On Centers of 2-blocks of Suzuki Groups by Gerald Cliff

Department of Mathematical Sciences University of Alberta Edmonton, Alberta Canada T6G 2G1

1. Introduction

Our main results are that the center of the principal block Z(kGē) of the Suzuki group G = Sz(q), where k is a field of characteristic 2, is isomorphic to the center of the Brauer corresponding block Z(kH), where H is the normalizer in G of a Sylow 2-subgroup, but over a discrete valuation ring R of characteristic 0 whose residue field has characteristic 2, the center of the principal block Z(RGe) is not R-isomorphic to the center Z(RH) of the Brauer corresponding block. These are the first examples known to the author of pairs of blocks whose centers are isomorphic over k but not over R. Here G = Sz(q) denotes the Suzuki group of order q^{2}(q − 1)(q^{2} + 1) where q is an odd power of 2, q ≥ 8, as defined in [Suz]. (There is only one block of kH.) We do not need to assume that R is complete or is large.

The motivation for studying this problem comes, in part, from conjectures of M. Broué [Br2]. Broué has conjectured that there is a perfect isometry between Brauer corresponding blocks if the defect group is abelian; this would imply that the centers of the blocks are isomorphic over a sufficiently large complete discrete valuation ring R, and it would also imply that the bilinear forms defined by the Cartan matrices of the corresponding blocks are equivalent over Z. Originally Broué [Br1] also made this conjecture, for principal blocks, in the case that the Sylow normalizer controls p-fusion, for example in the case of the principal 2-blocks of the Suzuki groups. This turned out not to hold for Sz(8), as in this case the bilinear forms defined by the Cartan matrices of the corresponding blocks are not equivalent over the integers [Br2]. I wrote some computer programs in the mid 1980’s to compute certain invariants of the centers of blocks; these invariants come from my joint work with W. Plesken and A. Weiss [CPW]. Following a suggestion of Broué, I used these programs to show that Z(RGe) and Z(RH) are not isomorphic, where G = Sz(8). The present paper, which does not involve the use of computers, is meant to give some explanation, as well as to extend this result from q = 8 to any odd power of 2 (≥ 8). It also shows that there is no perfect isometry between the principal blocks for G = Sz(q) and its Sylow 2-normalizer, for q equal to any odd power of 2 (≥ 8).

This research was supported in part by a grant from NSERC of Canada. email address: gcliff@math.ualberta.ca

The isomorphism Z(kGē) ≅ Z(kH) follows from the facts that Z(kGē) and Z(kH) have the same dimension over k, and, as we shall prove in Propositions 2.1 and 3.1 below, they both have radical squared equal to 0. The non-isomorphism over R can be explained as follows. Let η be a generator of the maximal ideal of R. (We use the more usual π to denote certain elements of G, following Suzuki.) Suppose that 2R = η^{a}R. Let Λ_{0} denote Z(RGe). Then Proposition 3.1 implies that (rad Λ_{0})^{2} ⊆ η rad Λ_{0}, so (1/η) rad Λ_{0} (which is contained in Z(KGe) where K is the field of fractions of R) is closed under addition and multiplication, and is a ring. Let Λ_{1} denote (1/η) rad Λ_{0}. Suppose that q = 2^{t}. If i is an integer such that i < a(t + 1)/2 and if Λ_{i} has been defined, then from Proposition 3.1, (rad Λ_{i})^{2} ⊆ η rad Λ_{i}, and we let Λ_{i+1} = (1/η) rad Λ_{i} ⊆ Z(KGe). We can form the same construction starting with Λ_{0}(H) = Z(RH), and we define Λ_{i}(H) in the same way as for Λ_{i}. We shall show that for i = a(t − 1)/2, then (rad Λ_{i})^{2} is not contained in η rad Λ_{i}, whereas (rad Λ_{i}(H))^{2} ⊆ η rad Λ_{i}(H). An R-isomorphism from Λ_{0} to Λ_{0}(H) could be extended to an R-isomorphism from Λ_{i} to Λ_{i}(H), and we would have a contradiction.

2. The Sylow Normalizer

From [Suz], §13 and §4, a Sylow 2-Subgroup Q of G has order q^{2}, and its normalizer H = N_{G}(Q) is the semi-direct product QA where A is a cyclic subgroup A of order q − 1; H is a Frobenius group of order q^{2}(q−1). The center C of Q, which is also the commutator subgroup of Q, has order q; C and Q/C are each isomorphic to the additive subgroup of the field GF (q) and A acts regularly on the sets of non-identity elements of both C and Q/C.

For an element τ of H, let C(τ) denote the conjugacy class containing τ. The classes of H are given as follows. There is a unique conjugacy class of involutions, consisting of the non-identity elements of C; for σ in this class, |C_{H}(σ)| = q^{2}. There are two classes of elements of order 4, C(ρ) and C(ρ^{−1}) where ρ is an element of Q not in C; |C_{H}(ρ^{±1})| = 2q. There are q − 2 classes of elements of odd order, namely C(π), π ∈ A, π ≠ 1; C_{H}(π) = A, and |C_{H}(π)| = q − 1. We shall let π denote a typical non-identity element of A. Let

A^{∗} = {π ∈ A : π ≠ 1}.

For τ ∈ H, let Ĉ(τ) denote the class sum of the class C(τ) in RH.

Let B = {1 + Ĉ(σ), Ĉ(ρ), Ĉ(ρ^{−1})} ∪ {Ĉ(π) : π ∈ A^{∗}} ⊂ Z(RH). Then B ∪ {1} is an R-basis of Z(RH).

Proposition 2.1. If b and b′ are in B, write

bb′ = n(b, b′, 1)1 + ∑_{b′′∈B} n(b, b′, b′′)b′′.

Then n(b, b′, 1) is an integer divisible by q^{2}/4, and for each b′′ ∈ B, n(b, b′, b′′) is an integer divisible by q/2.

Proof: For elements τ, τ′ of H, write Ĉ(τ)Ĉ(τ′) = ∑ a(τ, τ′, τ′′)Ĉ(τ′′) where the sum is over class representatives τ′′.

We start by calculating (1 + Ĉ(σ))b′, as b′ varies in B. We have 1 + Ĉ(σ) = ∑_{γ∈C} γ, and therefore

(1 + Ĉ(σ))^{2} = q(1 + Ĉ(σ)).

Thus n(1+Ĉ(σ), 1+Ĉ(σ), 1) = 0, n(1+Ĉ(σ), 1+Ĉ(σ), 1+Ĉ(σ)) = q, n(1+Ĉ(σ), 1+Ĉ(σ), b′′) = 0 for b′′ ∈ B, b′′ ≠ 1 + Ĉ(σ). We have Ĉ(σ)^{2} = (q − 1) · 1 + (q − 2)Ĉ(σ), and for use in the next section, we record that

(2.1)  a(σ, σ, σ) ≡ −2 mod q, a(σ, σ, ρ^{±1}) = 0.

Let Ĉ_{Q}(ρ) denote the class sum of the class of ρ in Q. Fix α ∈ A; for μ ∈ Q, then (ρ^{α})^{μ} = ρ^{α}γ_{μ} for some γ_{μ} ∈ C, since the derived group of Q is C. Since C is central in Q, for μ, ν ∈ Q, ρ^{α}^{μν} = ρ^{α}γ_{μ}γ_{ν}. So the set of γ_{μ}, as μ varies in Q, forms a subgroup of C which we denote by C_{α}. Since A acts regularly on the nontrivial elements of Q/C, then C_{Q}(ρ) = C_{H}(ρ) which has order 2q, so so the size of the class Ĉ_{Q}(ρ^{α}) is q^{2}/2q = q/2; since |C| = q, then |C : C_{α}| = 2. In particular, if α = 1, we see that half the elements γ ∈ C satisfy γĈ_{Q}(ρ) = Ĉ_{Q}(ρ), whereas the other half satisfy γĈ_{Q}(ρ) = Ĉ_{Q}(ρ^{−1}). Since Ĉ(ρ) = ∑_{α∈A} Ĉ_{Q}(ρ^{α}) it follows that (1 + Ĉ(σ))Ĉ(ρ) = (q/2)Ĉ(ρ) + (q/2)Ĉ(ρ^{−1}). Hence n(1 + Ĉ(σ), ρ, 1) = 0, n(1 + Ĉ(σ), ρ, ρ) = n(1 + Ĉ(σ), ρ, ρ^{−1}) = q/2, n(1 + Ĉ(σ), ρ, b′′) = 0 for b′′ ∈ B, b′′ ≠ C(ρ^{±1}).

(2.2)  a(σ, ρ, ρ) ≡ −1 mod \frac{q}{2}, a(σ, ρ, ρ^{−1}) = \frac{q}{2}, a(σ, ρ, σ) = 0.

The integer a(τ, τ′, τ′′) is the number of pairs (γ, γ′) such that γ ∈ C(τ), γ′ ∈ C(τ′) and γγ′ = τ′′. Also |C(τ′′)|a(τ, τ′, τ′′) is the number of triples (γ, γ′, γ′′) such that γ ∈ C(τ), γ′ ∈ C(τ′), γ′′ ∈ C(τ′′) with γγ′ = γ′′. Since γγ′ = γ′′ if and only if γ′ = γ^{−1}γ′′, then |C(τ′′)|a(τ, τ′, τ′′) = |C(τ′)|a(τ^{−1}, τ′′, τ′), so

(2.3)  |C_{H}(τ′)|a(τ, τ′, τ′′) = |C_{H}(τ′′)|a(τ^{−1}, τ′′, τ′).

In this last formula, take τ = ρ, τ′ = σ, τ′′ = ρ^{−1}. Then q^{2}a(ρ, σ, ρ^{−1}) = 2qa(ρ^{−1}, ρ^{−1}, σ), and since a(ρ, σ, ρ^{−1}) = q/2 from (2.2) we get

(2.4)  a(ρ^{−1}, ρ^{−1}, σ) = \frac{q^{2}}{4} = a(ρ, ρ, σ).

In (2.3), let τ = ρ, τ′ = σ, τ′′ = ρ, giving q^{2}a(ρ, σ, ρ) = 2qa(ρ^{−1}, ρ, σ), so

(2.5)  qa(ρ, σ, ρ) = 2a(ρ^{−1}, ρ, σ)
(2.6)  a(ρ^{−1}, ρ, σ) ≡ 0 ≡ a(ρ, ρ^{−1}, σ) mod q/2.

We now compute a(ρ, ρ, ρ). We want to count the number of pairs

(2.7)  (ρ^{αμ}, ρ^{βν}) such that μ, ν ∈ Q, α, β ∈ A, ρ^{αμ}ρ^{βν} = ρ.

Let ρ^{αμ} = ρ^{α}γ, where γ ∈ C_{α}, and ρ^{βν} = ρ^{β}δ, where δ ∈ C_{β}. Then (2.7) implies

(2.8)  ρ^{α}γρ^{β}δ = ρ,

and this in turn gives

(2.9)  ρ^{α}ρ^{β}C = ρC.

We first count solutions to (2.9). We cannot have α = 1 in (2.9) since ρ^{β}C ≠ C. Since A acts regularly on the nontrivial elements of Q/C, for each nonidentity α ∈ A there is exactly one β ∈ A such that (2.9) holds, so we have q − 2 choices for (α, β) in (2.9). For such a pair (α, β) pick any γ ∈ C_{α} and solve (2.8) for δ, so δ = γρ^{−α}ρ^{−β}ρ = γϵ, where ϵ = ρ^{−α}ρ^{−β}ρ which is in C by (2.9). Since δ = γϵ, then γ ∈ C_{α} ∩ ϵC_{β}. We cannot have α = β in (2.9), since (ρ^{2})^{α} ∈ C but ρ ∉ C. Since α ≠ β, then C_{α} ≠ C_{β}, since A acts regularly on the non-identity elements of C. We have 2 cases: if ϵ ∈ C_{β}, then γ ∈ C_{α} ∩ C_{β}; since C_{α} and C_{β} are distinct subgroups of index 2 in the elementary abelian group C, their intersection has index 4, and there are q/4 elements in C_{α} ∩ C_{β}. In the other case, ϵ ∉ C_{β}, and C = C_{β} ∪ ϵC_{β}, so C_{α} = (C_{α} ∩ C_{β}) ∪ (C_{α} ∩ ϵC_{β}), and since |C_{α}| = q/2 and |C_{α} ∩ C_{β}| = q/4, then again C_{α} ∩ ϵC_{β} has q/4 elements. Thus there are (q − 2)q/4 choices for (ρ^{αμ}, ρ^{βν}) in (2.7);

(2.10)  a(ρ, ρ, ρ) = (q − 2)\frac{q}{4} ≡ 0 mod \frac{q}{2}.

Similarly

(2.11)  a(ρ, ρ, ρ^{−1}) = a(ρ, ρ^{−1}, ρ^{±1}) ≡ 0 mod \frac{q}{2}.

\begin{aligned}Ĉ(ρ)^{2} & = a(ρ, ρ, σ)Ĉ(σ) + a(ρ, ρ, ρ)Ĉ(ρ) + a(ρ, ρ, ρ^{−1})Ĉ(ρ^{−1}) \\ & = −a(ρ, ρ, σ) · 1 + a(ρ, ρ, σ)(1 + Ĉ(σ)) + a(ρ, ρ, ρ)Ĉ(ρ) + a(ρ, ρ, ρ^{−1})Ĉ(ρ^{−1}). \end{aligned}

Then n(Ĉ(ρ), Ĉ(ρ), 1) = −n(Ĉ(ρ), Ĉ(ρ), 1 + Ĉ(σ)) = −a(ρ, ρ, σ) = q^{2}/4 from (2.4), and n(Ĉ(ρ), Ĉ(ρ), Ĉ(ρ^{±1}) = a(ρ, ρ, ρ^{±1}) ≡ 0 mod q/2 from (2.10), (2.11).

\begin{aligned}Ĉ(ρ)Ĉ(ρ^{−1}) & = (|C(ρ)| − a(ρ, ρ^{−1}, σ)) · 1 + a(ρ, ρ^{−1}, σ)(1 + Ĉ(σ)) \\ & + a(ρ, ρ^{−1}, ρ)Ĉ(ρ) + a(ρ, ρ^{−1}, ρ^{−1})Ĉ(ρ^{−1}). \end{aligned}

We have

|C(ρ)| = \frac{q(q − 1)}{2} ≡ −q/2 mod q^{2}/2.

From (2.5), a(ρ, ρ^{−1}, σ) = (q/2)a(ρ, σ, ρ), and from (2.2), a(ρ, σ, ρ) ≡ −1 mod q/2, so

(2.12)  a(ρ, ρ^{−1}, σ) ≡ −q/2 mod q^{2}/4.

Then n(Ĉ(ρ), Ĉ(ρ^{−1}), 1) = |C(ρ)| − a(ρ, ρ^{−1}, σ) ≡ 0 mod q^{2}/4. Also, a(ρ, ρ^{−1}, σ) ≡ a(ρ, ρ^{−1}, ρ) ≡ a(ρ, ρ^{−1}, ρ^{−1}) ≡ 0 mod q/2, so n(Ĉ(ρ), Ĉ(ρ^{−1}), b′′) ≡ 0 mod q/2 for b′′ ∈ B.

Since C_{H}(π) = A then Ĉ(π) = π ∑_{γ∈Q} γ = πQ̂. Then

(1 + Ĉ(σ))Ĉ(π) = ĈQ̂ = qĈ(π), Ĉ(ρ^{±1})Ĉ(π) = |C(ρ^{±1})|Ĉ(π) = \frac{q(q − 1)}{2}Ĉ(π).

If π′ ∈ A is not equal to π^{−1}, then

Ĉ(π)Ĉ(π′) = q^{2}Ĉ(ππ′).
Ĉ(π)Ĉ(π^{−1}) = (Q̂)^{2} = q^{2}Q̂ = q^{2}(1 + Ĉ(σ)) + q^{2}Ĉ(ρ) + q^{2}Ĉ(ρ^{−1}).

This completes the proof.

Let η be a generator of the maximal ideal of R. For b ∈ B, it follows from Proposition 2.1 that b^{2} ∈ ηZ(RH), so B ⊆ rad Z(RH), since the elements of B are nilpotent mod η. Further, rad Z(RH) has B ∪ {η · 1} as an R-basis; Z(RH)/ rad Z(RH) is isomorphic to k; hence RH has only one block. (This last statement also follows from the facts that H has a normal Sylow 2-subgroup and no subgroup of odd order.)

3. The Principal 2-Block of the Suzuki Groups

We first describe the characters and conjugacy classes of G. As in the introduction, q is an odd power of 2, q ≥ 8. We set r = \sqrt{2q}. As in [Suz] §17, there are irreducible characters of G called X, X_{i}, Y_{j}, Z_{k}, W_{l}, as well as the trivial character. In the following table, from [Suz], p. 141, the columns contain the characters, their degrees, and the numbers of characters of each degree, respectively.

\begin{matrix}X & q^{2} & 1 \\ X_{i} & q^{2} + 1 & q/2 − 1 \\ Y_{j} & (q − r + 1)(q − 1) & (q + r)/4 \\ Z_{k} & (q + r + 1)(q − 1) & (q − r)/4 \\ W_{l} & r(q − 1)/2 & 2 \end{matrix}

To present the character table, we recall some results and notation used by Suzuki. There are cyclic subgroups A_{0}, A_{1}, A_{2} of G, of orders q − 1, q + r + 1, q − r + 1, respectively. In the table below, π_{i} denotes representatives of the conjugacy classes of elements of nonidentity elements of A_{i}; From Proposition 16 of [Suz], C_{G}(π_{i}) = A_{i}, i = 0, 1, 2. The normalizer of A_{0} is a dihedral group of order 2(q − 1). For i = 1 or 2, the normalizers of A_{i} are Frobenius groups of orders 4(q ± r + 1); each π_{i} is conjugate to its inverse. The element σ is an involution, whose centralizer in G is a Sylow 2-subgroup Q of G; there is a unique conjugacy class of involutions. There are two classes of elements of order 4; ρ is an element of order 4, and ρ^{−1} is not conjugate in G to ρ. The centralizer in G of ρ has order 2q. The Sylow 2-subgroup Q of G is a trivial intersection set.

Let ε_{0} be a primitive (q − 1)st root of unity. If ξ_{0} is a generator of A_{0} the character ε_{0}^{i} of A_{0} is defined by

ε_{0}^{i}(ξ_{0}^{j}) = ε_{0}^{ij} + ε_{0}^{−ij} [Suz], eqn (43), p. 141.

Let ε_{1} be a primitive (q + r + 1)th root of unity, and let ε_{2} be a primitive (q − r + 1)th root of unity. If ξ_{j} is a generator of A_{j}, j = 1 or 2, the character ε_{j}^{i} of A_{i} is defined by

ε_{j}^{i}(ξ_{j}^{k}) = ε_{j}^{ik} + ε_{j}^{ikq} + ε_{j}^{−ik} + ε_{j}^{−ikq} [Suz], eqn (44),(45), p. 141, 142.

The character table, from [Suz] p. 143, is given as follows

1  σ  ρ, ρ^{−1}  π_{0}  π_{1}  π_{2}
X  q^{2}  0  0  1  −1  −1
X_{i}  q^{2} + 1  1  1  ε_{0}^{i}(π_{0})  0  0
Y_{j} (q − r + 1)(q − 1) r − 1  −1  0  −ε_{1}^{j}(π_{1})  0
Z_{k} (q + r + 1)(q − 1) −r − 1  −1  0  0  −ε_{2}^{k}(π_{2})
W_{l}  r(q − 1)/2  −r/2 ±\sqrt{−1}/2  0  1  −1

Here are the class sizes:

\begin{matrix}σ & ρ, ρ^{−1} & π_{0} & π_{1} & π_{2} \\ (q^{2} + 1)(q − 1) & \frac{q(q^{2}+1)(q−1)}{2} & q^{2}(q^{2} + 1) & q^{2}(q − 1)(q − r + 1) & q^{2}(q − 1)(q + r + 1) \end{matrix}

The character X belongs to a block of defect 0. Let e_{X} denote its primitive central idempotent, and let e = 1 − e_{X}. For a group element g, let C(g) denote its conjugacy class and Ĉ(g) its class sum. We have

e_{X} = \frac{X(1)}{|G|} ∑_{g∈G} X(g^{−1})g = \frac{1}{(q − 1)(q^{2} + 1)}(q^{2} + ∑_{π_{0}} Ĉ(π_{0}) − ∑_{π_{1}} Ĉ(π_{1}) − ∑_{π_{2}} Ĉ(π_{2}))

where for i = 0, 1, 2, ∑_{π_{i}} Ĉ(π_{i}) is the sum over the distinct classes of type π_{i}. Since ee_{X} = 0, then

(q^{2} + ∑_{π_{0}} Ĉ(π_{0}) − ∑_{π_{1}} Ĉ(π_{1}) − ∑_{π_{2}} Ĉ(π_{2}))e = 0 in Z(RGe).

Fix a non-identity element π_{0}^{∗} of A_{0}; then

(3.1)  Ĉ(π_{0}^{∗})e = (−q^{2} − ∑_{π_{0}≠π_{0}^{∗}} Ĉ(π_{0}) + ∑_{π_{1}} Ĉ(π_{1}) + ∑_{π_{2}} Ĉ(π_{2}))e.

Let S be a set of representatives of the conjugacy classes of G. Let

B = {Ĉ(g)e : g ∈ S, g ≠ 1, g ∉ C(σ), g ∉ C(π_{0}^{∗})} ∪ {(1 + Ĉ(σ))e} ⊂ Z(RGe).

Then B ∪ {e} is an R-basis of Z(RGe).

We first consider some results about structure constants. Write

Ĉ(g) Ĉ(g′) = ∑_{g′′∈S} a(g, g′, g′′)Ĉ(g′′), g, g′ ∈ G.

Lemma 3.1. Suppose that τ ∈ {σ, ρ^{±1}}, α ∈ G, and π is a non-identity element of A_{i}, i = 0, 1, 2. Then a(α, π, τ) ≡ 0 mod q.

Proof: Let P denote the set

P = {(β, γ) : β ∈ C(α), γ ∈ C(π), βγ = τ}.

Now C_{G}(τ) acts on P. No non-identity element of C_{G}(τ) fixes any element of C(π), since C_{G}(π) ≤ A_{i} and C_{G}(τ) ≤ Q. It follows that the size of P is divisible by |C_{G}(τ)|. The result follows.

Our next result says that some of the structure constants, mod q, are the same as those found in the previous section. If g, g′ are in a subgroup S of G, let C_{S}(g) denote the class of all S-conjugates of g, let Ĉ_{S}(g) be its class sum in RS, and write

Ĉ_{S}(g) Ĉ_{S}(g′) = ∑_{g′′} a_{S}(g, g′, g′′)Ĉ_{S}(g′′)

where the sum is over representatives of the classes of S.

Lemma 3.2. Suppose that α, β, γ are nonidentity elements of the Sylow 2-subgroup Q. Then a(α, β, γ) ≡ a_{H}(α, β, γ) mod q, where H = N_{G}(Q).

Proof: Let

P = {(α′, β′) : α ∈ C(α), β′ ∈ C(β), α′β′ = γ}.

Then C_{G}(γ) acts on P. For some (α′, β′) ∈ P suppose that α′ ∉ Q. Write α′ = α^{μ}. Since C_{G}(α) ≤ Q, C_{G}(α^{μ}) ≤ Q^{μ}, and Q ∩ Q^{μ} = 1, it follows that the size of a C_{G}(γ) orbit containing an element (α′, β′) where α′ ∉ Q is divisible by |C_{G}(γ)|, hence by q. The same result holds for orbits with β′ ∉ Q. Since every element of Q is conjugate in H to one of 1, σ, ρ^{±1}, it follows that mod q, the size of P is the same as that of

{(α′, β′) : α′ ∈ C_{H}(α), β′ ∈ C_{H}(β), α′β′ = γ}.

The result follows.

Proposition 3.1. For elements b, b′ of B, write

bb′ = n(b, b′, e)e + ∑_{b′′∈B} n(b, b′, b′′)b′′.

Then n(b, b′, e) is an integer divisible by q/2, and n(b, b, e) is divisible by q. For each b′′ ∈ B, n(b, b′, b′′) is an integer divisible by r/2, n(b, b, b′′) is divisible by r, and there exist b, b′, b′′ in B for which n(b, b′, b′′) is not divisible by r.

Proof: Suppose that x ∈ Z(RG),

(3.2) x = a_{1} +a_{σ}Ĉ(σ)+a_{ρ}Ĉ(ρ)+a_{ρ^{−1}} Ĉ(ρ^{−1})+∑_{π_{0}} a_{π_{0}} Ĉ(π_{0})+∑_{π_{1}} a_{π_{1}} Ĉ(π_{1})+∑_{π_{2}} a_{π_{2}} Ĉ(π_{1}).

Multiply by e and use equation (3.1) to substitute for Ĉ(π_{0}^{∗})e, giving

(3.3) xe = (a_{1} − a_{σ} − q^{2}a_{π_{0}^{∗}})e + a_{σ}(1 + Ĉ(σ))e + a_{ρ}Ĉ(ρ)e + a_{ρ^{−1}} Ĉ(ρ^{−1})e

+ ∑_{π_{0}≠π_{0}^{∗}}(a_{π_{0}} − a_{π_{0}^{∗}})Ĉ(π_{0})e + ∑_{π_{1}}(a_{π_{1}} + a_{π_{0}^{∗}})Ĉ(π_{1})e + ∑_{π_{2}}(a_{π_{2}} + a_{π_{0}^{∗}})Ĉ(π_{2})ε.

Suppose that b and b′ are in B and have the form b = Ĉ(τ)e and b′ = Ĉ(τ′)e. Let x on the left of (3.2) be Ĉ(τ)Ĉ(τ′), so the coefficients a_{τ′′} on the right in (3.2) are given by the structure constants a(τ, τ′, τ′′). Then from equation (3.3), we get

(3.4)  n(b, b′, e) = a(τ, τ′, 1) − a(τ, τ′, σ) − q^{2}a(τ, τ′, π_{0}^{∗})
≡ a(τ, τ′, 1) − a(τ, τ′, σ) mod q
n(b, b′, (1 + Ĉ(σ))e) = a(τ, τ′, σ),
n(b, b′, Ĉ(ρ^{±1})e) = a(τ, τ′, ρ^{±1}),
n(b, b′, Ĉ(π_{0})e) = a(τ, τ′, π_{0}) − a(τ, τ′, π_{0}^{∗}),
n(b, b′, Ĉ(π_{i})e) = a(τ, τ′, π_{i}) + a(τ, τ′, π_{0}^{∗}), i = 1, 2.

Let b = (1 + Ĉ(σ))e, and suppose that b′, b′′ ∈ B, b′ = Ĉ(τ′)e, b′′ = Ĉ(τ′′)e. Let x = (1 + Ĉ(σ))Ĉ(τ′) = Ĉ(σ)Ĉ(τ′) + Ĉ(τ′) in equation (3.2). Then a_{τ′} = a(σ, τ′, τ′) + 1, and for τ′′ ≠ τ′, a_{τ′′} = a(σ, τ′, τ′′). From equation (3.3), (3.5) n(1 + Ĉ(σ), b′, b′′) are as in the right of (3.4) where τ = σ, except that a(τ, τ′, τ′) must be increased by 1.

Let us start with b = b′ = (1 + Ĉ(σ))e. We have a(σ, σ, 1) = |C(σ)| = (q^{2} + 1)(q − 1) ≡ −1 mod q. From Lemma 3.2, a(σ, σ, σ) ≡ a_{H}(σ, σ, σ) mod q and from (2.1), a_{H}(σ, σ, σ) ≡ −2 mod q. Again from Lemma 3.2 and (2.1), a(σ, σ, ρ^{±1}) ≡ 0 mod q. We calculate a(σ, σ, π_{i}). Since |N_{G}(A_{i})| is divisible by 2 and since C_{G}(A_{i}) = A_{i} and A_{i} is cyclic, there is an involution which acts on A_{i} by inversion; we assume that this is σ. If σ′, σ′′ ∈ C(σ) and σ′σ′′ = π_{i}, then since σ′ and σ′′ are involutions, σ′ and σ′′ generate a dihedral group, and both invert π_{i}. Then σσ′ ∈ C_{G}(π_{i}) = A_{i}, so σ′ acts on A_{i} by inversion, and ⟨A_{i}, σ′⟩ is a dihedral group D_{i} of order 2|A_{i}|. Similarly ⟨A_{i}, σ′′⟩ is a dihedral group, and since σ′σ′′ ∈ C_{G}(A_{i}) = A_{i}, then this dihedral group is also D_{i}. Since D_{i} has one class of involutions, then a(σ, σ, π_{i}) = a_{D_{i}}(σ, σ, π_{i}). We have Ĉ_{D_{i}}(σ) = σÂ_{i}, (Ĉ_{D_{i}}(σ))^{2} = |A_{i}|Â_{i}, and a_{D_{i}}(σ, σ, π_{i}) = |A_{i}|.

The following table summarizes the values we have found for a(σ, σ, τ); the first three entries are modulo q:

\begin{matrix}τ & 1 & σ & ρ^{±1} & π_{0} & π_{1} & π_{2} \\ a(σ, σ, τ) & −1 & −2 & 0 & q − 1 & q + r + 1 & q − r + 1 \end{matrix}

For b = 1 + Ĉ(σ) we shall calculate n(b, b, b′′). In equation (3.2), let x = (1 + Ĉ(σ))^{2} = 1 + 2Ĉ(σ) + Ĉ(σ)^{2}. Then in the right hand side of (3.2), a_{1} = 1 + a(σ, σ, 1) ≡ 0 mod q, a_{σ} = 2 + a(σ, σ, σ) ≡ 0 mod q, a_{τ} = a(σ, σ, τ), τ ∉ C(σ), τ ≠ 1. From (3.3), we get

n(b, b, e) ≡ a_{1} − a_{σ} ≡ 0 mod q
n(b, b, b) = a_{σ} = 2 + a(σ, σ, σ) ≡ 2 − 2 ≡ 0 mod q
n(b, b, Ĉ(ρ^{±1})e) = a_{ρ^{±1}} = a(σ, σ, ρ^{±1}) ≡ 0 mod q
n(b, b, Ĉ(π_{0})e) = a(σ, σ, π_{0}) − a(σ, σ, π_{0}^{∗}) = 0
n(b, b, Ĉ(π_{i})e) = a(σ, σ, π_{i}) + a(σ, σ, π_{0}^{∗}) = (q ± r + 1) + (q − 1) = 2q ± r.

We next let b = (1 + Ĉ(σ))e, b′ = Ĉ(ρ)e, and we calculate n(b, b′, b′′). We have a(σ, ρ, 1) = 0. From Lemma 3.2 and (2.2), a(σ, ρ, σ) ≡ 0 mod q , a(σ, ρ, ρ) ≡ −1 mod q/2, a(σ, ρ, ρ^{−1}) ≡ 0 mod q/2. We compute a(σ, ρ, π_{0}) using the following formula:

(3.6)  a(τ, τ′, τ′′) = \frac{|C(τ)||C(τ′)|}{|G|} ∑_{χ∈Irr(G)} \frac{χ(τ)χ(τ′)χ(τ^{′′−1})}{χ(1)}.

In the sum ∑ χ(σ)χ(ρ)χ(π_{0}^{−1})/χ(1), the only non-zero terms come from χ = X_{i}; in this case, X_{i}(σ) = X_{i}(ρ), so the sum ∑ χ(σ)χ(ρ)χ(π_{0}^{−1})/χ(1) = ∑ χ(σ)χ(σ)χ(π_{0}^{−1})/χ(1), and it follows that a(σ, ρ, π_{0}) is the same as a(σ, σ, π_{0}) except that the class size |C(σ)| in formula (3.6) for a(σ, σ, π_{0}) has to be replaced by |C(ρ)|. Hence a(σ, ρ, π_{0}) = q(q − 1)/2. Using formula (3.6) for a(σ, ρ, π_{1}), we have |C(σ)||C(ρ)|/|G| = (q − 1)(q^{2} + 1)/2q, and ∑ χ(σ)χ(ρ)χ(π_{1}^{−1})/χ(1) = 1 + (r − 1) ∑_{j} ε_{1}^{j}(π_{1})/Y_{j}(1). Since ∑_{j} ε_{1}^{j}(π_{1}) is the sum of all non-trivial q + r + 1 roots of unity, it is −1, so

∑ χ(σ)χ(ρ)χ(π_{1}^{−1})/χ(1) = 1 − \frac{r − 1}{(q − r + 1)(q − 1)},

a(σ, ρ, π_{1}) = \frac{1}{2q} ((q − 1)(q^{2} + 1) − (q + r + 1)(r − 1)) = \frac{1}{2}(q^{2} − q − r).

Similarly a(σ, ρ, π_{2}) = (q^{2} − q + r)/2. To summarize, we have the following, where the first two entries are mod q and the rest are mod q/2:

\begin{matrix}τ & 1 & σ & ρ & ρ^{−1} & π_{0} & π_{1} & π_{2} \\ a(σ, ρ, τ) & 0 & 0 & −1 & 0 & 0 & −r/2 & r/2 \end{matrix}

From (3.6), using b = 1 + Ĉ(σ), b′ = Ĉ(ρ), we have

n(b, Ĉ(ρ)e, e) ≡ a(σ, ρ, 1) − a(σ, ρ, σ) ≡ 0 mod q
n(b, Ĉ(ρ)e, (1 + Ĉ(σ))e) = a(σ, ρ, σ) ≡ 0 mod q/2
n(b, Ĉ(ρ)e, Ĉ(ρ)e) = a(σ, ρ, ρ) + 1 ≡ 0 mod q/2
n(b, Ĉ(ρ)e, Ĉ(ρ^{−1})e) = a(σ, ρ, ρ^{−1}) ≡ 0 mod q/2
n(b, Ĉ(ρ)e, Ĉ(π_{0})e) = a(σ, ρ, π_{0}) − a(σ, ρ, π_{0}^{∗}) ≡ 0 mod q/2
n(b, Ĉ(ρ)e, Ĉ(π_{i})e) = a(σ, ρ, π_{i}) + a(σ, ρ, π_{0}^{∗}) ≡ ∓r/2 mod q/2, i = 1, 2.

Note that n(1 + Ĉ(σ))e, Ĉ(ρ)e, Ĉ(π_{i})e) is divisible by r/2 but not by r, for i = 1, 2. The calculations for b = (1 + Ĉ(σ))e, b′ = Ĉ(ρ^{−1})e are similar.

Continuing with b = (1 + Ĉ(σ))e, let b′ = Ĉ(π_{0})e. We have a(σ, π_{i}, 1) = 0. From Lemma 3.1, a(σ, π_{i}, τ) ≡ 0 mod q for τ ∈ {σ, ρ^{±1}}. We now compute a(σ, π_{0}, π′_{0}) where π_{0}, π′_{0} are nontrivial elements of A_{0}. We have

\frac{|C(π_{0})||C(σ)|}{|G|} = \frac{|C(π_{0})|}{|C_{G}(σ)|} = \frac{q^{2}(q^{2} + 1)}{q^{2}} = q^{2} + 1.

In the sum ∑ χ(σ)χ(π_{0})χ(π′_{0})/χ(1), the only non-zero terms come from the trivial character and χ = X_{i}, and X_{i}(σ) = 1; thus

a(σ, π_{0}, π′_{0}) = (q^{2} + 1)(1 + ∑_{i} X_{i}(π_{0})X_{i}(π′_{0})/(q^{2} + 1)).

Now ∑_{i} X_{i}(π_{0})X_{i}(π′_{0}) is the sum ∑ χ(π_{0})χ(π′_{0}) over all χ ∈ Irr(G) except for the trivial character and the character X. Then

∑_{i} X_{i}(π_{0})X_{i}(π′_{0}) = ∑_{χ∈Irr(G)} χ(π_{0})χ(π′_{0}) − X(π_{0})X(π′_{0}) − 1 = ∑_{χ∈Irr(G)} χ(π_{0})χ(π′_{0}) − 2.

If π_{0} and π′_{0} are in different classes, then ∑_{χ∈Irr(G)} χ(π_{0})χ(π′_{0}) = 0, so ∑_{i} X_{i}(π_{0})X_{i}(π′_{0}) = −2, and

a(σ, π_{0}, π′_{0}) = (q^{2} + 1) − 2 = q^{2} − 1.

If π_{0} = π′_{0}, then ∑_{χ∈Irr(G)} χ(π_{0})χ(π′_{0}) = |C_{G}(π_{0})| = q − 1, and then

a(σ, π_{0}, π_{0}) = q^{2} + 1 + (q − 1) − 2 = q^{2} + q − 2.

For i = 1, 2, χ(σ)χ(π_{0})χ(π_{i}) = 0 for all χ except the trivial character, so a(σ, π_{0}, π_{i}) = q^{2}+1. So we have the following values for a(σ, π_{0}, τ) mod q, where π′_{0} denotes a non-identity element of A_{0} not conjugate to π_{0}:

\begin{matrix}τ & 1 & σ & ρ^{±1} & π_{0} & π′_{0} & π_{1} & π_{2} \\ a(σ, π_{0}, τ) & 0 & 0 & 0 & −2 & −1 & 1 & 1 \end{matrix}

From (3.5), with b = 1 + Ĉ(σ), b′ = Ĉ(π_{0}), π_{0} ≠ π_{0}^{∗} we have
n(b, Ĉ(π_{0})e, e) ≡ a(σ, π_{0}, 1) − a(σ, π_{0}, σ) ≡ 0 mod q
n(b, Ĉ(π_{0})e, (1 + Ĉ(σ))e) = a(σ, π_{0}, σ) ≡ 0 mod q
n(b, Ĉ(π_{0})e, Ĉ(ρ^{±1})e) = a(σ, π_{0}, ρ^{±1}) ≡ 0 mod q
n(b, Ĉ(π_{0})e, Ĉ(π_{0})e) = (a(σ, π_{0}, π_{0}) + 1) − a(σ, π_{0}, π_{0}^{∗}) ≡ −2 + 1 + 1 ≡ 0 mod q
n(b, Ĉ(π_{0})e, Ĉ(π′_{0})e) = a(σ, π_{0}, π′_{0}) − a(σ, π_{0}, π_{0}^{∗}) ≡ 0 mod q
n(b, Ĉ(π_{0})e, Ĉ(π_{i})e) = a(σ, π_{0}, π_{i}) + a(σ, π_{0}, π_{0}^{∗}) ≡ 1 − 1 ≡ 0 mod q.

Continuing with b = (1 + Ĉ(σ))e, let b′ = Ĉ(π_{1})e. Similar to the case that b′ = Ĉ(π_{0})e above, we have

n(b, Ĉ(π_{1})e, e) ≡ n(b, Ĉ(π_{1})e, (1 + Ĉ(σ))e) ≡ n(b, Ĉ(π_{1})e, Ĉ(ρ^{±1})e) ≡ 0 mod q.

We compute a(σ, π_{1}, π_{0}) from (3.6).

\frac{|C(σ)||C(π_{1})|}{|G|} = \frac{|C(σ)|}{|C_{G}(π_{1})|} = \frac{(q − 1)(q^{2} + 1)}{q + r + 1} = (q − 1)(q − r + 1)

and for each χ, χ(σ)χ(π_{1})χ(π_{0}) = 0 except for the trivial character, so

a(σ, π_{1}, π_{0}) = (q − 1)(q − r + 1) ≡ −1 mod r.

We compute a(σ, π_{1}, π′_{1}) from (3.6). In the sum ∑ χ(σ)χ(π_{1})χ(π′_{1})/χ(1), the only non-zero terms correspond to the trivial character, χ = Y_{j}, and W_{l}. Then

a(σ, π_{1}, π′_{1}) = (q − 1)(q − r + 1)(1 + \frac{(r − 1) ∑ Y_{j}(π_{1})Y_{j}(π′_{1})}{(q − r + 1)(q − 1)} − \frac{r}{r/((q − 1)/2)}).

Since ∑ Y_{j}(π_{1})Y_{j}(π′_{1}) is the same as the sum ∑ χ(π)χ(π′) over all χ ∈ Irr(G) except the trivial character, X, and W_{l}, then

∑ Y_{j}(π_{1})Y_{j}(π′_{1}) = ∑ χ(π_{1})χ(π′_{1}) − 4.

If π′_{1} ∉ C(π_{1}), then ∑ χ(π_{1})χ(π′_{1}) = 0, so

a(σ, π_{1}, π′_{1}) = (q − 1)(q − r + 1) − 4(r − 1) − 2(q − r + 1) ≡ 1 mod r.

If π_{1} = π′_{1}, then ∑ χ(π_{1})χ(π_{1}) = |C_{G}(π_{1})| = (q + r + 1), which mod r is 1 greater than in the case that π′_{1} ∉ C(π_{1}), and hence ∑ Y_{j}(π_{1})Y_{j}(π_{1}) ≡ −3 mod r. So

a(σ, π_{1}, π_{1}) = (q − 1)(q − r + 1) − 3(r − 1) − 2(q − r + 1) ≡ 0 mod r.

Compute a(σ, π_{1}, π_{2}) using (3.6): we get

(q−1)(q−r+1)(1+2\frac{(−r/2)(1)(−1)}{r(q − 1)/2}) = (q−1)(q−r+1)+2(q−r+1) ≡ −1+2 ≡ 1 mod r.

So we have, mod r

\begin{matrix}τ & π_{0} & π_{1} & π′_{1} & π_{2} \\ a(σ, π_{1}, τ) & −1 & 0 & 1 & 1 \end{matrix}

For b = (1 + Ĉ(σ))e, b′ = Ĉ(π_{1})e, we have, from (3.6)

n(b, Ĉ(π_{1})e, Ĉ(π_{0})e) = a(σ, π_{1}, π_{0}) − a(σ, π_{1}, π_{0}^{∗}) = 0
n(b, Ĉ(π_{1})e, Ĉ(π_{1})e) = (a(σ, π_{1}, π_{1}) + 1) + a(σ, π_{1}, π_{0}^{∗}) ≡ 0 + 1 − 1 ≡ 0 mod r.
n(b, Ĉ(π_{1})e, Ĉ(π′_{1})e) = a(σ, π_{1}, π′_{1}) − a(σ, π_{1}, π_{0}^{∗}) ≡ 1 − 1 ≡ 0 mod r
n(b, Ĉ(π_{1})e, Ĉ(π_{2})e) = a(σ, π_{1}, π_{2}) − a(σ, π_{1}, π_{0}^{∗}) ≡ 1 − 1 ≡ 0 mod r

The calculations for π_{2} instead of π_{1} are similar.

Let b = Ĉ(ρ). Take b′ = Ĉ(ρ^{±1}). We have a(ρ, ρ, 1) = 0, a(ρ, ρ^{−1}, 1) = |C(ρ)| ≡ q/2 mod q. From Lemma 3.2 and (2.4), a(ρ, ρ, σ) ≡ a_{H}(ρ, ρ, σ) ≡ 0 mod q, and from (2.11) a(ρ, ρ^{−1}, σ) ≡ a_{H}(ρ, ρ^{−1}, σ) ≡ 0 mod q/2, From Lemma 4.2 and (2.10), a(ρ, ρ^{±1}, ρ) ≡ 0 mod q/2. We compute a(ρ, ρ^{±1}, π_{0}) using (3.6); we get the same answer as for a(σ, ρ^{±1}, π_{0}), except that |C(σ)| must be replaced by |C(ρ)| giving

a(ρ, ρ^{±1}, π_{0}) ≡ 0 mod q/2.

To compute a(ρ, ρ^{±1}, π_{1}) using (3.6), we have |C(ρ)||C(ρ^{±1})|/|G| = (q − 1)(q^{2} + 1)/4. Since W_{1}(ρ)W_{1}(ρ)W_{1}(π_{1}^{−1}) + W_{2}(ρ)W_{2}(ρ)W_{2}(π_{1}^{−1}) = 2(−r^{2}/4)/(r(q − 1)/2) = −r/(q − 1), we get

∑ χ(ρ)χ(ρ)χ(π_{1}^{−1})/χ(1) = 1 + \frac{1}{(q − r + 1)(q − 1)} − \frac{r}{q − 1},

a(ρ, ρ, π_{1}) = \frac{1}{4}((q − 1)(q^{2} + 1) + (q + r + 1) − r(q^{2} + 1))

which modulo q^{2}/4 is (1/4)(q − 1 + q + r + 1 − r) hence is 0 mod q/2. The calculation of a(ρ, ρ^{−1}, π_{1}) is similar, except that −r/(q −1) is replaced by +r/(q −1), so that mod q^{2}/4, a(ρ, ρ^{−1}, π_{1}) is (1/4)(q − 1 + q + r + 1 + r) which is ≡ r/2 mod q^{2}/4. Hence a(ρ, ρ^{−1}, π_{1}) is divisible by r/2 (but not by r). The calculations for a(ρ, ρ^{−1}, π_{2}) are similar.

To summarize, we have a(ρ, ρ^{±1}, τ) ≡ 0 mod q/2 for all τ, except that a(ρ, ρ^{−1}, π_{i}) is divisible by r/2 but not by r for i = 1, 2, and a(ρ, ρ, 1) ≡ a(ρ, ρ, σ) ≡ 0 mod q. Thus, from (3.4),

n(Ĉ(ρ)e, Ĉ(ρ)e, e) ≡ a(ρ, ρ, 1) − a(ρ, ρ, σ) ≡ 0 mod q,

n(Ĉ(ρ)e, Ĉ(ρ)e, b′′) ≡ 0 mod r for b′′ ∈ B, and n(Ĉ(ρ)e, Ĉ(ρ^{−1})e, b′′) ≡ 0 mod r/2 for b′′ ∈ B; note that n(Ĉ(ρ)e, Ĉ(ρ^{−1})e, Ĉ(π_{i})e) is not divisible by r, i = 1, 2.

Continuing with b = Ĉ(ρ)e, let b′ = Ĉ(π_{0})e. Then a(ρ, π_{0}, 1) = 0, and from Lemma 3.1, a(ρ, π_{0}, τ) ≡ 0 mod q if τ ∈ {σ, ρ^{±1}}. To compute a(ρ, π_{0}, π′_{0}) where π′_{0} ∈ A_{0}, use (3.6); we get the same answer as for a(σ, π_{0}, π′_{0}), except that |C(σ)| must be replaced by |C(ρ)|, so a(ρ, π_{0}, π′_{0}) ≡ 0 mod q/2. Similarly a(ρ, π_{0}, π_{i}) ≡ 0 mod q/2, i = 1, 2. So for b = Ĉ(ρ)e, b′ = Ĉ(π_{0})e and any b′′, n(b, b′, b′′) ≡ 0 mod q/2. This also holds for b′ = Ĉ(π_{i})e, i = 1, 2.

Let b = Ĉ(π)e and b′ = Ĉ(π′)e, where π and π′ are non-identity elements of odd order in G. Then a(π, π′, 1) is 0 or |C(π)|, which is divisible by q^{2}. From Lemma 3.1, a(π, π′, τ) ≡ 0 mod q if τ ∈ {σ, ρ^{±1}}. Then from (3.3), n(Ĉ(π)e, Ĉ(π′)e, b′′) ≡ 0 mod q if b′ is 1 or Ĉ(τ)e, τ ∈ {σ, ρ^{±1}}.

For the rest of the calculations, we use the following lemma.

Lemma 3.3. Let π, π′, and π′′ be non-identity elements of odd order in G. Then n(Ĉ(π), Ĉ(π′), Ĉ(π′′)) is an integer divisible by 4q^{2}/r.

Proof. Since π and π′ have centralizers of odd order, then

\frac{|C(π)||C(π′)|}{|G|} = \frac{|C(π)|}{|C_{G}(π′)|} = \frac{q^{2}m_{1}}{m_{2}} where m_{1} and m_{2} are odd integers.

Regard π and π′ as fixed, letting π′′ vary. In the sum ∑_{χ∈Irr(G)} χ(π)χ(π′)χ(π^{′′−1})/χ(1), let ζ(π′′)/m_{3}(π′′) be the contribution from χ = X_{i}, Y_{j}, Z_{k}, (which is 0 unless π, π′, π′′ are all in the same A_{i}) where m_{3}(π′′) is the degree of X_{i}, Y_{j} or Z_{k}, hence is odd, and let κ(π′′)/(4(q − 1)/2) be the contribution from χ = W_{l}. Then

a(π, π′, π′′) = \frac{q^{2}m_{1}}{m_{2}} (1 + \frac{X(π)X(π′)X(π′′)}{q^{2}} + \frac{ζ(π′′)}{m_{3}(π′′)} + 2\frac{κ(π′′)}{r(q − 1)/2}).

All numerators and denominators except possibly ζ(π′′) are in Z and a(π, π′, π′′) is an integer, so ζ(π′′) is in Q, hence in Z since it is an algebraic integer. (One can also see that ζ(π′′) ∈ Z from Galois theory.) From (3.4), n(Ĉ(π), Ĉ(π′), Ĉ(π′′)) = a(π, π′, π′′) ± a(π, π′, π_{0}^{∗}); the sign is negative if π′′ ∈ A_{0}, and positive if π′′ ∈ A_{1} or A_{2}. Since X(π_{0}) = 1 and X(π_{1}) = X(π_{2}) = −1, then

n(Ĉ(π), Ĉ(π′), Ĉ(π′′)) = a(π, π′, π′′) − X(π′′)a(π, π′, π_{0}^{∗})

= \frac{q^{2}m_{1}}{m_{2}} ((1 − X(π′′)) + \frac{ζ(π′′) − X(π′′)ζ(π_{0}^{∗})}{m_{3}(π′′)} + \frac{4}{r(q − 1)}(κ(π′′) − κ(π_{0}^{∗}))

+ \frac{X(π)X(π′)X(π′′) − X(π′′)X(π)X(π′)X(π_{0}^{∗})}{q^{2}})

Now 1−X(π′′) ∈ Z, (ζ(π′′)−X(π′′)ζ_{1}(π_{0}^{∗}))/m_{3}(π′′) is a rational number with odd denominator, and (4/r(q − 1))(κ(π′′) − κ(π_{0}^{∗})) is a rational number whose denominator has 2-part at most r/4. Since X(π_{0}^{∗}) = 1, then X(π)X(π′)X(π′′) − X(π′′)X(π)X(π′)X(π_{0}^{∗}) = 0. Thus n(Ĉ(π), Ĉ(π′), Ĉ(π′′)) is a quotient of an integer whose 2-part is q^{2} divided by an integer whose 2-part is at most r/4. The result follows.

The proof of Proposition 3.1 is complete

For b ∈ B, it follows from Proposition 3.1 that b^{2} ∈ ηZ(RGe), so the elements of B are nilpotent mod η; thus B ⊆ rad Z(RGe). Further, rad Z(RGe) has B ∪ {η · e} as an R-basis. Then Z(RGe)/ rad Z(RGe) is isomorphic to k; hence RGe is a block, the principal block. (This last statement could also be proved using the character table of G.)

4. Isomorphism over k and non-isomorphism over R

Proposition 4.1. Suppose that R is a discrete valuation ring of characteristic 0, with maximal ideal Rη, and that Λ is a commutative ring which is a free R module of rank n. Suppose that B is a subset of rad Λ of size n − 1, such that rad Λ has R-basis B ∪ {η · 1}. For b, b′ ∈ B, write

bb′ = n(b, b′, 1)1 + ∑_{b′′∈B} n(b, b′, b′′)b′′.

Suppose that B satisfies the following property P(m), for some positive integer m:

P(m) : n(b, b′, 1) ∈ Rη^{2m}, n(b, b, 1) ∈ Rη^{2m+1}, n(b, b′, b′′) ∈ Rη^{m}, n(b, b, b′′) ∈ Rη^{m+1}, b, b′, b′′ ∈ B.

Then Λ_{1} = (1/η) rad Λ (which is contained in KΛ, where K is the field of fractions of R) is a ring, and the set (1/η)B = {(1/η)b : b ∈ B} is contained in rad Λ_{1} and rad Λ_{1} has R-basis (1/η)B ∪ {η · 1}. Moreover (1/η)B satisfies P(m − 1).

Proof: It follows from P(m) that (rad Λ)^{2} ⊆ η rad Λ, so Λ_{1} = (1/η) rad Λ is closed under addition and multiplication, hence is a ring. For b ∈ B,

b^{2} ∈ Rη^{2m+1} · 1 + ∑_{b∈B} Rη^{m+1}b, (\frac{1}{η} b)^{2} ∈ Rη^{2m−1} · 1 + ∑_{b∈B} Rη^{m} \frac{1}{η} b ⊆ ηΛ_{1}.

Hence modulo ηΛ_{1}, (1/η)b is a nilpotent element of Λ_{1}/ηΛ_{1}, and so (1/η)b ∈ rad Λ_{1}. Since rad Λ has R-basis B ∪ {η · 1}, then rad Λ_{1} has R-basis (1/η)B ∪ {η · 1}. We have

(\frac{1}{η} b)(\frac{1}{η} b′) = (\frac{1}{η^{2}} )n(b, b′, 1) · 1 + ∑_{b′′∈B}(\frac{1}{η} n(b, b′, b′′))(\frac{1}{η} b′′).

It follows that (1/η)B satisfies P(m − 1).

Here is our main result.

Theorem 4.1. For the Suzuki group G and its Sylow 2-normalizer H, the centers of the principal blocks Z(kGē) and Z(kH) are isomorphic, but the centers over R, Z(RGe) and Z(RH) are not R-isomorphic.

Proof: We first compute the k-dimensions of Z(kGē) and Z(kH). There are q/2 − 1 characters of type X_{i}, (q + r)/4 of type Y_{j}, (q − r)/4 of type Z_{k}, 2 of type W_{l}, the character X, and the trivial character, for a total of q + 4 irreducible characters of G. All of these except X are in the principal block, so the center of the principal block Z(kGe) has dimension q + 3 over k. For the Sylow 2-normalizer H, there are q − 1 classes of type π_{0}, the 2 classes C(ρ^{±1}), the class C(σ), and the trivial element, so the dimension of Z(kH) is also q + 3.

From Propositions 2.1 and 3.1, the radical squared of both Z(kGē) and Z(kH) are 0. So an isomorphism between Z(kGē) and Z(kH) is given by sending a basis of the radical of Z(kGē) to a basis of the radical of Z(kH). (Note that there are many isomorphisms.)

Suppose that 2R = Rη^{a} and that q = 2^{t} where t is odd. Then r = (2q)^{1/2} = (2 · 2^{t})^{1/2} = 2^{(t+1)/2}. We shall show that the set B of Section 3 satisfies P(m) where m = a(t − 1)/2. For b, b′, b′′ ∈ B, from Proposition 3.1, n(b, b′, b′′) ∈ Z(r/2) = Z2^{(t−1)/2} ⊆ Rη^{a(t−1)/2} = Rη^{m}. Similarly, since n(b, b, b′′) is divisible by r, then n(b, b, b′′) ∈ Rη^{m+1}. Since n(b, b′, e) is divisible by q/2 = 2^{t−1}, then n(b, b′, e) ∈ Rη^{a(t−1)} and because the integer m = a(t − 1)/2, then n(b, b′, e) ∈ Rη^{2m}. Similarly, since n(b, b, e) is divisible by q, then n(b, b′, e) ∈ Rη^{2m+1}. So B satisfies P(m). Starting with Λ_{0} = Z(RGe), let Λ_{1} = (1/η) rad Λ_{0}. Then (1/η)B satisfies P(m − 1). If i ≤ m and Λ_{i−1} has been defined, let Λ_{i} = (1/η) rad Λ_{i−1}. So (1/η^{i})B satisfies P(m − i). This continues until i = m; (1/η^{m})B satisfies P(0). Proposition 4.1 applied to Λ_{m−1} tells us that (1/η^{m})B ⊆ rad Λ_{m} and (1/η^{m})B ∪ {η · 1} is an R-basis of rad Λ_{m}. For b, b′ ∈ B,

(3.7)  (\frac{1}{η^{m}} b)(\frac{1}{η^{m}} b′) = (\frac{1}{η^{2m}} )n(b, b′, 1) · 1 + ∑_{b′′∈B}(\frac{1}{η^{m}} n(b, b′, b′′))(\frac{1}{η^{m}} b′′).

The coefficient of (1/η^{m})b′′ is n(b, b′, b′′)/η^{m}. From Proposition 3.1, there exist b, b′, b′′ ∈ B for which n(b, b′, b′′) is divisible by r/2 but not by r so n(b, b′, b′′) = (r/2)j where j is an odd integer. Then

\frac{n(b, b′, b′′)}{η^{m}} = \frac{(r/2)j}{η^{m}} = \frac{2^{(t−1)/2}j}{η^{m}} .

Since 2 = η^{a}u where u is a unit of R,

\frac{n(b, b′, b′′)}{η^{m}} = \frac{η^{a(t−1)/2}uj}{η^{m}} = \frac{η^{m}ju}{η^{m}} = ju

which is a unit of R. Since (1/η^{m})B ∪ {η · 1} is an R-basis of rad Λ_{m}, it follows that (rad Λ_{m})^{2} is not contained in η rad Λ_{m}.

Now let Λ(H) = Z(RH), and let B(H) be the set B defined in Section 2. From Proposition 2.1, B(H) also satisfies P(m) where m = a(t − 1)/2, and we define Λ_{i}(H) as in the previous paragraph, starting with Λ_{0} = Z(RH). Since n(b, b′, b′′) of Section 2 are divisible by q/2 for all b, b′, b′′ ∈ B(H), it follows from (3.7) that (rad Λ_{m}(RH))^{2} is contained in η rad Λ_{m}(RH).

An R-isomorphism between Λ = Z(RGe) and Λ(H) = Z(RH) could be extended to an R-isomorphism between Λ_{1} = (1/η) rad Λ and Λ_{1}(H) = (1/η) rad Λ(H), and then, recursively, to an R-isomorphism between Λ_{i} and Λ_{i}(H), for 1 ≤ i ≤ a(t − 1)/2. But for i = a(t − 1)/2, (rad Λ_{i})^{2} is not contained in η rad Λ_{i}, whereas (rad Λ_{i}(H))^{2} is contained in η rad Λ_{i}(H). Therefore Λ = Z(RGe) and Λ(H) = Z(RH) are not isomorphic over R. The proof is complete.

The construction of the series Λ_{i} was inspired by a similar construction in [CPW]; we shall denote the [CPW] series by Λ′_{i} to distinguish it from the series defined in the proof of Theorem 4.1. In [CPW], starting with an R-order Λ, Λ′_{0} defined as Λ, and recursively, Λ′_{i+1} is defined as

Λ′_{i+1} = {x ∈ KΛ_{i} : x rad Λ′_{i} ⊆ rad Λ′_{i}}.

Since η · 1 ∈ rad Λ′_{i}, then ηΛ′_{i+1} ⊆ rad Λ′_{i}, so Λ′_{i+1} ⊆ (1/η) rad Λ′_{i}. For Λ = Z(RGe) of Theorem 4.1, if i < m, since (rad Λ_{i})^{2} ⊆ η rad Λ_{i}, then ((1/η) rad Λ_{i}) rad Λ_{i} ⊆ rad Λ_{i}. It follows that for i ≤ m, the Λ_{i} in the proof of Theorem 4.1 is the same as Λ′_{i} of [CPW].

REFERENCES

[Br1] M. Broué, Isométries parfaites, types de blocs, catégories derivées, Astérisque 181-182 (1990), 61–92.

[Br2] M. Broué, Blocs, isométries parfaites, catégories dérivées, C. R. Acad. Sci. Paris Sér. I Math. 307 (1988), 13–18.

[CPW] G. H. Cliff, W. Plesken, A. Weiss, Order-theoretic properties of the center of a block, The Arcata Conference on Representations of Finite Groups (Arcata, Calif., 1986), 413–420, Proc. Sympos. Pure Math., 47, Part 1, Amer. Math. Soc., Providence, RI, 1987.

[Suz] M. Suzuki, On a class of doubly transitive groups. Ann. of Math. (2) 75 (1962), 105–145.