Further Reductions for the k(GV)-Problem

by

GEOFFREY R. ROBINSON DEPARTMENT OF MATHEMATICS & COMPUTER SCIENCE UNIVERSITY OF LEICESTER LEICESTER LE1 7RH ENGLAND

1 December 1996 INTRODUCTION: The aim of this note is to make further reductions towards an affirmative answer to the “k(GV)-problem” for all primes p. We extend somewhat the methods and results of [R-T]. Our main result here is THEOREM1: Let p be a prime greater than 211. Suppose that there is a finite p′-group G and a faithful irreducible GF (p)G-module V such that k(GV) > |V |. Then there is a finite field F of characteristic p, a finite group H with F^{∗}(H) = Z(H)E(H) and E(H) quasi-simple, a faithful F H-module U on which E(H) acts absolutely irreducibly, such that for no vector u ∈ U does Res_{C_{H}(u)}^{H}(U) have a faithful self-dual submodule.

In particular, there is no u ∈ U such that C_{H}(u) is an elementary Abelian 2-group. REMARK: In conjunction with work of D.Goodwin and the earlier results of Liebeck [L], and [R-T], the above theorem has the consequence that the k(GV)-problem has an affirmative answer for all primes p > 311.

SECTION 1: PRELIMINARIES

Lemma 1.1: Let P be a finite group of the form Q < α >, where Q ◁ P and Q is a q-group for some prime q, with Q′ ≤ Z(Q). Suppose that χ is a faithful complex irreducible character of P such that Res_{Q}^{P}(χ) remains irreducible. Then if χ(α) ≠ 0, we have |χ(α)|^{2} = [C_{Q}(α) : Z(Q)]. PROOF: Since P = QC_{P} (α), we have [P : C_{P} (α)] = [Q : C_{Q}(α)]. Suppose that χ(α) ≠ 0. Consideration of a matrix representation affording χ readily yields that

∑_{x∈P/C_{P} (α)} χ(α^{−1}α^{x}) = [P : C_{P} (α)]\frac{|χ(α)|^{2}}{χ(1)} .

Since χ vanishes on Q\Z(Q), we have χ(1)^{2} = [Q : Z(Q)].

Now for any x ∈ Q, α^{−1}α^{x} ∈ P′ ≤ Q. Hence χ(α^{−1}α^{x}) = 0 unless α^{−1}α^{x} ∈ Z(Q). If α^{−1}α^{x} ∈ Z(Q), then α^{−1}α^{x} = 1_{P}, for otherwise, if α^{−1}α^{x} = z ∈ Z(Q)^{#}, we have χ(α^{x}) = \frac{χ(z)χ(α)}{χ(1)}, forcing χ(α) = 0, contrary to assumption. Hence we obtain [P : C_{P} (α)]\frac{|χ(α)|^{2}}{χ(1)} = χ(1), so |χ(α)|^{2} = [C_{Q}(α) : Z(Q)]. COROLLARY 1.2: Suppose that P, Q, χ, α are as above and that q ≠ 3. Suppose further that α has order q if q is odd, or order 4 if q = 2. Let χ(1) = q^{n}. Then ⟨Res_{⟨α⟩}^{P}(χ), 1⟩ ≤ \frac{q^{n} + 3q^{n−1}}{4} . PROOF: Suppose first that χ(α) = 0. If q is odd, we see that ⟨Res_{<α>}^{P}(χ), 1⟩ = q^{n−1} < \frac{q^{n} + 3q^{n−1}}{4} . If α has order 4, then we apply Lemma 1.1 to α^{2} to obtain either α^{2} ∈ Z(P)^{#} or |χ(α^{2})| ≤ q^{n−1}. In either case ⟨Res_{⟨α⟩}^{P}(χ), 1⟩ ≤ \frac{1}{4}[q^{n} + q^{n−1}]. Hence we may suppose that χ(α) ≠ 0, (and that α ∉ Z(Q)). CASE q = 2 q = 2 q = 2: Then |χ(α)|^{2} ≤ q^{2n−1} (and is an integer power of q), and χ(α) ∈ ℤ[i]. If |χ(α)|^{2} = q^{2m−1} for some integer m, then there are signs ε_{1}, ε_{2} such that χ(α) = q^{m−1}(ε_{1} + ε_{2}i). If α^{2} ∈ Q, then χ(α^{2}) ∈ {0, −q^{n}}, while if α^{2} ∉ Q, then Lemma 1.1 yields χ(α^{2})^{2} ≤ q^{2n−1} (so |χ(α^{2})| ≤ q^{n−1}, as χ(α^{2}) ∈ ℤ). Hence ⟨Res_{S⟨α⟩}^{P}(χ), 1⟩ ≤ \frac{1}{4}[q^{n} + 3q^{n−1}].

If |χ(α)|^{2} = q^{2m} for some integer m, then m ≤ n−1, and χ(α^{2}) ≤ q^{n−1}, so that ⟨Res_{⟨α⟩}^{P}(χ), 1⟩ ≤ \frac{1}{4}[q^{n} + 3q^{n−1}]. CASE q > 3 q > 3 q > 3: Let ω = e^{\frac{2πi}{q}} , ρ = (1 − ω) (the unique prime ideal of ℤ[ω] containing q). We have |χ(α)|^{2} = q^{m} for some integer m ≤ 2n − 1. Since ν_{ρ}(q) = q − 1, and ρ is invariant under complex conjugation, we have ν_{ρ}(χ(α)) = ν(\overline{χ(α)}) = m (\frac{q − 1}{2}). If m = 2a is an even integer, then χ(α) = εq^{a}ω^{j} for some j ∈ ℤ, some sign ε. Then ⟨Res_{⟨α⟩}^{P}(χ), 1⟩ ≤ \frac{1}{q} [q^{n} + (q − 1)q^{n−1}].

If m = 2a + 1 is an odd integer, then there is an integer j and there are signs ε_{1}, ε_{2} such that χ(α) = ε_{1}q^{a}ω^{j}\sqrt{ε_{2}q}, where ℚ[\sqrt{ε_{2}q}] is the quadratic subfield of ℚ[ω]. Now if r + qℤ is a generator of the multiplicative group of squares in (ℤ/qℤ)^{×}, we have:

\sqrt{ε_{2}q} = ε_{3} [2 ∑_{k=0}^{\frac{q−3}{2}} ω^{r^{k}} + 1]

for some sign ε_{3}, so it follows that the maximum possible multiplicity of an eigenvalue of α is

2q^{a} ≤ 2q^{n−1} ≤ \frac{q^{n} + 3q^{n−1}}{4} .

LEMMA 1.3: Let ⟨h⟩ be a finite cyclic group, V be a finite dimensional F ⟨h⟩-module for some field F. Suppose that V has a tensor decomposition V = U ⊗ Uh … ⊗ Uh^{m−1}, where U is a semi-simple F ⟨h^{m}⟩-module. Then Trace_{V} (h) = Trace_{U}(h^{m}). PROOF: We may suppose that there is a basis {u_{i} : 1 ≤ i ≤ n} of U which consists of eigenvectors of h^{m}. Then {u_{i_{0}} ⊗ u_{i_{1}}h … ⊗ u_{i_{m−1}}h^{m−1} : 1 ≤ i_{j} ≤ n for each j} is a basis for V under which the matrix of h is monomial.

The only basis vectors in the above basis which are eigenvectors of h are those of the form v_{i} = u_{i} ⊗ u_{i}h … ⊗ u_{i}h^{m−1}(1 ≤ i ≤ n), and if u_{i}h^{m} = λ_{i}u_{i}, we have v_{i}h = λ_{i}v_{i}.

SECTION 2: CLIFFORD THEORY

HYPOTHESIS 2.1: p is a prime with the property that whenever H is a finite p′-group, F is a finite field of characteristic p, U is a faithful F H-module such that EITHER:

(a) For some prime q ≠ p, Q = O_{q}(H) acts absolutely irreducibly on U and all characteristic Abelian subgroups of Q are central, but Q is non-Abelian, OR:

(b) E = E(H) is a quasi-simple and acts absolutely irreducibly on U, then there is a vector u ∈ U such that Res_{C_{H}(u)}^{H}(U) has a faithful self-dual submodule.

THEOREM 2.2: Let p be a prime which satisfies hypothesis 2.1, k be a finite field of characteristic p, G be a finite p′-group, V be a faithful irreducible kG-module of dimension at least 2. Then for some v ∈ V, Res_{C_{G}(v)}^{G}(V) has a faithful self-dual submodule of dimension at least 2. PROOF: Suppose that the theorem is false, and that (G, V, k) have been chosen to violate result with dim_{k}(V) minimal.

We first note that we may suppose that V is absolutely irreducible. Let F =End_{kG}(V), a finite extension of k. Then V ⊗_{k} F is a direct sum of Galois conjugate irreducible modules, and if U is any one of these, then the permutation actions of G on vectors of U, V respectively are equivalent. If dim_{F} (U) = 1, then G is Abelian and there is a vector v ∈ V with C_{G}(v) = 1_{G}, so the result clearly holds. If 1 < dim_{F} (U) < dim_{k}(V), then there is a vector u ∈ U such that Res_{C_{G}(u)}^{G}(U) has a faithful self-dual submodule, W say. Let X =Gal(F/k), Y =Stab_{X}(U), v = ∑_{σ∈X/Y} uσ. Then C_{G}(v) = C_{G}(u) and ∑_{σ∈X/Y} W σ is a faithful self-dual submodule of Res_{C_{G}(v)}^{G}(V) (up to isomorphism).

Next, we claim that V is primitive. For if V ≅Ind_{H}^{G}(W) for some H < G, let T be a right transversal to H in G. If dim_{k}(W) = 1, then H/C_{H}(W) is Abelian, so there is a vector w ∈ W such that C_{H}(w) = C_{H}(W). Let u = ∑_{t∈T} wt. Then {wt : t ∈ T } spans a permutation module of dimension [G : H] which is faithful for C_{G}(u) ( and is certainly self-dual).

If dim_{k}(W) > 1, then there is a vector w ∈ W such that ResH̄ ; CH̄(w)(W) has a faithful self-dual submodule, X say, of dimension at least 2, where H̄ = H/C_{H}(W). Let u = ∑_{t∈T} wt. Then ⊕_{t∈T} Xt is easily checked to be a faithful self-dual submodule for C_{G}(u) (of dimension at least 2).

Next, we claim that whenever N ◁ G, all irreducible summands of Res_{N}^{G}(V) are absolutely irreducible. For otherwise, we may argue as in [R-T] that there is M ◁ G with G/M cyclic of order t > 1 such that F =End_{kM}(V) is a field of order |k|^{t} with V ⊗_{k} F ≅ Ind_{M}^{G}(U) where U is an absolutely irreducible F M-module, and Res_{M}^{G}(V) is irreducible.

If U is 1-dimensional, then there is a vector u ∈ U with C_{M}(u) = C_{M}(U) = 1, and there is a vector v ∈ V with C_{M}(v) = C_{M}(u). Then Res_{C_{G}(v)}^{G}(V ⊗_{k} F) has a direct summand isomorphic to Ind_{C_{M} (v)}^{C_{G}(v)}(Res_{C_{M} (v)}^{M}(U)), so has a regular summand, as C_{M}(v) = 1. On the other hand, if dim_{F} (U) > 1, then there is a vector u ∈ U such that Res_{C_{M} (u)}^{M}(U) has a faithful self-dual submodule, X say, of dimension at least 2, and C_{M}(u) = C_{M}(v) for some v ∈ V. Then Res_{C_{G}(v)}^{G}(V ⊗_{k} F) has a summand isomorphic to Ind_{C_{M} (v)}^{C_{G}(v)}(X), which is faithful and self-dual. Hence Res_{C_{G}(v)}^{G}(V) has a faithful self-dual submodule which is isomorphic to a sum of certain Galois conjugates of Ind_{C_{M} (v)}^{C_{G}(v)}(X), and has dimension at least 2.

We next claim that whenever N ◁ G with N ≰ Z(G), Res_{N}^{G}(V) is irreducible. Otherwise, as in [R-T], we may argue that there is a central extension of G, say G̃, such that G̃ = {(g, z : z ∈ k^{×}} and V ≅ U ⊗ W as kG̃-modules, where U, W both have dimension greater than one. Then there are vectors u, w ∈ U, W respectively such that ResG̃ ; CG̃(u)(U) has a self-dual summand, X say, (of dimension at least two) with CG̃(u)∩CG̃(X) = CG̃(U) and ResG̃ ; CG̃(w)(W) has a self-dual summand, Y say, (of dimension at least two) with CG̃(w) ∩ CG̃(Y) = CG̃(W). We may, and do, suppose that u ∈ X, w ∈ Y (otherwise, we replace X by X ⊕ ku, for example). Now we set v = u ⊗ w. If (g, 1) ∈ G̃ fixes v, then for some λ ∈ k^{×}, λ^{−1}(g, 1) fixes u, λ(g, 1) fixes w. Hence (g, 1) acts on X ⊗ Y. Furthermore, if (g, 1) acts trivially on X ⊗ Y, then (g, 1) acts as a scalar on X and acts as a scalar on Y. Since u ∈ X, (g, 1) acts as λ on U, λ^{−1} on W, so (g, 1) acts trivially on U ⊗ W ≅ V.

Now we may proceed to a final contradiction. Suppose that F (G) ≰ Z(G). Then for some prime q, Q = O_{q}(G) is non-central in G, and acts absolutely irreducibly on V. Since V is primitive, every characteristic Abelian subgroup of Q is central in G. Also, Q is non-Abelian and dim_{k}(V) ≥ 2. By a) of hypothesis 2.1, there is a vector v ∈ V such that Res_{C_{G}(v)}^{G}(V) has a faithful self-dual submodule, X say, and we may suppose that v ∈ X, as above. If dim_{k}(X) = 1, then C_{G}(v) = 1 and we may replace X by V. Otherwise, dim_{k}(X) ≥ 2. Either eventuality contradicts the fact that (G, V, k) is a counterexample to the theorem. Hence F (G) ≤ Z(G). A similar argument, using b) of hypothesis 2.1, shows that E(G) is not quasi-simple.

Hence F^{∗}(G) = Z(G)E(G), where E(G)(= E) is a central product of t(> 1) G-conjugate components. Let L_{1} be one of these components, M_{1} = N_{G}(L_{1}). As a projective representation of G, the given representation is tensor induced from a projective representation of M_{1} on one of the t tensor factors of Res_{E}^{G}(V), say U_{1}. Let M̃_{1} be the central extension of M_{1} constructed as in [R-T]. Let {g_{i} : 1 ≤ i ≤ t} be a right transversal to M_{1} in G (with g_{1} = 1_{G}), and let U_{i} = U_{1}g_{i} for each i. Since dim_{k}(U_{1}) ≥ 2, there is a vector u_{1} ∈ U_{1}, and a faithful self-dual submodule X_{1} of

ResM̃ ; C_{M̃(u_{1})}(U_{1}).

We set X_{i} = X_{1}g_{i}, u_{i} = u_{1}g_{i} for 2 ≤ i ≤ t. As before, we may, and do, suppose that u_{1} ∈ X_{1}. We set u = u_{1} ⊗ u_{2} … ⊗ u_{t}, X = X_{1} ⊗ X_{2} … ⊗ X_{t}.

Now if h ∈ G satisfies uh = u, then there are scalars λ_{1}, λ_{2}, …, λ_{t} ∈ k such that Π_{i=1}^{t}λ_{i} = 1, and a permutation γ ∈ S_{t} such that u_{i}h = λ_{i}u_{iγ} for each i. Hence λ_{i}^{−1}(g_{i}hg_{iγ}^{−1}, 1) ∈ CM̃_{1}(u_{1}) for each i, and h acts on X_{1}⊗X_{2} …⊗X_{t} = X. If h acts trivially on X, (possibly repeated) applications of Lemma 1.3 allow us to conclude that γ is the identity permutation. Furthermore, for each i, λ_{i}^{−1}(g_{i}hg_{i}^{−1}, 1) acts as a scalar on X_{1}. Since λ_{i}^{−1}(g_{i}hg_{i}^{−1}, 1) fixes u_{1} ∈ X_{1}, h acts as λ_{i} on U_{i}, so h acts trivially on V = U_{1} ⊗ U_{2} … ⊗ U_{t}(as Π_{i=1}^{t}λ_{i} = 1). Hence X is a faithful kC_{G}(u)-module. It remains to check that X is self-dual. We consider the contribution to the (Brauer) character of < h > afforded by X from a single (M_{1}, < h >)-double coset. The corresponding part of the tensor factorization of X has the form Y_{j} = X_{gj} ⊗ X_{g_{jγ}} ⊗ … ⊗ X_{g_{jγ^{m−1}}} , where g_{j}h^{m}g_{j}^{−1} ∈ M_{1} and m is the length of a cycle of γ (the permutation induced by h on the right cosets of M_{1} in G). Lemma 1.3 tells us that Trace_{V_{j}}(h) = Trace_{X_{1}}(g_{j}h^{m}g_{j}^{−1}). Now we have g_{j}h^{m}g_{j}^{−1} = Π_{i=0}^{m−1}g_{jγ^{i}}hg_{jγ^{i+1}}^{−1} (in that order), and λ_{jγ^{i}}^{−1}(g_{jγ^{i}}hg_{jγ^{i+1}}^{−1}, 1) ∈ CM̃_{1}(u_{1}) for each i. Hence the value of the Brauer character of ⟨h⟩k^{×} afforded by Y_{j} takes a real value at hΠ_{i=0}^{m−1}λ_{jγ^{i}}^{−1}. Since Π_{r=1}^{t}λ_{r} = 1, we see that the Brauer character of < h > afforded by X takes a real value at h. Hence X is self-dual as C_{G}(u)-module, the final contradiction.

SECTION 3: GENERIC BOUNDS IN THE SYMPLECTIC CASE

In this section, G is a finite group, q is a prime, F is a finite field of characteristic p ≠ q, V is faithful F G-module. Also, F (G) = Z(G)Q where Q = O_{q}(G) acts absolutely irreducibly on V, and all characteristic Abelian subgroups of Q are central in Q (so also in G under the other hypotheses). We suppose also that Z(G) ≅ F^{×}, that |F | = r, and that [Q : Z(Q)] = q^{2n} = χ(1)^{2}, where χ is the Brauer character of G afforded by V. Then G/F (G) is isomorphic to a subgroup of Sp(2n, q).

We let δ = {g ∈ G : o(g) is 4 or an odd prime}. We note that if |∪_{g∈δ\Z(G)}C_{V} (g)| < 2|V |, then there is some v ∈ V such that C_{G}(v) is an elementary Abelian 2-group (possibly trivial). For if w ∈ V and C_{G}(w) is not an elementary Abelian 2-group, then w is fixed by at least two elements of δ\Z(G). CASE 1: q ≠ 3 q ≠ 3 q ≠ 3: We claim that for each s ∈ δ\Z(G), ⟨Res_{<s>}^{G}(χ), 1⟩ ≤ \frac{q^{n} + 2q^{n−1}}{3} . If q |̸o(s), then it follows by Glauberman correspondence that χ(s) = εq^{m}α for some sign ε, some integer m ≤ n − 1 and some root of unity α. If s has order 4, we may suppose that s^{2} ∉ Z(G), in which case χ(s^{2}) has a similar form. In any case, we certainly have ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{q^{n} + 2q^{n−1}}{3} . If q|o(s), then the claim follows from Corollary 1.2.

Since each element of G has at most q^{n} eigenvalues in F, we see that each coset of Z(G) in G contains at most q^{n} elements g for which |C_{V} (g)| > 1. If V = ∪_{g∈δ\Z(G)}C_{V} (g), it follows that 2|V | = 2r^{q^{n}} ≤ r \frac{q^{n}+2q^{n−1}}{3} q^{n}q^{2n}|Sp(2n, q)| < r^{\frac{q^{n}+2q^{n−1}}{3}} q^{2n^{2}+4n}. This leads easily to r <

(\frac{q^{n^{2}+2n}}{\sqrt{2}} )^{\frac{3}{q^{n}−q^{n−1}}} .

If q ≥ 5, this yields r < (\frac{125}{\sqrt{2}})^{\frac{3}{4}} < (90)^{\frac{3}{4}} , so r ≤ 29. If q = 2 and n ≥ 5, then r < 2^{\frac{207}{32}} , so r ≤ 89. REMARK The results of this section tell us, then, that if q ≥ 5 and r > 29, or if q = 2, n > 5 and r > 89, then the triple (G, V, F) does not provide an obstacle to part a) of hypothesis 2.1.

SECTION 4: THE CASES q = 2, n ≤ 4.

The hypotheses of this section are as in the previous section, but we assume in addition that q = 2 and n ≤ 4. We also suppose that V = ∪_{g∈δ\Z(G)}C_{V} (g). CASE n = 1 n = 1 n = 1: Let C be a non-trivial coset of Z(G) in G. Then | ∪_{g∈C} C_{V} (g)| ≤ 2r. Since [G : Z(G)] ≤ 24 in this case V = ∪_{g∈δ\Z(G)}C_{V} (g) yields 2r^{2} ≤ 46r, so r ≤ 23.

CASE n = 2 n = 2 n = 2: For each s ∈ δ\Z(G), we have ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{2^{3} + 2^{2}}{2} = \frac{8}{3}, so that ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ 2 in this case. For each non-trivial coset C of Z(G) in G, this inequality easily yields | ∪_{g∈δ∩C} C_{V} (g)| ≤ 2r^{2}.

The number of cosets of Z(G) in G is at most 16|Sp(4, 2)| = 2^{8}3^{2}.5. Thus V = ∪_{g∈δ\Z(G)}C_{V} (g) yields 2r^{4} ≤ 2^{9}.3^{2}.5r^{2}, so r ≤ 48\sqrt{5}. Hence r ≤ 107 in this case. CASE n = 3 n = 3 n = 3: In this case, G/F (G) is isomorphic to a subgroup of Sp(6, 2). By a result of Steinberg, the number of 2-elements of Sp(6, 2) is 2^{18}. Hence at most 2^{24} cosets of Z(G) in G contain elements of order 4. For each element s ∈ δ\Z(G) of order 4 we have ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{8 + 12}{4} = 5 by corollary 1.2. For each non-trivial coset C of Z(G) in G which contains such an s ∈ δ, we see easily that | ∪_{g∈δ∩C} C_{V} (g)| ≤ r^{5} + r^{3}.

Now if s ∈ δ\Z(G) has odd prime order l ≥ 5, we see easily (as in earlier arguments) that ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{q^{n} + (l − 1)q^{n−1}}{l} which is at most \frac{8 + 16}{5}, so at most 4. If s ∈ δ\Z(G) has order 3 then χ(s) ∈ ℤ and χ(s) ≡ 8(mod3) give χ(s) ∈ {2, −1, −4}, while χ(s) ∉ ℤ yields χ(s) + χ(s^{−1}) = ±|χ(s)| ≡ 1(mod3) which forces χ(s) + χ(s^{−1}) ∈ {1, −2, 4}, so ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ 4 in all cases. Hence ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ 4 for all s ∈ δ of odd prime order.

Hence as V = ∪_{s∈δ\Z(G)}C_{V} (s), we certainly have 2r^{8} < 2^{24}(r^{5} + r^{3}) + 2r^{4}2^{6}|Sp(6, 2)| (for if C is a non-trivial coset of Z(G) in G which contains an element s ∈ δ of odd prime order,

| ∪_{g∈δ∩C} C_{V} (g)| ≤ 2r^{4}). Now |Sp(6, 2)| < 2^{21}, so we obtain (\frac{r}{2^{7}} )^{3} < 4 + \frac{4}{r^{2}} + \frac{64}{r}. If r > 2^{7} we

obtain (\frac{r}{2^{7}} ) < 4 + \frac{1}{2} + \frac{1}{2^{12}} < (\frac{5}{3})^{3}, so r < \frac{640}{3}. Hence r ≤ 211, as r is a prime power. CASE n = 4 n = 4 n = 4: In this case, Corollary 1.2 and earlier arguments yield that for s ∈ δ not of order 3, we have ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{q^{n} + 3q^{n−1}}{4} = 10. If s ∈ δ has order 3, then an argument like that of the previous case yields ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{q^{n} + q^{n−1}}{3} = 8. Thus for each s ∈ δ, we have ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ 10, and it easily follows that V = ∪_{g∈δ\Z(G)}C_{V} (g) yields 2r^{16} ≤ (r^{10} + r^{6})2^{8}|Sp(8, 2)|. This leads easily to r^{6} ≤ 2^{44}, r ≤ 128 ^{3} \sqrt{2} < 167, so r ≤ 163 as r is a prime power.

SECTION 5: THE CASE q = 3:

In this case, we retain the hypotheses of section 3, and assume now that q = 3, V = ∪_{s∈δ\Z(G)}C_{V} (s). Corollary 1.2 does not apply in this case, and we must revert to the bound ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ \frac{q^{n} + q^{n−1}}{2} for each s ∈ δ (which was obtained in [R-T]).

An elementary argument shows that for each non-trivial coset C of Z(G) in G, we have | ∪_{g∈C} C_{V} (g)| ≤ 2r^{\frac{q^{n}+q^{n−1}}{2}} . Now V = ∪_{s∈δ\Z(G)}C_{V} (s) yields 2r^{q^{n}} ≤ 2r^{\frac{q^{n}+q^{n−1}}{2}} q^{2n}|Sp(2n, q)| so that r < q^{\frac{4n^{2}+6n}{q^{n}−q^{n−1}}} . When n ≥ 2, this yields r < 81 ^{3} \sqrt{3}, so r ≤ 121, as r is a prime power.

When n = 1 we have ⟨Res_{⟨s⟩}^{G}(χ), 1⟩ ≤ 2 for each s ∈ δ, so that for each non-trivial coset C of Z(G) in G, we have | ∪_{g∈δ∩C} C_{V} (g)| ≤ r^{2} + r. Hence V = ∪_{s∈δ\Z(G)}C_{V} (s) yields 2r^{3} < (r^{2} + r) × 9 × 24, so r < 108 + \frac{108}{r}, and r ≤ 107, as r is a prime power.

References

[L] .Liebeck, Regular Orbits of Linear Groups, J. Algebra 184, (1996), 1136-1142.

[R-T] .R.Robinson and J.G. Thompson, On Brauer’s k(B)-problem, J.Algebra 184 (1996), 1143-1160.