G. Assume that p≠q. By Conjecture C, we know that [P,Q] = 1 for some P ∈Syl_{p}(G) and Q ∈Syl_{q}(G).
Step 0. If \(1\neq N \lhd G\), then p divides N or q divides N.
Otherwise by [23, Theorem 9.9(c)]
and by induction we are done.
Step 1. Let L be a proper normal subgroup of G. Then G/L has order divisible by p or q.
Suppose that G/L has \(p^{\prime }\) and \(q^{\prime }\)order. We claim that \(\text {Irr}_{p^{\prime }}(B_{p}(L))=\text {Irr}_{q^{\prime }}(B_{q}(L))\). Indeed, let \(\theta \in \text {Irr}_{q^{\prime }}(B_{q}(L))\). Then there exists χ ∈Irr(B_{q}(G)) over 𝜃. Then χ has \(q^{\prime }\)degree by [15, Theorem 11.29] and therefore \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(G))\), by hypothesis. Therefore \(\theta \in \text {Irr}_{p^{\prime }}(B_{p}(L))\). By symmetry, the claim is proved. Therefore, Step 1 follows by using the inductive hypothesis.
Step 2. Let N be a minimal normal subgroup of G and suppose that N is an elementary abelian pgroup. Then G = C_{G}(N)C_{G}(Q).
Write M = C_{G}(N) and L = MC_{G}(Q). We have that \(Q\subseteq \mathbf {C}_{G}(P) \subseteq \mathbf {C}_{G}(N)=M\) and G/M is a \(q^{\prime }\)group. Also by the Frattini argument G = MN_{G}(Q) and \(L \lhd G\). Notice that G/L is a \(q^{\prime }\)group and a \(p^{\prime }\)group (because \(P \subseteq \mathbf {C}_{G}(Q) \subseteq L\)). Then we use Step 1.
Step 3. Let N be a minimal normal subgroup of G and suppose that N is abelian. Then G = C_{G}(N).
By Step 0, we may assume that N is a pgroup or a qgroup. By symmetry, assume that N is a pgroup. Write M = C_{G}(N). We prove first that B_{p}(G) is the only pblock of G covering B_{p}(M). Let B be a pblock of G covering B_{p}(M) and let D be a defect group of B. Then \(N\subseteq D\) by [23, Theorem 4.8] and \(\mathbf {C}_{G}(D)\subseteq M=\mathbf {C}_{G}(N)\). Then B is regular with respect to M ([23, Lemma 9.20]) and hence by [23, Theorem 9.19] we have that B = B_{p}(M)^{G} = B_{p}(G) by the third main theorem (see [23, Theorem 6.7]). Hence B_{p}(G) is the only pblock of G covering B_{p}(M). In particular, we have that \(\text {Irr}(G/M)\subseteq \text {Irr}(B_{p}(G))\).
Next, we prove that G = M. Recall that G/M is a \(q^{\prime }\)group, because [Q,N] = 1. By Step 2, we have G = MC_{G}(Q). Hence by Theorem 2.3, we have that the restriction map
is a bijection. We claim that \(\text {Irr}_{p^{\prime }}(G/M)=\{1_{G}\}\). Indeed, if \(\chi \in \text {Irr}_{p^{\prime }}(G/M)\), then \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(G))=\text {Irr}_{q^{\prime }}(B_{q}(G))\). Therefore, we have that χ_{M} is irreducible and since χ lies over 1_{M} we have χ_{M} = 1_{M}. Hence χ = 1_{G} by the injectivity of the restriction map. Thus \(\text {Irr}_{p^{\prime }}(G/M)=\{1_{G/M}\}\) and G = M by [26, Lemma 2.2].
Step 4. Let N be a minimal normal subgroup of G, then N is not abelian.
Suppose the contrary and assume without loss of generality that N is an elementary abelian pgroup, so by Step 3 we have G = C_{G}(N) and \(N\subseteq \mathbf {Z}(G)\). By [23, Theorem 9.10] we have that B_{p}(G/N) is the unique pblock of G/N contained in B_{p}(G). We claim that \(\text {Irr}_{p^{\prime }}(B_{p}(G/N))=\text {Irr}_{q^{\prime }}(B_{q}(G/N))\). Indeed, we have that
where we have used [23, Theorem 9.9] in the last equality. On the other hand, let \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(G/N))\), so \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(G))=\text {Irr}_{p^{\prime }}(B_{p}(G))\) and \(N\subseteq \text {Ker}(\chi )\). Since B_{p}(G/N) is the only pblock of G/N contained in B_{p}(G), we have that \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(G/N))\) and the claim is proven. By using the inductive hypothesis, we have that p does not divide G/N. Therefore, \(\{1_{G}\}=\text {Irr}(B_{p}(G/N))=\text {Irr}_{p^{\prime }}(B_{p}(G/N))=\text {Irr}_{q^{\prime }}(B_{q}(G/N))\) and \(q\nmid G/N\) by [26, Lemma 2.1]). Hence \(q\nmid G\) and this is a contradiction.
Step 5. Let N be a minimal normal subgroup of G, then pq divides N.
Suppose that N is a \(p^{\prime }\)group. We claim first that NQ does not have a normal qcomplement. Indeed, suppose the contrary and let \(X\lhd NQ\) be a normal qcomplement. Then N ∩ X is a normal qcomplement of N and by the minimality of N we have that either N ∩ X = 1 or N ∩ X = N. If N ∩ X = N, N is \(q^{\prime }\) and \(p^{\prime }\), contradiction with Step 0. If N ∩ X = 1 then N≅XN/X is a qgroup, which is a contradiction with Step 4. Therefore NQ does not have a normal qcomplement. By [13, Corollary 3] there is τ ∈Irr(B_{q}(QN)) nonlinear of \(q^{\prime }\)degree. Therefore \(1\ne \tau _{N}\in \text {Irr}_{q^{\prime }}(B_{q}(N))\). By [22, Lemma 4.3] we have that there is some \(\gamma \in \text {Irr}_{q^{\prime }}(B_{q}(G))\) lying over τ_{N}. By hypothesis, we have that γ is in the principal pblock of G, and therefore τ_{N} is in the principal pblock of N, which is a contradiction since N is a \(p^{\prime }\)group and τ_{N}≠ 1.
Final Step. If N is a minimal normal subgroup of G, then N is semisimple by Step 4. Suppose that N is a direct product of all the different Gconjugates of a simple group S of order divisible by pq. Suppose that (a) of Theorem 2.1 holds and let α be the character in \(\text {Irr}_{p^{\prime }}(B_{p}(S))\) (not in Irr(B_{q}(S))) which is Aut(S)_{p}invariant. Notice that any Gconjugate or Aut(G)conjugate of α is in the principal pblock and not in the principal qblock of S. By Lemma 2.2, there exists \(\tau \in \text {Irr}_{p^{\prime }}(B_{p}(N))\) which is Pinvariant, and such that each of its factors does not belong to the principal qblock. In particular, τ does not belong to the principal qblock of N. Now τ extends to PN by [14, Corollary 8.16]. By [22, Lemma 4.3] there is \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(G))\) lying over τ. Then \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(G))\) and thus \(\tau \in \text {Irr}_{q^{\prime }}(B_{q}(N))\), which is a contradiction. Assuming (b) in Theorem 2.1 and reasoning analogously we get again a contradiction. □
3 Conjecture C and pSolvable Groups
As of the writing of this article, we can only prove Conjecture C for psolvable groups by assuming the so called Inductive Alperin–McKay condition (for the prime q). We shall need the following.
Theorem 3.1
Suppose that N is normal in G, and let P ∈Syl_{p}(N). Assume that all the nonabelian simple groups involved in N satisfy the inductive Alperin–McKay condition. Then there is a bijection
such that for each \(\theta {\in } \text {Irr}_{p^{\prime }}(B_{p}(N))\), there is a bijection \(f_{\theta }: \text {Irr}(B_{p}(G)\theta ) \rightarrow \text {Irr}(B_{p}(\mathbf {N}_{G}(P))\theta ^{\ast })\) such that χ(1)/𝜃(1) = f_{𝜃}(χ)/𝜃^{∗}(1) for all χ ∈Irr(B_{p}(G)𝜃).
Proof
This is Theorem B and Theorem 7.1 of [27]. □
Let us remark that we shall only need Theorem 3.1 in the case where G/N is a \(p^{\prime }\)group.
In our proof, we shall also need a McKay divisibility theorem, which was made possible after M. Geck proved a remarkable conjecture on Glauberman correspondents [9].
In the following, we follow the proof of [30, Theorem A], and then use Geck’s result.
Theorem 3.2
Let G be a psolvable group and let P ∈Syl_{p}(G). Then there is a bijection
such that χ^{∗}(1) divides χ(1) and χ(1)/χ^{∗}(1) divides G : N_{G}(P).
Proof
We argue by induction on G. As in the proof of [30, Theorem A] we may assume that O_{p}(G) = 1, and hence \(K=\mathbf {O}_{p^{\prime }}(G) >1\).
Let S/K = O_{p}(G/K), and notice that P_{0} = P ∩ S is a Sylow psubgroup of S. By the Frattini argument we have that G = KN_{G}(P_{0}) and N_{G}(P_{0}) < G since O_{p}(G) = 1. Let 𝜃_{1},…,𝜃_{s} be a complete set of representatives of the orbits of the action of N_{G}(P) on the Pinvariant irreducible characters of K. By [25, Lemma 9.3], we have that
is a disjoint union. Fix 𝜃_{i}∈Irr(K) and observe that 𝜃_{i} is also P_{0}invariant. Let \(\theta _{i}^{\ast }\in \text {Irr}(\mathbf {C}_{K}(P_{0}))\) be the Glauberman correspondent of 𝜃_{i} (see [25, Theorem 2.9], for instance) and let T_{i} be the stabilizer of 𝜃_{i} in G. Since the Glauberman correspondence and the action of N_{G}(P_{0}) commute (see [25, Lemma 2.10]), it follows that \(\mathbf {N}_{T_{i}}(P_{0}) =T_{i}\cap \mathbf {N}_{G}(P_{0})\) is the stabilizer of \(\theta _{i}^{\ast }\) in N_{G}(P_{0}).
Again as in the proof of [30, Theorem A] we obtain a bijection
satisfying \(\psi (1)/\psi ^{\ast }(1)=\theta _{i}(1)/\theta _{i}^{\ast }(1)\). Hence we have that ψ^{∗}(1) divides ψ(1) by the main result of [9]. Moreover, since G = N_{G}(P_{0})K we have that \(K:\mathbf {C}_{K}(P_{0}) = T_{i}:\mathbf {N}_{T_{i}}(P_{0})\) and since \(\theta _{i}(1)/\theta _{i}^{\ast }(1)\) divides K : C_{K}(P_{0}) we conclude that ψ(1)/ψ^{∗}(1) divides \(T_{i}:\mathbf {N}_{T_{i}}(P_{0})\).
The remaining part of the proof proceeds exactly as in the proof of [30, Theorem A]. □
Theorem 3.3
Let G be a finite psolvable group. Assume that the inductive Alperin–McKay condition (for q) holds for every nonabelian simple group involved in G. Then Conjecture C is true for G. In particular, Conjecture C holds for {p,q}solvable groups.
Proof
By [19], we only need to prove that if all \(\text {Irr}_{p^{\prime }}(B_{p}(G))\) have \(q^{\prime }\)degree and all \(\text {Irr}_{q^{\prime }}(B_{q}(G))\) have \(p^{\prime }\)degree, then [P,Q] = 1 for some P ∈Syl_{p}(G) and Q ∈Syl_{q}(G). Let N be a normal subgroup of G. Since the hypothesis is satisfied by G/N, by induction, we know that \([P,Q] \subseteq N\) for some P ∈Syl_{p}(G) and Q ∈Syl_{q}(G).
Suppose that \(\mathbf {O}_{p^{\prime }}(G)=1\). Then we know that Irr(B_{p}(G)) = Irr(G), by Theorem 10.20 of [23]. Hence, all the irreducible characters in \(\text {Irr}_{p^{\prime }}(G)\) have \(q^{\prime }\)degree. Let L = O_{p}(G) and let P ∈Syl_{p}(G) and Q ∈Syl_{q}(G) such that \([P,Q]\subseteq L\). Therefore Q normalizes P and G : N_{G}(P) is not divisible by q. By Theorem 3.2, we have that all characters in \(\text {Irr}(\mathbf {N}_{G}(P)/P^{\prime })\) have \(q^{\prime }\)degree. By the Itô–Michler theorem we have that \(QP^{\prime }\) is normal in N_{G}(P), and hence \([QP^{\prime },P]\subseteq P^{\prime }\). Then Q acts trivially on \(P/P^{\prime }\), and therefore Q acts trivially on P by coprime action (see [15, Corollary 3.29]). Thus [Q,P] = 1 and we are done in this case.
Suppose that \(L=\mathbf {O}_{p^{\prime }}(G)>1\) and let P ∈Syl_{p}(G) and Q ∈Syl_{q}(G) with \([Q,P] \subseteq L\). By Hall–Higman 1.2.3 Lemma (see [15, Theorem 3.21]) we have that \(\mathbf {C}_{G/L}(\mathbf {O}_{p}(G/L))\subseteq \mathbf {O}_{p}(G/L)\). Since \(QL/L\subseteq \mathbf {C}_{G/L}(PL/L)\) we conclude that \(QL/L\subseteq \mathbf {O}_{p}(G/L)\) and hence \(Q\subseteq L\). Thus G/L is \(q^{\prime }\) and G = LN_{G}(Q), in particular G : N_{G}(Q) is not divisible by p. We claim that all the irreducible characters in \(\text {Irr}_{q^{\prime }}(B_{q}(\mathbf {N}_{G}(Q)))\) have \(p^{\prime }\)degree. Indeed, let \(\chi ^{\ast }\in \text {Irr}_{q^{\prime }}(B_{q}(\mathbf {N}_{G}(Q)))\) and let \(\theta ^{\ast }\in \text {Irr}_{q^{\prime }}(B_{q}(\mathbf {N}_{L}(Q)))\) under χ^{∗}. Let \(\theta \in \text {Irr}_{q^{\prime }}(B_{q}(L))\) be the preimage of 𝜃^{∗} given by the bijection in Theorem 3.1. Again using Theorem 3.1, let χ ∈Irr(B_{q}(G)𝜃) be such that f_{𝜃}(χ) = χ^{∗}, so we know that
Then χ(1)/𝜃(1) is not divisible by q and thus \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(G))\). By hypothesis, χ(1) is not divisible by p. Hence χ^{∗}(1)/𝜃^{∗}(1) is not divisible by p, and therefore, since 𝜃^{∗} is of \(p^{\prime }\)degree we have that χ^{∗}(1) is not divisible by p, and the claim follows.
Let \(X=\mathbf {O}_{q^{\prime }}(\mathbf {N}_{G}(Q))\). Then all the elements in \(\text {Irr}(\mathbf {N}_{G}(Q)/Q^{\prime }X)=\text {Irr}_{q^{\prime }}(B_{q}(\mathbf {N}_{G}(Q)))\) have degree not divisible by p. By the Itô–Michler theorem, we have that this group has a normal Sylow psubgroup (which is a Sylow psubgroup of G, since G : N_{G}(Q) is \(p^{\prime }\)), and therefore P centralizes \(Q/Q^{\prime }\). By coprime action, [P,Q] = 1. □
We thank the referee for pointing out the “in particular” in Theorem 3.3 above. In fact, this observation by the referee gives a different proof of Theorem E in [26], something which we have not noticed before.
4 Conjecture C and Simple Groups
As we have mentioned in the Introduction, we note that the “if” direction of Conjecture C follows from the work of Malle–Navarro in [19], namely [19, Theorem 4.1].
Hence we focus on the “only if” direction. First, we consider the cases easily dealt with in GAP:
Proposition 4.1
Conjecture C holds for sporadic simple groups, alternating and symmetric groups \(\mathfrak {A}_{n}\) and \(\mathfrak {S}_{n}\) with n ≤ 8, the Tits group \(^{2}\mathrm {F}_{4}(2)^{\prime }\), \(\mathrm {G}_{2}(2)^{\prime }\), and groups of Lie type with exceptional Schur multipliers.
Proof
This can be seen using [8] and its Character Table Library. □
4.1 Conjecture C for Alternating and Symmetric Groups
Here we prove Conjecture C in the case of alternating groups \(\mathfrak {A}_{n}\) and symmetric groups \(\mathfrak {S}_{n}\). Note that it follows from [19, Proposition 3.3] that [P,Q]≠ 1 for every Sylow psubgroup P and Sylow qsubgroup Q of \(\mathfrak {S}_{n}\) or \(\mathfrak {A}_{n}\).
Proposition 4.2
Let G be an alternating or symmetric group \(\mathfrak {A}_{n}\) or \(\mathfrak {S}_{n}\) with n ≥ 9 and let p, q be primes dividing G. Then either there exists \(\chi \in \textup {Irr}_{p^{\prime }}(B_{p}(S))\) with degree divisible by q or there exists \(\chi \in \textup {Irr}_{q^{\prime }}(B_{q}(S))\) with degree divisible by p.
The Strategy
We first recall some facts and give the basic idea of the proof. The set \(\text {Irr}(\mathfrak {S}_{n})\) is indexed by partitions of n, and two characters χ_{λ}, χ_{μ} corresponding to partitions λ, μ lie in the same pblock if λ, μ have the same pcore (and similar for q). In particular, writing n = pm + b with 0 ≤ b < p, the set \(\text {Irr}(B_{p}(\mathfrak {S}_{n}))\) consists of the characters χ_{λ} such that λ has pcore (b). Furthermore, recall that the degree of the character χ_{λ} is given by the hooklength formula \(\chi _{\lambda }(1)=\frac {n!}{\prod h_{\lambda }}\), where the denominator is the product of all hooklengths in the tableau corresponding to the partition λ. Furthermore, if λ is not selfconjugate, then the corresponding character restricts irreducibly to \({\mathfrak {A}_{n}}\).
So, our strategy will be (up to switching p and q) to illustrate a nonselfconjugate partition λ of n with pcore (b) such that the numerator in the hooklength formula has the same ppart as the denominator and larger qpart than the denominator. Our proof will require several technical cases and analysis of the degrees given by the hooklength formula.
Setting Notation
Throughout our proof, we will assume without loss that q < p. We will write mp = wq + r with 0 ≤ r < q, and let n = mp + b with 0 ≤ b < p. Note that we may assume that m > 1, since otherwise a Sylow psubgroup is abelian, and the result follows from [10, Theorem 3.5], together with the principal block version of Brauer’s height zero conjecture [20]. In studying the degrees of the characters that we construct, several expressions will appear repeatedly. Hence we define once and for all:
which we see are each relatively prime to p.
It will also be useful to set the padic and qadic expansions of mp: Let
with t_{i} < t_{i+ 1}, s_{j} < s_{j+ 1}, 0 < a_{i} < q, and 0 < b_{j} < p for each appropriate value of i,j. With these established, we further define
Note that \(mp\neq a_{1}q^{t_{1}}\), since \(p\nmid a_{1}\). In the situation that the expression \(X^{\prime }\) becomes relevant, we will see that also \(mp\neq b_{1}p^{s_{1}}\). Finally, for an integer x, we will write (x)_{p} (or just x_{p} if it is clear) for the ppart of x.
Proof Proof of Proposition 4.2
Keep the notation above.
(I) First, suppose that r > 0. If b = 0, so that n = mp, then consider λ = (1^{mp−r− 1},1 + r). Then \(\chi _{\lambda }\in \text {Irr}(B_{p}(\mathfrak {S}_{n}))\) and \(\chi _{\lambda }(1)=Z^{\prime }\), which is \(p^{\prime }\) but divisible by q since mp − r = wq and r < q. If b≠ 0 and 0 < r < b, consider λ = (1^{mp−r− 1},1 + r,b). If 0 < b < r, consider λ = (1^{mp−r− 1},1 + b,r). In these cases, \(\chi _{\lambda }\in \text {Irr}(B_{p}(\mathfrak {S}_{n}))\) and \(\chi _{\lambda }(1)= Y\cdot Z^{\prime }\cdot \frac {br}{mpr+b}\). Note that the qpart of the numerator of Y must be at least as large as the qpart of the denominator. (Each remainder modulo b appears once as a factor in the numerator.) Hence, this character still has degree that is \(p^{\prime }\) and divisible by q, with the possible exception of if q∣b and b!(wq + b) has larger qpart than (mp + 1)⋯(mp + b). In the latter case, (1^{mp},b), giving degree \(Y^{\prime }\), works instead.
If b = r > 0, note that r + 1 < wq, as otherwise we would have r = q − 1 and w = 1 = m, contradicting our assumption that m > 1. Then let λ = (1^{r},r + 1,wq − 1), so that \(\chi _{\lambda }(1)=Z\cdot Z^{\prime }\cdot (wqr1)/(2r+1)\), which is \(p^{\prime }\) and divisible by q unless 2r + 1 = p and p∣(m − 1) or if 2r + 1 = q and \(q\nmid w\). In the latter cases, the partition (1^{wq− 2},1 + r,1 + r) works, unless we were in the case 2r + 1 = p with p∣(m − 1) and r + 1 = q with \(q\nmid w\). In this case, if q≠ 2 (and hence r≠ 1), take λ = (1^{mp},r), which corresponds to a character χ_{λ} that lies in \(B_{p}(\mathfrak {S}_{n})\) and has degree \(Y^{\prime }=(mp+1)\cdots (mp+r1)/(q2)!\). This is relatively prime to p, and is divisible by q since there must be a number between mp and mp + q − 1 = mp + r divisible by q, but neither mp = (w + 1)q − 1 nor n = mp + r = wq + 2q − 2 can be divisible by q. If q = 2, we have r = 1, q = 2, p = 3, and the character corresponding to (1^{n− 2},2) lies in \(B_{2}(\mathfrak {S}_{n})\) and has degree n − 1 = 3m = 2w + 1, which is odd and divisible by 3.
From now on, we may therefore assume that r = 0, so that mp = wq.
(II) First, assume that \(a_{1}q^{t_{1}}<b_{1}p^{s_{1}}\). If b = 0, so n = mp = wq, consider \(\lambda =(1^{mpa_{1}q^{t_{1}}1}, 1+a_{1}q^{t_{1}})\). Then the corresponding degree is X, which we see is equal to
Note that the qpart of this is \(q^{t_{2}}/q^{t_{1}}\), which is divisible by q. Furthermore, the ppart is 1, since \(a_{1}q^{t_{1}}<b_{1}p^{s_{1}}\) implies that the ppart of \(i+b_{1}p^{s_{1}}+\cdots +b_{k^{\prime }}p^{s_{k^{\prime }}}\) is the same as that of i for \(1\leq i\leq a_{1}q^{t_{1}}\).
Now suppose that b > 0, so n = mp + b = wq + b. If \(a_{1}q^{t_{1}}\neq b\), consider either \((1^{mpa_{1}q^{t_{1}}1}, 1+a_{1}q^{t_{1}}, b)\) or \((1^{mpa_{1}q^{t_{1}}1}, 1+b, a_{1}q^{t_{1}})\), depending on whether b is larger or smaller than \(a_{1}q^{t_{1}}\). Then the corresponding character lies in \(B_{p}(\mathfrak {S}_{n})\) and has degree \(X\cdot Y \cdot \frac {ba_{1}q^{t_{1}}}{na_{1}q^{t_{1}}}\). Note that since \(a_{1}q^{t_{1}}<b_{1}p^{s_{1}}\), the ppart of \(ba_{1}q^{t_{1}}_{p}\leq p^{s_{1}}\), and from this we see \(ba_{1}q^{t_{1}}\) and \(na_{1}q^{t_{1}}=mp+ba_{1}q^{t_{1}}\) have the same ppart. Hence this character is a member of \(\text {Irr}_{p^{\prime }}(B_{p}(\mathfrak {S}_{n}))\). Furthermore, its degree is still divisible by q, except possibly if \(ba_{1}q^{t_{1}}_{q}<(na_{1}q^{t_{1}})_{q}\). This can only happen if \((b)_{q}\geq q^{t_{2}}\). In the latter case, consider again the partition (1^{mp},b). The degree is \(Y^{\prime }\) and hence we have removed \(\frac {mp+b}{b}\) from the expression Y, which is \(p^{\prime }\) and, from before, has qpart of the numerator at least as large as that of the denominator. Since in our situation \((mp+b)_{q}=q^{t_{1}}<q^{t_{2}}\leq (b)_{q}\), this degree \(Y^{\prime }\) is also divisible by q.
Now assume \(b=a_{1}q^{t_{1}}\). If b + 1≠p or \(p\nmid (m1)\), we take λ = (1^{mp−b− 2},b + 1,b + 1), with \(\chi _{\lambda }(1)= Y\cdot \frac {(mp2)\cdots (mpb1)}{(b+1)!}\). Then \(\chi _{\lambda }\in \text {Irr}(B_{p}(\mathfrak {S}_{n}))\) and \(p\nmid \chi _{\lambda }(1)\) due to the assumption b + 1≠p or m − 1 is not divisible by p. Furthermore, χ_{λ}(1) is divisible by q since the qpart of the numerators of each of the two fractions is at least that of the denominators, as before, and in this case, the factor (mp − b) is divisible by \(q^{t_{2}}\), but no factor in the denominator is. Now, if \(p=b+1=1+a_{1}q^{t_{1}}\) and p∣(m − 1), this forces also \(b_{1}p^{s_{1}}=p\), as \(mpp=\sum {b_{i}p^{s_{i}}}  p\) must be divisible by p^{2}. Here consider the partition (1^{mp−p},b + p), which gives a character with degree (mp + b − 1)⋯(mp − p + 1)/(p + b − 1)! in \(B_{p}(\mathfrak {S}_{n})\). Note that the only factor in the numerator divisible by p is mp, which is divisible by p exactly once. Then since the denominator is divisible by p, we see this character lies in \(\text {Irr}_{p^{\prime }}(B_{p}(\mathfrak {S}_{n}))\). Furthermore, since \(mpp+1={\sum }_{i\geq 2}a_{i}q^{t_{i}}\) in this case, we see (mp − p + 1 + j)_{q} ≥ (j)_{q} for 1 ≤ j ≤ p + b − 2, and that \(\frac {mpp+1}{p+b1}=\frac {{\sum }_{i\geq 2}a_{i}q^{t_{i}}}{2a_{1}q^{t_{1}}}\) is divisible by q except possibly if q = 2 and t_{2} = t_{1} + 1. In the latter case, the character corresponding to (1^{mp− 1},b + 1), which has degree \(\frac {mp\cdot {\prod }_{i=1}^{2^{t_{1}}1} (mp+i)}{2^{t_{1}}!}\) lies in \(\text {Irr}_{2'}(B_{2}(\mathfrak {S}_{n}))\) and has degree divisible by p.
(III) Finally, suppose that \(a_{1}q^{t_{1}}>b_{1}p^{s_{1}}\). Note here that \(mp\neq b_{1}p^{s_{1}}\), as \(mp\geq a_{1}q^{t_{1}}\). If b = 0, then reversing the roles of p and q in the corresponding case in (II) above yields a character in \(B_{q}(\mathfrak {S}_{n})\) with degree \(X^{\prime }\), which is relatively prime to q but divisible by p.
Hence we assume b > 0, so n = mp + b = wq + b. Note here that \(b=w^{\prime }q+b^{\prime }\) for some integers \(w^{\prime }\), \(b^{\prime }\) with \(0\leq b^{\prime }<q\), and \(B_{q}(\mathfrak {S}_{n})\) consists of those characters whose corresponding partitions have qcore \((b^{\prime })\).
Now, the partition \((1^{mpb_{1}p^{s_{1}}1}, b+1, b_{1}p^{s_{1}})\) gives a character in \(B_{q}(\mathfrak {S}_{n})\) with degree \(Y\cdot X^{\prime }\cdot \frac {b_{1}p^{s_{1}}b}{nb_{1}p^{s_{1}}}\). Note that the third factor is not divisible by p nor q, since \(p\nmid b\) and \(b_{1}p^{s_{1}}b<a_{1}q^{t_{1}}\) so \(b_{1}p^{s_{1}}b\) and \(nb_{1}p^{s_{1}}=mp(b_{1}p^{s_{1}}b)\) have the same qpart. Furthermore, \(p\nmid Y\), and also \(q\nmid Y\) as long as \(b<(qa_{1})q^{t_{1}}\). So if \(b<(qa_{1})q^{t_{1}}\), this character lies in \(\text {Irr}_{q^{\prime }}(B_{q}(\mathfrak {S}_{n}))\) with degree divisible by p.
So, we now assume that q∣Y, so \(b\geq (qa_{1})q^{t_{1}}\). Then the partition (b + 1,mp − 1) corresponds to a character in \(B_{p}(\mathfrak {S}_{n})\) with degree \(Y\cdot \frac {mp(b+1)}{b+1}\), which is divisible by q and is relatively prime to p if b + 1≠p or \(p\nmid (m1)\). Hence we may now assume that further p = b + 1 and p∣(m − 1). Then setting λ = (1^{mp},b) yields \(\chi _{\lambda }\in \text {Irr}(B_{p}(\mathfrak {S}_{n}))\) and \(\chi _{\lambda }(1)=Y^{\prime }\), which is prime to p. Since q∣Y, we have \(q\mid Y^{\prime }\) unless q∣b and (mp + b)_{q} > (b)_{q}, which forces \(b=(qa_{1})q^{t_{1}}\).
So we are reduced to the case \(q\nmid Y^{\prime }\), \(b=(qa_{1})q^{t_{1}}=p1\), and p∣(m − 1). Then the partition (1^{mp− 1},1 + b) gives a character in \(B_{q}(\mathfrak {S}_{n})\) with degree \(Y^{\prime }\cdot {mp}/{b}\), which is divisible by p but is not divisible by q since the qpart of both b and mp are \(q^{t_{1}}\).
Finally, note that in all cases, the λ described is not selfconjugate, and hence the characters restrict irreducibly to \(\mathfrak {A}_{n}\), completing the proof. □
4.2 Conjecture C for Simple Groups of Lie Type
Let r_{0} be a prime and \(r:={r_{0}^{a}}\) be some power of r_{0}. For p another prime, we denote by d_{p}(r) the order of r modulo p, respectively modulo 4, if p is odd, respectively p = 2. Here we will prove the remaining direction of Conjecture C for simple groups of Lie type S defined in characteristic r_{0}. (For the case of Suzuki and Ree groups, we let r be 2^{2n+ 1} or 3^{2n+ 1}, as appropriate.)
Several cases here also follow quickly from [19]:
Proposition 4.3
Let S be a simple group of Lie type defined over \(\mathbb {F}_{r}\), and let p, q be primes such that [P,Q]≠ 1 for every Sylow psubgroup P and Sylow qsubgroup Q of S but that at least one of the following conditions holds:

1.
r_{0} ∈{p,q};

2.
a Sylow psubgroup or a Sylow qsubgroup of S is abelian; or

3.
d_{p}(r)≠d_{q}(r);
Then either there exists \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(S))\) with degree divisible by q or there exists \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(S))\) with degree divisible by p.
Proof
First suppose that r_{0} ∈{p,q} and without loss, say r_{0} = p. Then Irr(B_{p}(S)) = Irr(S) ∖{St_{S}} contains all characters of degree divisible by q. Let G be a quasisimple group of Lie type of simply connected type such that G/Z(G) = S. Suppose first that there is a semisimple element s of the dual group G^{∗} that does not centralize a Sylow qsubgroup of G^{∗} and which lies in \(O^{p^{\prime }}(G^{\ast })\). (See e.g. [5, Chapter 4] for a discussion of the dual group G^{∗}.) Considering a semisimple character χ_{s} corresponding to s (in particular, we may fix χ_{s} to correspond to the trivial character of \(\mathbf {C}_{G^{\ast }}(s)\) under Jordan decomposition), we see that \(\chi _{s}(1)=[G^{\ast }:\mathbf {C}_{G^{\ast }}(s)]_{p^{\prime }}\) (see e.g. [17, (2.1)]) and χ_{s} is trivial on Z(G) (see e.g. [32, Proposition 2.7]), so χ_{s} has degree divisible by q but lies in \(\text {Irr}_{p^{\prime }}(B_{p}(S))\) as a character of S.
Now suppose that every semisimple element of G^{∗} that lies in \(O^{p^{\prime }}(G^{\ast })\) centralizes a Sylow qsubgroup of G^{∗}. In particular, this means that \(O^{p^{\prime }}(G^{\ast })\) has abelian Sylow qsubgroups, so S also has abelian Sylow qsubgroups. (Indeed, in many cases, \(S\cong O^{p^{\prime }}(G^{\ast })\), and otherwise the claim can be seen from the observations in [18, Section 2.1].) In this case, \(\text {Irr}_{q^{\prime }}(B_{q}(S))=\text {Irr}(B_{q}(S))\) by Brauer’s height zero conjecture for principal blocks [20]. Then let χ be any nontrivial unipotent character in Irr(B_{q}(S)). This character will have degree divisible by p by [17, Theorem 6.8], unless p = r∈{2,3} and S is one of the exceptions given in loc. cit. Using Proposition 4.1, this leaves only the case S = B_{n}(2) = C_{n}(2) = Sp_{2n}(2) for n ≥ 4. Let e be the order of r^{2} = 4 modulo q and write n = me + b with b < e. Now, following the proof in [17, Theorem 6.8], we see the nontrivial unipotent characters with degree not divisible by 2 are those whose corresponding symbols contain only the numbers 0, 1, n. Using the theory of ecore and ecocore partitions (the relevant details of which we have summarized in the section on B_{n} and C_{n} in the proof of [26, Proposition 3.7]), we have one of the characters indexed by symbols \(0, b+1\choose me\), \(0, me\choose b+1\), or \(b+1, me\choose 0\) lies in B_{q}(S). These characters are not the odddegree characters, unless b = 0, i.e. e∣n. In the latter case, the Steinberg character has degree a power of 2 and lies in B_{q}(S).
Next, suppose that r_{0}∉{p,q} and that a Sylow psubgroup of S is abelian. Then \(\text {Irr}_{p^{\prime }}(B_{p}(S))=\text {Irr}(B_{p}(S))\) by Brauer’s height zero conjecture for principal blocks [20], and hence the proof of [19, Theorem 5.1] yields the desired character.
Finally, assume that r_{0}∉{p,q}, that no Sylow p or q subgroup of S is abelian, and that d_{p}(r)≠d_{q}(r). Note that the assumption [P,Q]≠ 1 for every choice of Sylow p and Sylow qsubgroups of S (and hence analogously for G) implies that also [P^{∗},Q^{∗}]≠ 1 for any Sylow p and q subgroups of the dual group G^{∗}. (Again, this is pointed out already in [19, Theorem 5.1].) Now, let d_{p}(r) ≤ d_{q}(r) and suppose that there exists 1≠s ∈Z(Q^{∗}) for some \(Q^{\ast }\in \text {Syl}_{q}(G^{\ast })\) such that s centralizes a Sylow psubgroup of G^{∗}. Then the same argument as in the first two paragraphs of [19, Proposition 3.5], but now applied to G^{∗}, yields that d_{p}(r) = d_{q}(r), a contradiction. Hence, given 1≠s ∈Z(Q^{∗}), we havestop \(\mathbf {C}_{G^{\ast }}(s)\) contains Q^{∗} but does not contain a Sylow psubgroup P^{∗} of G^{∗}. Then the corresponding semisimple character χ_{s} of G has degree divisible by p but lies in \(\text {Irr}_{q^{\prime }}(B_{q}(G))\), since \(\chi _{s}(1)=[G^{\ast }:\mathbf {C}_{G^{\ast }}(s)]_{r_{0}^{\prime }}\) and using [12, Corollary 3.4]. Furthermore, arguing as in [11, Theorem 3.5] shows that such s can be chosen so that χ_{s} is trivial on Z(G), completing the proof. □
Our task is now to prove Conjecture C in the case that neither p nor q is the defining characteristic, d_{p}(r) = d_{q}(r), and no Sylow p or qsubgroup of S is abelian. We begin with the case of exceptional groups of Lie type, by which we mean the groups S = G_{2}(r), F_{4}(r), \(\mathrm {E}_{6}^{\epsilon }(r)\), E_{7}(r), E_{8}(r), \(^{3}\mathrm {D}_{4}(r)\), \(^{2}\mathrm {G}_{2}(r)\), \(^{2}\mathrm {F}_{4}(r)\), and \(^{2}\mathrm {B}_{2}(r)\).
Proposition 4.4
Let S be an exceptional simple group of Lie type defined over \(\mathbb {F}_{r}\), and let p, q be primes such that [P,Q]≠ 1 for every Sylow psubgroup P and Sylow qsubgroup Q of S. Then either there exists \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(S))\) with degree divisible by q or there exists \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(S))\) with degree divisible by p.
Proof
By Proposition 4.3, we may assume that r is not a power of p nor q, that no Sylow p or q subgroup of S is abelian, and that d_{p}(r) = d_{q}(r). Let d := d_{p}(r) = d_{q}(r). With these constraints, we see that S is not of Suzuki or Ree type, that p and q are at most 7, and that d is a regular number in the sense of Springer [34] (see also [33, Definition 2.5]). Hence we see that the principal blocks B_{p}(S) and B_{q}(S) are the unique blocks of S containing \(p^{\prime }\), respectively, \(q^{\prime }\)degree unipotent characters (see, e.g., [31, Lemma 3.6]). Under these conditions, we see by observing the explicit list of unipotent character degrees in [5, Section 13.9] that there exists a unipotent character χ satisfying either \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(S))\) and q∣χ(1) or \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(S))\) and p∣χ(1). □
We next consider the case of linear and unitary groups.
Proposition 4.5
Let \(S=\text {PSL}_{n}^{\epsilon }(r)\) with n ≥ 2, and let p≠q be primes such that [P,Q]≠ 1 for every Sylow psubgroup P and Sylow qsubgroup Q of S. Then either there exists \(\chi \in \text {Irr}_{p^{\prime }}(B_{p}(S))\) with degree divisible by q or there exists \(\chi \in \text {Irr}_{q^{\prime }}(B_{q}(S))\) with degree divisible by p.
Proof
By Proposition 4.1, we may assume S is not isomorphic to a sporadic or alternating group. Furthermore, by Proposition 4.3, we may assume that r is not a power of p nor q, that no Sylow p or q subgroup of S is abelian, and that d_{p}(r) = d = d_{q}(r).
Write \(G:=\text {SL}_{n}^{\epsilon }(r)\), \(\widetilde {G}:=\text {GL}_{n}^{\epsilon }(r)\), and e := d_{p}(𝜖r) = d_{q}(𝜖r). Note that \(\widetilde {G}^{\ast }\cong \widetilde {G}\), so we will make this identification. Let \(\widetilde {P}\in \text {Syl}_{p}(\widetilde {G})\) and \(\widetilde {Q}\in \text {Syl}_{q}(\widetilde {G})\). It suffices to show that (up to switching p and q) there is some \(s\in \mathbf {Z}(\widetilde {P}) \cap G\) such that \(\mathbf {C}_{\widetilde {G}}(s)\) is not divisible by \(\widetilde {Q}\) and sz is not \(\widetilde {G}\)conjugate to s for any \(1\neq z\in \mathbf {Z}(\widetilde {G})\). (Indeed, that \(s\in G=\mathrm {O}^{r_{0}^{\prime }}(\widetilde {G})\) implies that the corresponding semisimple character χ_{s} is trivial on \(\mathbf {Z}(\widetilde {G})\) and that sz is not conjugate to s for nontrivial \(z\in \mathbf {Z}(\widetilde {G})\) implies that χ_{s} is irreducible on restriction to G using e.g. [32, Proposition 2.7] and [31, Lemma 1.4]; the remaining conditions imply \(\chi _{s}\in \text {Irr}(B_{p}(\widetilde {G}))\) using [12, Corollary 3.4] and χ_{s}(1) is \(p^{\prime }\) but divisible by q since \(\chi _{s}(1)=[\widetilde {G}:\mathbf {C}_{\widetilde {G}}(s)]_{r_{0}^{\prime }}\).)
Now, if 2∉{p,q}, we see using the results of Weir [35] that \(\widetilde {P}\) and \(\widetilde {Q}\) are naturally isomorphic to the corresponding Sylow subgroups of \(\text {GL}_{we}^{\epsilon }(r)\), embedded naturally into \(\text {GL}_{we}^{\epsilon }(r)\times \text {GL}_{b}^{\epsilon }(r)\leq \text {GL}_{n}^{\epsilon }(r)\) where n = we + b with 0 ≤ b < e. The results of Carter–Fong [6] yield the same when p or q is 2, except that if (e,b) = (2,1), then \(\text {GL}_{b}^{\epsilon }(r)\) is divisible by 2 exactly once, and the Sylow 2subgroup of \(\text {GL}_{n}^{\epsilon }(r)\) in this case is that of \(\text {GL}_{we}^{\epsilon }(r)\times \text {GL}_{b}(r)\).
Let \(w=a_{1}p^{t_{1}}+a_{2}p^{t_{2}}+\cdots +a_{k}p^{t_{k}}=b_{1}q^{m_{1}}+\cdots +b_{k^{\prime }}q^{m_{k^{\prime }}}\) be the padic and qadic expansions of w, with t_{1} < ⋯ < t_{k}; \(m_{1}<{\cdots } < m_{k^{\prime }}\); 1 ≤ a_{i} < p for each 1 ≤ i ≤ k; and 1 ≤ b_{j} < q for each \(1\leq j\leq k^{\prime }\). By [6, 35], we have \(\widetilde {P}\cong P_{1}^{a_{1}}\times \cdots \times P_{k}^{a_{k}} \times X\), where P_{i} is a Sylow psubgroup of \(\text {GL}_{ep^{t_{i}}}^{\epsilon }(r)\) (which, if p is odd, is a Sylow psubgroup of \(\text {GL}^{\epsilon }_{p^{t_{i}}}(r^{e})\) embedded naturally) for each 1 ≤ i ≤ k and \(X\in \text {Syl}_{p}(\text {GL}_{b}^{\epsilon }(r))\) is isomorphic to C_{2} if (p,e,b) = (2,2,1) and is trivial otherwise. Here we view \(\prod \text {GL}_{ep^{t_{i}}}^{\epsilon }(r)\) as the natural diagonallyembedded subgroup. Similarly, \(\widetilde {Q}\cong Q_{1}^{b_{1}}\times \cdots \times Q_{k^{\prime }}^{b_{k^{\prime }}}\times Y\) where \(Q_{j}\in \text {Syl}_{q}(\text {GL}_{eq^{m_{j}}}^{\epsilon }(r))\) for \(1\leq j\leq k^{\prime }\) and \(Y\in \text {Syl}_{q}(\text {GL}_{b}^{\epsilon }(r))\).
Without loss of generality, assume \(a_{1}p^{t_{1}}<b_{1}q^{m_{1}}\). (Note that we cannot have \(a_{1}p^{t_{1}}=b_{1}q^{m_{1}}\), as this would contradict that either a_{1} < p < q or b_{1} < q < p.) Let x ∈Z(P_{1}) have no eigenvalues equal to 1 (indeed, this can be done by taking x as an element of a Sylow psubgroup of \(\mathbf {Z}(\text {GL}_{p^{t_{1}}}^{\epsilon }(r^{e}))\) embedded naturally into \(\text {GL}_{ep^{t_{1}}}^{\epsilon }(r)\)) and consider the element \(s=\text {diag}(x, \ldots , x, I_{nea_{1}p^{t_{1}}})\in \mathbf {Z}(\widetilde {P})\), with a_{1} copies of x. In fact, taking x (and hence its eigenvalues) to have order p, we obtain \(\det (x)=1\) and hence s ∈ G. Here \(\mathbf {C}_{\widetilde {G}}(s)=\mathbf {C}_{\text {GL}_{a_{1}ep^{t_{1}}}^{\epsilon }(r)}(\text {diag}(x,\ldots , x))\times \text {GL}_{nea_{1}p^{t_{1}}}^{\epsilon }(r)\). Now, by considering the structure, and hence size, of \(\widetilde {Q}\) (namely, each Q_{j} is a wreath product \(Q^{\prime } \wr C_{q}\) where \(Q^{\prime }\) is a Sylow qsubgroup of \(\text {GL}_{eq^{m_{j}1}}^{\epsilon }(r)\)), we see \(\text {GL}_{a_{1}ep^{t_{1}}}^{\epsilon }(r)\times \text {GL}_{nea_{1}p^{t_{1}}}^{\epsilon }(r)\) is not divisible by \(\widetilde {Q}\). Furthermore, by considering the block sizes, we see that sz and z cannot have the same eigenvalues (and hence they cannot be \(\widetilde {G}\)conjugate) for any nontrivial scalar matrix \(z\in \mathbf {Z}(\text {GL}_{n}^{\epsilon }(r))\). This completes the proof. □
We next consider the remaining classical types, for which the proof is very similar to the linear and unitary case.
Proposition 4.6
Let S = PSp_{2n}(r) with n ≥ 2, PΩ_{2n+ 1}(r) with n ≥ 3, or \(\mathrm {P}{{\Omega }}_{2n}^{\epsilon }(r)\) with n ≥ 4. Let p≠q be primes such that [P,Q]≠ 1 for every Sylow psubgroup P and Sylow qsubgroup Q of S. Then either there exists \(\chi \in \textup {Irr}_{p^{\prime }}{B_{p}(S)}\) with degree divisible by q or there exists \(\chi \in \textup {Irr}_{q^{\prime }}{B_{q}(S)}\) with degree divisible by p.
Proof
As before, we may assume S is not isomorphic to a sporadic or alternating group, r is not a power of p nor q, no Sylow p or q subgroup of S is abelian, and that d_{p}(r) = d = d_{q}(r).
First we set some notation. We define H_{n} := Sp_{2n}(r), SO_{2n+ 1}(r), and \(\text {SO}_{2n}^{\epsilon }(r)\) in the cases S = PSp_{2n}(r), PΩ_{2n+ 1}(r), and P\({{\Omega }}_{2n}^{\epsilon }(r)\), respectively. Let H := H_{n} and let \({\Omega }:=\mathrm {O}^{r_{0}^{\prime }}(H)\) so that Ω is perfect and S = Ω/Z(Ω). Note that the dual groups are \(H_{n}^{\ast } = \text {SO}_{2n+1}(r)\), Sp_{2n}(r), and \(\text {SO}_{2n}^{\epsilon }(r)\), respectively, and we will write \(H^{\ast }:=H_{n}^{\ast }\). Note that Z(Ω) ≤Z(H) and that H/Ω and Z(H) are 2groups. Let \(\widetilde {P}\) and \(\widetilde {Q}\) be Sylow p and qsubgroups of H^{∗}.
In this situation, it suffices to show that (up to switching p and q) there is some \(s\in \mathbf {Z}(\widetilde {P})\) such that \(\mathbf {C}_{\widetilde {G}}(s)\) is not divisible by \(\widetilde {Q}\), using similar reasoning to the above case. Indeed, if p is odd, then \(\widetilde {P}\) may be considered as a Sylow psubgroup of \(\mathrm {O}^{r_{0}^{\prime }}(H^{\ast })\) and sz cannot be H^{∗}conjugate to s for any 1≠z ∈Z(H^{∗}) since Z(H^{∗}) is a 2group, so a corresponding semisimple character χ_{s} of H is trivial on Z(H) and restricts irreducibly to Ω. If instead p = 2, then such a character χ_{s} would have odd degree, and therefore restrict irreducibly to Ω since H/Ω is a 2group. Then since Ω is perfect, Z(Ω) is a 2group, and χ_{s} has odd degree, this forces χ_{s} to be trivial on Z(Ω). Furthermore, as before, χ_{s} ∈Irr(B_{p}(H)) in either case.
Assume first that p and q are odd. In these cases, the work of Weir [35] again describes the structure of \(\widetilde {P}\) and \(\widetilde {Q}\), building off of the case of linear groups. If \(H^{\ast }=\text {SO}_{2n+1}(r)\) or Sp_{2n}(r), we have Sylow p and qsubgroups are already Sylow subgroups of GL_{2n+ 1}(r) (and hence of GL_{2n}(r)) when d is even, and are Sylow subgroups of the naturallyembedded GL_{n}(r) if d is odd. For these cases, let e := d_{p}(r^{2}) = d_{q}(r^{2}), write n = ew + b with 0 ≤ b < e, and let \(w=a_{1}p^{t_{1}}+a_{2}p^{t_{2}}+\cdots +a_{k}p^{t_{k}}=b_{1}q^{m_{1}}+\cdots +b_{k^{\prime }}q^{m_{k^{\prime }}}\) be the padic and qadic expansions of w as before. Again without loss, we assume \(a_{1}p^{t_{1}}<b_{1}q^{m_{1}}\). In particular, \(\widetilde {P}\) and \(\widetilde {Q}\) are again isomorphic to Sylow subgroups of \(H_{ew}^{\ast }\) and of the form \(\widetilde {P}\cong P_{1}^{a_{1}}\times \cdots \times P_{k}^{a_{k}}\), where each P_{i} is a Sylow psubgroup of \(\text {GL}_{dp^{t_{i}}}(r)\) and can be identified with a Sylow psubgroup of \(H_{ep^{t_{i}}}^{\ast }\), and similar for \(\widetilde {Q}\). As before, let x ∈Z(P_{1}) with no eigenvalues equal to 1 and let \(s=(x,\dots , x, 1, {\ldots } 1)\in \mathbf {Z}(\widetilde {P})\) with a_{1} copies of x. Then we can see from the centralizer structure of semisimple elements that \(\mathbf {C}_{H^{\ast }}(s)\cong \mathbf {C}_{H^{\ast }_{a_{1}ep^{t_{1}}}}{(x,\ldots , x)}\times H^{\ast }_{nea_{1}p^{t_{1}}}\). Since the Sylow qsubgroups of \(H_{a_{1}ep^{t_{1}}}^{\ast }\) and \(H_{nea_{1}p^{t_{1}}}^{\ast }\) can be identified with Sylow subgroups of linear groups in an analogous way as for H^{∗}, depending on whether d was even or odd, we have \(\widetilde {Q}\nmid \mathbf {C}_{H^{\ast }}(s)\) for the same reason as in the case of linear groups above.
If \(H^{\ast }=\text {SO}_{2n}^{\epsilon }(r)\), then we have embeddings \(\text {SO}_{2n1}(r)\leq H^{\ast }\leq \text {SO}_{2n+1}(r)\), and \(\widetilde {P}\) and \(\widetilde {Q}\) are both Sylow subgroups of either SO_{2n− 1}(r) or SO_{2n+ 1}(r). In this case, letting m ∈{n,n − 1} so that \(\widetilde {P}\), \(\widetilde {Q}\) are Sylow subgroups of SO_{2m+ 1}(r) and now writing m = ew + b with w written with p and qadic expansions as before, \(\widetilde {P}\) can again be written \(\widetilde {P}\cong P_{1}^{a_{1}}\times \cdots \times P_{k}^{a_{k}}\) with each P_{i} a Sylow subgroup of \(\text {GL}_{dp^{t_{i}}}(r)\), which in this case can also be identified with a Sylow psubgroup of either \(\text {SO}_{2ep^{t_{i}}}^{+}(r)\) or \(\text {SO}_{2ep^{t_{i}}}^{}(r)\). From here, arguing similar to before, we obtain an element \(s\in \mathbf {Z}(\widetilde {P})\) with \(\mathbf {C}_{H^{\ast }}(s)\) isomorphic to a subgroup of \(\mathbf {C}_{\text {GO}_{2a_{1}ep^{t_{1}}}^{\pm }(r)}(\text {diag}(x,\ldots , x)) \times \text {GO}^{\pm }_{2(nea_{1}p^{t_{1}})}(r)\). Since q is odd, we again see in the same way as above that \(\widetilde {Q}\) does not divide \(\mathbf {C}_{H^{\ast }}(s)\).
We are finally left with the case that 2 ∈{p,q}. Let \(\hat H^{\ast }\) denote the group GO_{2n+ 1}(r), Sp_{2n}(r), or \(\text {GO}_{2n}^{\epsilon }(r)\) respectively, so that \([\hat H^{\ast }: H^{\ast }]\) divides 2. Note that if p = 2 and \(H^{\ast }\neq \hat H^{\ast }\), then \(\widetilde {P}\) is index2 in a Sylow 2subgroup \(\hat P\) of \(\hat H^{\ast }\), which are again described by Carter–Fong [6]. Here in the case of GO_{2n+ 1}(r) or Sp_{2n}(r), writing \(n=2^{t_{1}}+{\cdots } + 2^{t_{k}}\) for the 2adic expansion with t_{1} < ⋯ < t_{k}, we have \(\hat P\cong P_{1}\times \cdots \times P_{k}\), where P_{i} is a Sylow 2subgroup of \(\text {GO}_{2\cdot 2^{t_{i}}+1}(r)\), respectively, \(\text {Sp}_{2\cdot 2^{t_{i}}}(r)\). In the case \(\hat H^{\ast }=\text {GO}_{2n}^{\epsilon }(r)\), we have \(\hat {P}\) is either a Sylow 2subgroup of GO_{2n+ 1}(r), embedded as before, or of the form P_{0} × C_{2} × C_{2}, where P_{0} is a Sylow 2subgroup of GO_{2n− 1}(r). From here, we may argue analogously to before, keeping in mind that when p = 2, choosing x ∈Z(P_{1}) to have \(2^{t_{1}+1}\) eigenvalues − 1 yields an element of determinant 1, and hence an element of \(\widetilde {P}=\hat P\cap H^{\ast }\). □
Conjecture C for simple groups (and Theorem E) now follows from Propositions 4.1–4.6.
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Acknowledgements
The first and second authors are partially supported by Grant PID2019103854GBI00 funded by MCIN/AEI/10.13039/501100011033. The second author also acknowledges support by Generalitat Valenciana AICO/2020/298 and Grant PID2020118193GAI00 funded by MCIN/AEI/10.13039/501100011033. The third author is partially supported by a grant from the National Science Foundation, Award No. DMS2100912. We thank the referees for many helpful comments and D. Rossi for useful conversations on Section refsec3.
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Navarro, G., Rizo, N. & Schaeffer Fry, A.A. Principal Blocks for Different Primes, II. Vietnam J. Math. 51, 589–604 (2023). https://doi.org/10.1007/s1001302200594z
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DOI: https://doi.org/10.1007/s1001302200594z