The minimal polynomials of unipotent elements in irreducible representations of the classical groups in

characteristic p with p-large highest weights

Irina D. Suprunenko Institute of Mathematics Academy of Sciences of Belarus Minsk, 220072, Belarus e-mail: suprunenko@im.bas-net.by

Abstract

Irreducible representations of the classical groups over an algebraically closed field of odd characteristic p are considered. The main goal of the paper is to show that the degree of the minimal polynomial of the image of a unipotent element of a classical group G in an irreducible representation φ is equal to the order of this element provided the highest weight of φ is large enough with respect to p. More exactly, a function l is defined on the set of irreducible representations of G; it is proved that the degree of the minimal polynomial of a unipotent element is equal to its order if l(φ) ≥ p, and for all four types of classical groups, arbitrarily large ranks and every p examples of irreducible representations φ with l(φ) < p where the regular unipotent elements have the minimal polynomials of degrees less than their orders are given. For the special linear group the relevant results follow from the algorithms for computing the minimal polynomials of unipotent elements in irreducible representations announced by the author earlier. If φ is p-restricted, then l(φ) is equal to the value of its highest weight on the maximal root of the group. Some other results on the minimal polynomials of unipotent elements in irreducible representations of the symplectic group are obtained.

1 Introduction

Let K be an algebraically closed field of characteristic p > 2, and let G be a semisimple algebraic group of a classical type over K. This article is one of the series of the author’s papers on the minimal polynomials of unipotent elements in irreducible representations of semisimple algebraic groups over K. Our main goal is to show that the degree of the minimal polynomial of the image of a unipotent element of G in its irreducible representation φ is equal to the order of this element provided the highest weight of φ is large enough with respect to p. Throughout the text Irr is the set of irreducible rational representations of G, ω(φ) is the highest weight of a representation φ, r is the rank of G, ω_{i}, 1 ≤ i ≤ r, are the fundamental weights, and β is the maximal root of G. Let also ⟨μ, α⟩ be the value of the canonical pairing on a weight μ and a root α. Recall that ⟨μ, α⟩ = 2(μ, α)/(α, α) where (,) is a nondegenerate symmetric bilinear form on the set of weights invariant under the action of the Weyl group of G. For any φ ∈ Irr one can write ω(φ) = ∑_{j=0}^{l} p^{j}λ_{j} where λ_{j} are p-restricted dominant weights, i.e. linear combinations of the fundamental weights with integer nonnegative coefficients less than p. Set ω̄(φ) = ∑_{j=0}^{l} λ_{j} (obviously, this sum is uniquely determined) and put l(φ) = ⟨ω̄(φ), β⟩. A representation φ ∈ Irr is called p-restricted if ω(φ) is p-restricted. Unless otherwise stated, we assume that r > 2 for G = B_{r}(K), r > 1 for G = C_{r}(K) and r > 3 for G = D_{r}(K).

Definition. A representation φ is called p-large if l(φ) ≥ p.

Theorem 1.1 Let p > 2 and φ ∈ Irr be p-large. Then for a unipotent element x ∈ G the degree of the minimal polynomial of φ(x) is equal to |x|.

The following theorem shows that the condition l(φ) = p is threshold with respect to the degree of the minimal polynomial of a unipotent element. Here and throughout the article k_{φ}(x) denotes the degree of the minimal polynomial of φ(x), k_{M}(x) is the similar degree for a module M.

Theorem 1.2 Let n be the dimension of the standard representation of G. Suppose that n = p^{l} + 1 for G = A_{r}(K) and C_{r}(K), n = p^{l} + 2 for G = B_{r}(K) and n = p^{l} + 3 for G = D_{r}(K); l > 0 in all cases. Let ω(φ) = (p − 1)ω_{1} for G = A_{r}(K) and C_{r}(K) and ω(φ) = \frac{1}{2}(p − 1)ω_{2} for G = B_{r}(K) and D_{r}(K). Then l(φ) = p − 1 and for a regular unipotent element x ∈ G the degree k_{φ}(x) = (p − 1)p^{l} + 1 < |x| = p^{l+1}.

Recall that n = r + 1 for G = A_{r}(K), 2r + 1 for G = B_{r}(K) and 2r for G = C_{r}(K) or D_{r}(K). Let ω̄(φ) = ∑_{i=1}^{r} b_{i}ω_{i}. Using the formulae for the maximal roots in [2, Tables 1-4], one can easily deduce that l(φ) = ∑_{i=1}^{r} b_{i} for G = A_{r}(K) and C_{r}(K), l(φ) = b_{1} + 2(∑_{i=1}^{r−1} b_{i}) + b_{r} for G = B_{r}(K) and l(φ) = b_{1} + 2(∑_{i=1}^{r−2} b_{i}) + b_{r−1} + b_{r} for G = D_{r}(K).

Theorem 1.3 below yields that for some elements of the symplectic groups the conclusion of Theorem 1.1 holds under much weaker assumptions on ω(φ).

Theorem 1.3 Let G = C_{r}(K), φ ∈ Irr be p-restricted, ω(φ) = ∑_{i=1}^{r} a_{i}ω_{i} and a_{1} + 2 ∑_{i=2}^{r} ≥ p. Let x ∈ G be a unipotent element which lies in the subgroup A_{r−1}(K) naturally embedded into G. Then k_{φ}(x) = |x|.

Corollary 1.4 Let G and φ be as in 1.3, n = 2r. Assume that r ≥ p^{l} + 1, x ∈ G is a unipotent element of order p^{l+1} and in the standard realization of G the element x has either a Jordan block of size ≥ 2(p^{l} + 1), or two Jordan blocks of equal size t ≥ p^{l} + 1. Then k_{φ}(x) = |x|.

Due to Steinberg’s results [16, Theorem 1.1], any irreducible K-representation of a finite classical group in characteristic p is the restriction of some representation of the relevant classical G. This allows to transfer our results to K-representations of finite classical groups in describing characteristic. So the article may be regarded as a contribution to the programme of extending the fundamental results of Hall and Higman [6] on the minimal polynomials of p-elements in finite irreducible p-solvable groups in characteristic p to groups which are not p-solvable. According to [6, Theorem B], the degree of the minimal polynomial of an element of order p^{s} in such group is at least (p − 1)p^{s−1} for p > 2 and 3p^{s−2} for p = 2, s > 1. If p is odd and is not a Ferma prime, then this degree is always p^{s}. Shult [12] has extended these results to the complex case. A.E. Zalesskii has investigated the minimal polynomials of elements of order p in complex irreducible projective representations of finite groups of Lie type over fields of characteristic p [21, 22, 23, 24] and symmetric and alternating groups [26] as well as those of arbitrary p-elements both in complex and p-modular irreducible representations of finite quasisimple groups with cyclic p-Sylow subgroups [25]. (The article [21] has been written together with P.H. Tiep.) In all these situations usually the degree of the minimal polynomial is equal to the order of an element, the exceptional cases are described. The author has found the minimal polynomials of elements of order p in irreducible representations of semisimple algebraic groups over fields of characteristic p [18] and the minimal polynomials of arbitrary unipotent elements in irreducible representations of the group SL_{n}(K) for p > 2 [20]. Theorem 4 [20] yields Theorem 1.1 for G = A_{r}(K). These results can be easily transferred to finite groups of Lie type in describing characteristic.

The notion of a p-large representation has been introduced in [19]. It is proved there that the image of an element of order p in a p-large representation of G has at least f(r) Jordan blocks of size p where f is an explicitly determined linear function. For all types of classical groups examples of irreducible φ with l(φ) = p − 1 such that the image of a long root element in φ has just one Jordan block of size p are given in [19]. Thus, the condition l(φ) = p is threshold with respect to the number of Jordan blocks of size p in images of elements of order p as well.

Notation

Throughout the text C is the field of complex numbers, Z and Z^{+} are the sets of integers and nonnegative integers, E_{l} is the identity matrix of degree l, J_{l} is the upper Jordan block of degree l with the identity on the diagonal.

For a semisimple algebraic group S denote by L(S), W (S), X(S) and R(S) the Lie algebra, the Weyl group, the weight and root systems of S, respectively. X^{+}(S) and R^{+}(S) are the sets of dominant weights and positive roots (with respect to some fixed maximal torus), R^{−}(S) is the set of negative roots, and Π(S) is a basis in R(S). For x ∈ S denote by cl(x) the conjugacy class containing x, and let C̄ be the Zariski closure of a conjugacy class C. If α ∈ R(S), t ∈ K, then X_{α}, X_{α} and x_{α}(t) are the root subgroup and the root elements in L(S) and S associated with α and t ∈ K. For subgroups S_{1}, …, S_{t} of S and vectors v_{1}, …, v_{t} of some vector space we denote by ⟨S_{1}, …, S_{t}⟩ and ⟨v_{1}, …, v_{t}⟩ the subgroup generated by S_{1}, …, S_{t} and the linear span of v_{1}, …, v_{t}. If β_{1}, …, β_{l} ∈ R^{+}(S), the subgroup ⟨X_{β_{i}}, X_{−β_{i}} | 1 ≤ i ≤ l⟩ is denoted by H(β_{1}, …, β_{l}). Put U^{+}(S) = ⟨X_{α} | α ∈ R^{+}(S)⟩ and U^{−}(S) = ⟨X_{α} | α ∈ R^{−}(S)⟩.

Irr S is the set of irreducible rational representations of S, I_{p}(S) ⊂ Irr S is the set of p-restricted representations ; φ|Γ is the restriction of a representation φ to a subgroup Γ. If S is a group over K, then S_{C} is the group of the same type over C. In this case for a representation φ ∈ Irr S we write ”a p-adic expansion” ω(φ) = ∑_{i=0}^{j} p^{i}λ_{i} where λ_{i} are p-restricted, set ω̄(φ) = ∑_{i=0}^{j} λ_{i} and let φ_{C} ∈ Irr S_{C} and ω(φ_{C}) = ω̄(φ). In what follows dim φ (dim M) is the dimension of a representation φ (a module M). For ω ∈ X^{+}(S) denote by M(ω) and V (ω) the irreducible module and the Weyl module with highest weight ω. For an S-module V and a weight vector v ∈ V the symbol ω(v) denotes the weight of v. If H = H(β_{1}, …, β_{k}), then ω_{H}(v) is the weight of v regarded as an element of the H-module V. If β_{1}, …, β_{k} ∈ R^{+}(S) and form a basis of R(H), we assume that ⟨ω_{i}, β_{j}⟩ = δ_{ij} (the Kronecker symbol) for the fundamental weight ω_{i} ∈ X(H). For a vector v ∈ V and an element y of S set V_{y} = (y − 1)^{p−1}V. Only finite dimensional rational representations and modules are considered. Denote by X(φ) (X(V)) the set of weights of a representation φ (a module V). Since the value k_{φ}(x) is the same for a fixed conjugacy class C, one can set k_{φ}(C) = k_{φ}(x) for x ∈ C. For α ∈ R(S), t ∈ K and v ∈ V we have

x_{α}(t)v = ∑_{d=0}^{∞} t^{i}X_{α,d}v  (1)

where X_{α,d} ∈ End V, X_{α,0} = 1, X_{α,d} = 0 for d > d_{0} = d_{0}(V), X_{α,d}V_{μ} ⊂ V_{μ+dα} and X_{α,d} = X_{α}^{d}/d! for d < p ([1, Proposition 5.13]).

If S = G, we often omit the indication of a group in the notation above and write L, X, R, W, etc. instead of L(G), X(G), R(G) and W (G). Set Π = {α_{1}, …, α_{r}}, H(i_{1}, …, i_{s}) = H(α_{i_{1}}, …, α_{i_{s}}), X_{±i} = X_{±α_{i}}, X_{±i} = X_{±α_{i}}, x_{±i}(t) = x_{±α_{i}}(t), X_{±i,d} = X_{±α_{i},d}. For a vector v ∈ V put v(i_{1} · d_{1}, …, i_{t} · d_{t}) = X_{i_{1},d_{1}} … X_{i_{t},d_{t}}v. The fundamental representations and the simple roots of G are labelled as in [2]. Denote by the symbols π_{i} the fundamental representations of G_{C} with highest weight ω_{i}. In what follows ε_{i} with 1 ≤ i ≤ r + 1 for G = A_{r}(K) and 1 ≤ i ≤ r for other classical groups are weights of the standard realization of G labelled as in [3, ch.VIII, §13]. Let n be the dimension of the standard G-module. For a unipotent element x ∈ G denote by J(x) the canonical Jordan form of x in the standard realization of G. We write J(x) = diag(J_{k_{1}}, …, J_{k_{s}}) with k_{1} ≥ … ≥ k_{s}. Recall that |x| = p^{l} if and only if p^{l−1} < k_{1} ≤ p^{l}. The element x is regular if it is conjugate to ∏_{i=1}^{r} x_{i}(1) (and to ∏_{i=1}^{r} x_{−i}(1)). It is well known that a regular element has a single Jordan block in the standard realization of A_{r}(K), B_{r}(K) and C_{r}(K). The end of a proof is marked by the symbol □.

2 Preliminary facts

Theorem 2.1 ([18, Theorem 1.1]) Let S be a semisimple algebraic group over K. Assume that p > 2 if not all simple components of S are groups of types A_{r} or E_{6} and p > 3 if there are groups of types G_{2} or E_{8} among these components. Then there exists a bijection f from the set of unipotent conjugacy classes of S to the similar set for S_{C} such that

k_{φ}(C) = min{p, k_{φ_{C}}(f(C))}

for all classes C which consist of elements of order p and representations φ ∈ Irr S. For the classical groups J(x) = J(y) for every x ∈ C and y ∈ f(C).

We need some notation in order to state results on the minimal polynomials of unipotent elements in irreducible representations of the group SL_{n}(K). Let G = SL_{n}(K), and let x ∈ G be a unipotent element of order p^{s} > 1 with J(x) = diag(k_{1}, ..., k_{d}). Then G_{C} = SL_{n}(C). For φ ∈ Irr set ω = ω(φ_{C}). Denote by y_{0}, y_{1}, …, y_{s} = 1 unipotent elements of G_{C} with J(y_{l}) = J(x^{p^{l}}) for 0 ≤ l ≤ s. Recall that x is regular if and only if d = 1. Then k_{1} ≤ p^{s}. If d > 1, put H_{t} = SL_{k_{t}}(C) with 1 ≤ t ≤ d, and let H = H_{1}×…×H_{d} ⊂ G_{C} (the natural embedding). One can regard y_{l} as an element of H. A weight λ ∈ X(H) can be written in the form (λ_{1}, …, λ_{d}) where λ_{t} ∈ X(H_{t}) is the restriction of λ to a relevant maximal torus of H_{t}; the weights λ_{1}, …, λ_{d} completely determine λ. Set c_{t}(λ) = λ_{t}. In the statement of Theorem 2.2 we use the symbols ε̄_{j} to denote the weights of the standard realization of H_{t}. This causes no confusion because it is clear which t is considered. For ε̄_{j} ∈ X(H_{t}) set f(ε̄_{j}) = k_{t} − 2j + 1. The following notation is introduced for k_{1} < p^{s} only. Write down the p-adic expansions k_{t} = b_{s−1,t}p^{s−1} + … + b_{0,t} for 1 ≤ t ≤ d. Let m = max{t | k_{t} ≥ b_{r−1,1}p^{r−1} +…+b_{1,1}p}, m_{1} = max{t | k_{t} ≥ b_{r−1,1}p^{r−1} +…+(b_{1,1} −1)p}, b = ∑_{t=1}^{m} b_{0,t}.

Theorem 2.2 ([20, Theorems 1-3, Proposition 1])

I. Let d = 1. Then

k_{φ}(x) = min{p^{l}k_{φ_{C}}(y_{l}) | 0 ≤ l ≤ s},

except the case where n = p^{s−1} + p, s > 1 and ω = ω_{p} or ω_{n−p}. In the exceptional case k_{φ}(x) = p^{s} − p + 2.

II. Let s > 1 and k_{1} ≥ p^{s−1} + p. Assume that ω = ∑_{i=1}^{n−1} a_{i}ω_{i} and a_{i} ≠ 0 for some i with p ≤ i ≤ n − p. Then k_{φ}(x) = |x|, except the cases where k_{1} = p^{s−1} + p, ω = ω_{p} or ω_{n−p} and one has s > 2 and k_{2} ≤ p^{s−1} − p provided d > 1. In the exceptional cases k_{φ}(x) is given by Item I if d = 1 and

k_{φ}(x) = max{p^{s} − p + 2, p^{s−1}(p − 1) + p + k_{2} − 1}

for d > 1.

III. If k_{1} = p^{s} and ω ≠ 0, then k_{φ}(x) = p^{s}.

IV. Let d > 1 and k_{1} < p^{s}. Assume also that ω = ∑_{i=1}^{p−1}(a_{i}ω_{i}+a_{n−i}ω_{n−i}) if k_{1} ≥ p^{s−1}+p with s > 1. Define a homomorphism τ : X → X(H) using the rules 1)-7) below. Set τ_{t} = c_{t}τ.

1). For each i, 1 ≤ i ≤ n, there exists j such that τ_{j}(ε_{i}) = ε̄_{q} and τ_{t}(ε_{i}) = 0 for all t ≠ j with 1 ≤ t ≤ d; put j(i) = j and q(i) = q.

2). If u > i and j(i) = j(u), then q(u) > q(i).

3). j(i) = j(n + 1 − i) and q(n + 1 − i) = k_{j(i)} + 1 − q(i).

4). If i ≤ b, then j(i) ≤ m and q(i) ≤ b_{0,j(i)}.

5). If k_{1} ≥ p^{s−1} + p and i < p, then j(i) ≤ m_{1}; if in this situation j(i) > m, then q(i) ≤ b_{0,j(i)}.

6). j(1) = q(1) = 1.

7). If 1 < i ≤ (n + 1)/2 and τ(ε_{u}) are already determined for u < i, then denote by A the set of all pairs (j, ε̄_{k}) with ε̄_{k} ∈ X(H_{j}) which can be taken for (j(i), τ_{j}(ε_{i})) under assumptions 1)-6), and let A_{1} ⊂ A consist of all pairs with the maximal value of f(ε̄_{k}); now choose the (unique) pair from A_{1} with minimal j to determine j(i) and τ_{j}(ε_{i}).

Let ψ ∈ Irr H and ω(ψ) = τ(ω).Then

k_{φ}(x) = min{p^{l}k_{ψ}(y^{l}) | 0 ≤ l ≤ s}.

If k_{1} ≥ p^{s−1} + p, then k_{ψ}(y_{l}) = k_{φ_{C}}(y_{l}) for l > 0.

Remark. In fact, Theorem 2.2 will be applied in this article for regular unipotent elements only, but we want to emphasize that the computation of the minimal polynomials in the general case is much more complicated, though quite explicit. Examples of fundamental representations are available where the minimal polynomials of some unipotent elements with two Jordan blocks cannot be computed with the help of the formula in Item I of Theorem 2.2.

Proposition 2.3 ([18, Propositions 1.3 and 2.6]) Let S be a semisimple group over C, C be a unipotent conjugacy class in S, π, ψ_{1}, ψ_{2} ∈ Irr S, ω(π) = ω(ψ_{1}) + ω(ψ_{2}). Then k_{π}(C) = 1 + k_{ψ_{1}}(C) + k_{ψ_{2}}(C). In particular, if ω(π) = ∑_{i=1}^{r} a_{i}ω_{i}, we have k_{π}(C) = 1 + ∑_{i=1}^{r} a_{i}(k_{π_{i}}(C) − 1).

Algorithm 2.4 ([18, Algorithms 1.4]). Computing the indices k_{π_{i}}(C) for the classical groups. Let i ≤ r for G_{C} = A_{r}(C) and C_{r}(C), i < r for G_{C} = B_{r}(C) and i < r −1 for G_{C} = D_{r}(C), let x ∈ C and J(x) = diag(J_{k_{1}}, …, J_{k_{m}}). Let N(C) be the following sequence of integers:

k_{1} − 1, k_{1} − 3, ..., 1 − k_{1}, k_{2} − 1, ..., 1 − k_{2}, ..., k_{m} − 1, ..., 1 − k_{m}.

Then k_{π_{i}}(C) − 1 is equal to the sum of i largest members of the sequence N(C). For the representation π_{r} of the group B_{r}(C) one has to take the half of such sum. The situation is slightly more complicated for the spinor representations π_{r−1} and π_{r} of the group D_{r}(C). If there is an odd integer among k_{1}, …, k_{m}, we act such as in the case of the representation π_{r} of B_{r}(C). However, if k_{1}, …, k_{m} are all even, there exist two distinct unipotent classes C_{1}, C_{2} ⊂ G_{C} with N(C_{1}) = N(C_{2}). Here k_{π_{r}}(C_{1})−1 is equal to the halfsum of the r largest members of N(C_{1}), k_{π_{r}}(C_{2}) = k_{π_{r}}(C_{1}) − 1, and k_{π_{r−1}}(C_{i}) = k_{π_{r}}(C_{j}) for {i, j} = {1, 2}.

Corollary 2.5 Theorem 1.1 holds for elements of order p.

Proof. It follows from the description of unipotent elements of the classical groups in [15, ch. IV, Exercise 2.15] that the standard realizations of the groups B_{r}(K) and D_{r}(K) do not contain elements x with J(x) = diag(J_{2}, E_{l}) (transvections). Let f be the bijection mentioned in Theorem 2.1, x ∈ G, |x| = p and C = f(cl(x)). It follows from Algorithm 2.4 that k_{π_{i}}(C) − 1 can be less than 2 in the following cases only: G = A_{r}(K) or C_{r}(K); G = B_{r}(K), i = 1 or r and G = D_{r}(K), i ∈ {1, r − 1, r}. Now Proposition 2.3 and the formulae for l(φ) given in the Introduction yield that k_{φ_{C}}(C) > p if φ is p-large. Hence, by Theorem 2.1, k_{φ}(x) = p. □

Corollary 2.6 Let G = A_{r}(K), φ ∈ Irr G_{C}, ω(φ) = ∑_{i=1}^{r} a_{i}ω_{i}. For a unipotent element x ∈ G_{C} we have k_{φ}(x) ≥ k_{π_{3}}(x) if a_{i} ≠ 0 for some i with 3 ≤ i ≤ r − 2, k_{φ}(x) ≥ k_{π_{2}}(x) if a_{i} ≠ 0 for some i with 2 ≤ i ≤ r − 1, and k_{φ}(x) ≥ k_{π_{1}}(x) if ω ≠ 0. If ω(φ) = μ + λ where μ ∈ {ω_{2}, 2ω_{1}, ω_{r−2}, 2ω_{r−1}, ω_{1} + ω_{r−1}}, λ ∈ X^{+}, λ ≠ 0 or ω(φ) = ν + τ where ν ∈ {3ω_{1}, 3ω_{r−1}}, τ ∈ X^{+}, then k_{φ}(x) ≥ k_{π_{3}}(x).

Proof. This follows immediately from Results 2.3 and 2.4. □

Lemma 2.7 Let l = ap^{s} + b, a, b, s ∈ Z^{+}, 0 < b ≤ p^{s}, and let x ∈ SL_{l}(K) be a unipotent element with the single Jordan block . Set y = x^{p^{s}}. Then y has b Jordan blocks of size a+1 and (p^{s}−b) Jordan blocks of size a. Let V be the standard SL_{l}(K)-module, V_{1} = (y−1)^{a}V. Then k_{V_{1}}(x) = b.

Proof. Let e_{1}, …, e_{l} be a basis of V where x acts as the upper Jordan block J_{l}. Observe that (x − 1)^{p^{s}}e_{i} = 0 for i ≤ p^{s} and (x − 1)^{p^{s}}e_{i} = e_{i−p^{s}} for i > p^{s} and that (x − 1)^{p^{s}} = y − 1. For s = 0 the lemma is trivial. Let s > 0. Now it is clear that the subspaces

⟨e_{1}, e_{p^{s}+1}, …, e_{ap^{s}+1}⟩, …, ⟨e_{b}, e_{p^{s}+b}, …, e_{n}⟩, ⟨e_{b+1}, e_{p^{s}+b+1}, …, e_{(a−1)p^{s}+b+1}⟩, …,
⟨e_{p^{s}}, …, e_{ap^{s}}⟩

are invariant with respect to y, the element y has a single Jordan block on each of them and that V_{1} = ⟨e_{1}, …, e_{b}⟩. This implies all our claims. □

Lemma 2.8 Let H = Sp_{l}(K) or SO_{l}(K), l ≥ p^{s}+1, and let x ∈ H be a unipotent element of order p^{s+1}. Then \overline{cl(x)} contains an element y with J(y) = diag(J_{p^{s}+1}, E_{l−p^{s}−1}) for H = Sp_{l}(K) and J(y) = diag(J_{p^{s}+2}, E_{l−p^{s}−2}) or J(y) = diag(J_{p^{s}+1}, J_{p^{s}+1}, E_{l−2(p^{s}+1)}) for H = SO_{l}(K). If H = Sp_{l}(K) and J(x) has either a block of size ≥ 2(p^{s} + 1), or two blocks of size t ≥ p^{s} +1, then \overline{cl(x)} contains an element y with J(y) = diag(J_{p^{s}+1}, J_{p^{s}+1}, E_{l−2(p^{s}+1)}).

Proof. Let k be the maximal size of blocks in J(x). Since |x| = p^{s+1}, we have k ≥ p^{s} + 1. We shall call k proper for H if k is even for H = Sp_{l}(K) and odd for H = SO_{l}(K), otherwise k is called unproper for H. It follows from [15, ch. IV, Exercise 2.15] that x has at least two blocks of size k if k is unproper. The description of the inclusion relations between the conjugacy classes of H in [14, Theorem II.8.2] together with [15, ch. IV, Exercise 2.15] implies the following facts: if J(x) has a block of a proper size t, then for each u ≡ t(mod 2) and u ≤ t the set \overline{cl(x)} contains an element y with J(y) = diag(J_{u}, E_{l−u}); if J(x) has two blocks of size m, then for every positive integer q ≤ m this set contains an element y with J(y) = diag(J_{q}, J_{q}, E_{l−2q}) and if H = Sp_{l}(K) and J(x) has a block of size 2f, then \overline{cl(x)} contains an element y with J(y) = diag(J_{f}, J_{f}, E_{l−2f}). Obviously, \overline{cl(y)} ⊂ \overline{cl(x)} if y ∈ \overline{cl(x)}. Several applications of these facts yield our claims. □

Lemma 2.9 ([18, Lemma 2.20]) Let S be a semisimple algebraic group over K or C, x, y ∈ S be unipotent elements and y ∈ \overline{cl(x)}. Then k_{φ}(y) ≤ k_{φ}(x) for each rational representation φ of S.

Proposition 2.10 Let x ∈ G be unipotent, y = x^{p^{s}} ≠ 1. Let M be a G-module, k_{M}(y) = a + 1, M_{1} = (y − 1)^{a}M, k_{M_{1}}(x) = b. Then b ≤ p^{s} and k_{M}(x) = ap^{s} + b.

Proof. Write M = ⊕_{j=1}^{l}A_{j} where A_{j} are x-invariant subspaces and x acts on each A_{j} as a single Jordan block. Label the subspaces A_{j} such that dim A_{1} ≥ dim A_{j} for all j. Then dim A_{1} = k_{M}(x). Now use 2.7. Set N = M_{1}∩A_{1}. Since x commutes with y, the subspaces M_{1} and N are x-invariant. Put k_{M}(x) = a_{1}p^{s} + b_{1} with a_{1}, b_{1} ∈ Z^{+} and 0 < b_{1} ≤ p^{s}. The arguments of 2.7 applied to all A_{j} show that k_{M}(y) = k_{A_{1}}(y) = a + 1 = a_{1} + 1 and k_{M_{1}}(x) = k_{N}(x) = b = b_{1}. This yields our claim. □

Lemma 2.11 Let s > 1, n = p^{s} − 2 or p^{s} − 1, H = SL_{n}(K), and let x ∈ H be a regular unipotent element. Let φ ∈ Irr H and ω(φ) = ω_{i} with 2 ≤ i ≤ n − 1. Then k_{φ}(x) = p^{s}.

Proof. Recall that J(x) = J_{n}. Set x_{l} = x^{p^{l}}, 0 ≤ l < s, and let y_{l} ∈ H_{C} be a unipotent element with the same canonical Jordan form as x_{l}. For l > 0 Lemma 2.7 yields that the element x_{l} has p^{l}−2 Jordan blocks of size p^{s−l} and 2 blocks of size p^{s−l}−1 if n = p^{s}−2 and this element has p^{l} − 1 Jordan blocks of size p^{s−l} and 1 block of size p^{s−l} − 1 if n = p^{s} − 1. Then Algorithm 2.4 forces that k_{φ_{C}}(y_{0}) ≥ 2p^{s} − 7 > p^{s} and p^{l}(k_{φ_{C}}(y_{l}) ≥ 2p^{s} − 2p^{l} > p^{s} for l > 0. Hence k_{φ}(x) = p^{s} by Theorem 2.2. □

Lemma 2.12 Let s > 1 and s > 2 if p = 3, d = (p^{s} − 3)/2, H = SL_{d}(K), and let x ∈ H be a regular unipotent element. Let φ ∈ Irr H and

ω(φ) ∉ {0, p^{l}ω_{1}, (p^{l} + p^{m})ω_{1}, p^{l}ω_{1} + p^{m}ω_{d−1}, p^{l}ω_{d−1}, (p^{l} + p^{m})ω_{d−1}, p^{l}ω_{2}, p^{l}ω_{d−2}}.

where l, m ∈ Z^{+}. Then either k_{φ}(x) = p^{s}, or p = 3, p^{s} = 27 and ω(φ) = 3^{l}ω_{3} or 3^{l}ω_{d−3}.

Proof. Set a = (p − 1)/2. We have d = ap^{s−1} + … + ap + (p − 3)/2. So Lemma 2.7 implies that for 2 ≤ u < s the element x^{p^{u}} has ≥ p ≥ 3 Jordan blocks of size ≥ ap^{s−1−u} + 1; for p > 3 the element x^{p} has a − 1 blocks of size ap^{s−2} + … + a + 1 and blocks of size ap^{s−2} + … + a; if p = 3, this element has 3 blocks of size 3^{s−2} + 3^{s−3} + … + 1. One can also observe that d = p^{k} + p with k > 0, k ∈ Z^{+} if and only if p = s = 3. Now apply Theorem 2.2 taking the elements y_{j}, 0 ≤ j ≤ s, as in that theorem. Let ω̄(φ) = ∑_{i=1}^{d−1} a_{i}ω_{i}. According to the assumptions of the lemma, one of the following holds:

a) a_{j} ≠ 0 for some j with 2 < j < d − 2,

b) a_{2} + a_{d−2} ≠ 0 and ω̄(φ) ∉ {ω_{2}, ω_{d−2}},

c) a_{1} + a_{d−1} > 2. Proposition 2.3, Algorithm 2.4, Corollary 2.6 and our assumptions on s yield that in this situation k_{φ_{C}}(y_{0}) ≥ 3(d − 3) + 1 ≥ 3ap^{s−1} − 5 for p > 3, k_{φ_{C}}(y_{0}) ≥ 3ap^{s−1} if p = 3, p^{u}k_{φ_{C}}(y_{u}) ≥ 3ap^{s−1} + p^{u} for 1 ≤ u < s, except the case u = 1, s = 2, p < 11 and pk_{φ_{C}}(y_{1}) ≥ p^{2} in the exceptional case. Since s > 1, we have 3ap^{s−1} − 5 = p^{s} − p^{s−1} + (ap^{s−1}/2) − 5 ≥ p^{s} for p > 3. So one easily observes that p^{u}k_{φ_{C}}(y_{u}) ≥ p^{s} for 0 ≤ u < s. Now our claim follows immediately from Theorem 2.2. □

Lemma 2.13 Let p ≥ 11, t = (p − 3)/2, H = B_{t}(K), φ ∈ Irr H and ω(φ) ∉ {0, p^{l}ω_{1}}. Let x ∈ H be a regular unipotent element. Then k_{φ}(x) = p.

Proof. Write ω̄(φ) = ∑_{i=1}^{t} a_{i}ω_{i}. Let z ∈ H_{C} be a regular unipotent element. Applying Proposition 2.3 and Algorithm 2.4, we get that k_{φ_{C}}(z) ≥ 2(p − 4) + 1 = 2p − 7 > p if ∑_{i=2}^{t−1} a_{i} ≠ 0, k_{φ_{C}}(z) ≥ 1 + (p − 1)(p − 3)/8 ≥ p if a_{t} ≠ 0 and k_{φ_{C}}(z) ≥ 2p − 5 > p if a_{1} > 1. Now Theorem 2.1 implies our lemma. □

Lemma 2.14 Let s > 1 and s > 2 if p = 3, d = (p^{s} − 1)/2, H = SL_{d}(K), and let x ∈ H be a regular unipotent element. Let φ ∈ Irr H and

ω(φ) ∉ {0, p^{l}ω_{1}, (p^{l} + p^{m})ω_{1}, p^{l}ω_{1} + p^{m}ω_{d−1}, p^{l}ω_{d−1}, (p^{l} + p^{m})ω_{d−1}, p^{l}ω_{2}, p^{l}ω_{d−2}}.

where l, m ∈ Z^{+}. Then k_{φ}(x) = p^{s}.

Proof. Our arguments are based on Theorem 2.2, Proposition 2.3, Algorithm 2.4, Lemma 2.7 and Corollary 2.6 and are quite similar to that of 2.12. One easily observes that now d cannot be equal to p^{u} + p. Define the elements y_{j}, 0 ≤ j ≤ s, and the coefficients a_{i}, 1 ≤ i ≤ d − 1, as in 2.12. Then the coefficients a_{i} satisfy one of the conditions a)-c) of 2.12, and we can obtain the estimates for k_{φ_{C}}(y_{j}), 0 ≤ j < s, similar to those of 2.12 and show that k_{φ_{C}}(y_{j}) > p^{s}. Now Theorem 2.2 forces our claim. □

Lemma 2.15 Let d = p − 1, H = Sp_{d}(K), φ ∈ Irr H, x ∈ H be a regular unipotent element and ω(φ) ∉ {0, p^{l}ω_{1}} for l ∈ Z^{+}. Then k_{φ}(x) = p.

Proof. Let y ∈ H_{C} be a regular unipotent element. We have ω̄(φ) ∉ {0, ω_{1}}. Then Proposition 2.3 and Algorithm 2.4 yield k_{φ_{C}}(y) ≥ p and Theorem 2.1 implies our lemma. □ Lemma 2.16 Let I ⊂ Π(S) be a proper subset, V be an S-module, and let v ∈ V be a weight vector. Let R_{I} ⊂ R^{+}(S) be the set of all roots which are linear combinations of the simple roots in I and R′ = R^{+}(S)\R_{I}. Assume that x′ ∈ ⟨X_{α} | α ∈ R′⟩, x_{I} ∈ ⟨X_{α} | α ∈ R_{I}⟩, x = x′x_{I} and (x_{I} − 1)^{m}v ≠ 0. Then (x − 1)^{m}v ≠ 0. In particular, k_{V} (x) ≥ k_{V} (x_{I}).

Proof. Let ω = ω(v),

V_{1} = ⟨V_{μ} | μ = ω + ∑_{i∈I} b_{i}α_{i}, b_{i} ∈ Z^{+}⟩, V_{2} = ⟨V_{μ} | V_{μ} ⊄ V_{1}⟩.

Using [17, Lemma 72], one easily concludes that (x_{I} − 1)^{m}v ∈ V_{1} and (x − 1)^{m}v = (x_{I} − 1)^{m}v + v′ with v′ ∈ V_{2}. Since V_{1} ∩ V_{2} = 0 and (x_{I} − 1)^{m}v ≠ 0, this yields the lemma. □

Lemma 2.17 ([5, Lemma 6.14] for s > 1 and [4, ch. VIII, Theorem 2.7] for s = 1). Let 0 < a, b ≤ p^{s} and a + b > p^{s}. Then the degree of the minimal polynomial of J_{a} ⊗ J_{b} = p^{s}. If a + b ≤ p, this degree is equal to a + b − 1.

Proposition 2.18 Let k_{φ}(x) = |x| for all unipotent x ∈ G and all p-large p-restricted representations of G. Then Theorem 1.1 holds for G.

Proof. Let φ be tensor decomposable. Using Steinberg’s tensor product theorem [16, Theorem 1.1], write φ = ⊗_{j=1}^{t}φ_{j} Fr^{i_{j}} where the superscripts i_{j} are all distinct, φ_{j} ∈ I_{p} and ω(φ_{j}) ≠ 0 for all j. Apply induction on t. Let t = 1. As the Frobenius morphism does not influence the minimal polynomial, one can assume that φ ∈ I_{p} in the latter case. Then Theorem 1.1 holds by our assumptions.

Let t > 1. Then we can take φ = φ_{1} ⊕ φ_{2} where φ_{1}, φ_{2} ∈ Irr are nontrivial and ω̄(φ) = ω̄(φ_{1}) + ω̄(φ_{2}). Corollary 2.5 implies that it suffices to consider elements of order p^{s+1} with s > 0. Let |x| = p^{s+1} and put y = x^{p^{s}}. Then |y| = p. Let C = f(cl(y)) ⊂ G_{C} where f is the canonical bijection of Theorem 2.1. Set ψ_{i} = (φ_{i})_{C}, l_{i} = k_{ψ_{i}}(C) − 1 for i = 1, 2. By Proposition 2.3, k_{φ_{C}}(C) = l_{1} + l_{2} + 1. Arguing as in the proof of 2.5, we deduce that k_{φ_{C}}(C) > p whenever φ is p-large. Hence l_{1} +l_{2} ≥ p. Set m_{i} = min{l_{i}, p−1}, i = 1, 2. By Theorem 2.1, k_{φ_{i}}(y) = m_{i} + 1. Hence (y − 1)^{m_{i}} = (x − 1)^{p^{s}m_{i}} ≠ 0 and we get k_{φ_{i}}(x) > p^{s}m_{i}. So φ_{i}(x) has a Jordan block of size q_{i} > p^{s}m_{i}. Since φ_{i} is nontrivial, l_{i} > 0. Now it is clear that q_{1} + q_{2} > p^{s+1}. Then, by 2.17, φ(x) has a Jordan block of size p^{s+1}, i.e. k_{φ}(x) = p^{s+1} as required. □

Lemma 2.19 Let N be a G-module, μ ∈ X, u ∈ N_{μ}\{0}, α ∈ R, X_{α,b}u = 0 for b > 0 and ⟨μ, α⟩ = c. Then X_{α}X_{−α,b}u = (c − b + 1)X_{−α,b−1}u and X_{−α,c}u ≠ 0. In particular, X_{−α,d}u ≠ 0 for d ≤ c if 0 < c < p.

Proof. In fact, this is well known. Relevant comments can be found in [19, Lemma 2.1]. □

In the definition and Lemmas 2.20 and 2.21 below M is a p-restricted G-module with highest weight ω = ∑_{i=1}^{r} a_{i}ω_{i}, v ∈ M is a nonzero highest weight vector.

Definition. Let 1 ≤ i, j ≤ r, and let all the roots α_{t} with t in the interval with the ends i and j form a chain. Set x_{k} = −⟨α_{k+1}, α_{k}⟩, y_{k} = −⟨α_{k−1}, α_{k}⟩. For an integer d with 0 < d ≤ a_{j} define the vector v(i, j, d) as follows. Put d_{j} = d. If i < j, set d_{k} = a_{k} + d_{k+1}x_{k} for i ≤ k < j. If i > j, put d_{k} = a_{k} + d_{k−1}y_{k} for i ≥ k > j. Now take v(i, j, d) = v(i · d_{i}, …, k · d_{k}, …, j · d). For i = j put v(i, j, d) = X_{−i,d}v.

Lemma 2.20 We have v(i, j, d) ≠ 0 and X_{l,b}v(i, j, d) = 0 for positive l ≠ i and b > 0. Hence X_{l} fixes v(i, j, d).

Proof. One can obtain the inequality applying 2.19 several times. Other assertions of the lemma are contained in [19, Lemma 2.9]. □

Lemma 2.21 ([19, Lemma 2.10]) Put m = v(i, j, d). If i < j, suppose that i > 1 and ⟨α_{k}, α_{k−1}⟩ = −1 for i ≤ k ≤ j. If i > j, assume that i < r and ⟨α_{k}, α_{k+1}⟩ = −1 for j ≤ k ≤ i. Put l = i − 1 for i < j and l = i + 1 for i > j. If i = j, suppose that l ∈ {i − 1, i + 1} and ⟨α_{i}, α_{l}⟩ = −1. Let ⟨ω(m), α_{l}⟩ = p. Then X_{−l}m ≠ 0.

Proposition 2.22 ([18, Proposition 2.15]) Let ω = λ_{1} + λ_{2}, λ_{j} ∈ X^{+} for j = 1, 2, and x ∈ U^{−}. Put M^{j} = M(λ_{j}) and for each l ∈ Z^{+} set

X_{l} = {μ ∈ X(M) | ω − μ = ∑_{i=1}^{r} c_{i}α_{i}, ∑_{i=1}^{r} c_{i} ≥ l}.

Let v_{j} ∈ M_{λ_{j}}^{j} and v ∈ M_{ω} be nonzero, and let (x − 1)^{k_{j}}v_{j} ≠ 0 for some k_{j} ∈ Z^{+}, j = 1, 2. Assume that k = k_{1} +k_{2}, \binom{k}{k_{1}} ≢ 0(mod p) and dim V (ω)_{μ} = dim M_{μ} for all μ ∈ X_{k}. Then (x − 1)^{k}v ≠ 0. In particular, this inequality holds if X_{k} consists of the lowest weight of M and in the case where the module V (ω) is irreducible, |x| = p and m < p.

Proposition 2.23 Let S_{1} and S_{2} be semisimple algebraic groups, S = S_{1} × S_{2}, and let U be an indecomposable S-module generated by a nonzero highest weight vector u. Set μ = ω(u), μ_{i} = ω_{S_{i}}(u) for i = 1, 2 and U_{1} = KS_{1}u. Assume that U_{1} ≅ V (μ_{1}). Then U has a quotient isomorphic to V (μ_{1}) ⊗ M(μ_{2}).

Proof. Put V = V (μ), V_{i} = V (μ_{i}), M = M(μ_{2}). By the universal property of the Weyl modules [8, Part II, Lemma 2.13], U = V/N where N ⊂ V is an S-submodule. Using [17, §12, Corollary a) of Lemma 68] and dimension arguments, one also concludes that V = V_{1} ⊗ V_{2} and a nonzero highest weight vector v ∈ V can be identified with v_{1} ⊗ v_{2} where v_{i} ∈ V_{i} are nonzero highest weight vectors. We have M = V_{2}/N_{2} where N_{2} ⊂ V_{2} is the maximal S_{2}-submodule. Set V′_{1} = V_{1} ⊗ v_{2}. It is clear that U_{1} ≅ V′_{1}/V′_{1} ∩ N. Since U_{1} ≅ V_{1}, we get V′_{1} ∩ N = 0. It suffices to prove that N ⊂ V_{1} ⊗ N_{2}. Indeed, in this case V_{1} ⊗ M ≅ V_{1} ⊗ V_{2}/V_{1} ⊗ N_{2} ≅ V_{1} ⊗ V_{2}/N/(V_{1} ⊗ N_{2}/N). Suppose N ⊄ V_{1} ⊗ N_{2}.

Let x_{1}, …, x_{d} and y_{1}, …, y_{f} be bases of V_{1} and V_{2}, respectively, which consist of weight vectors. Assume also that for all μ ∈ X(V_{2}) the sequence y_{1}, …, y_{f} contains bases of (N_{2})_{μ}. Then there exists a vector z ∈ N such that z = ∑_{i,j} b_{ij}x_{i} ⊗ y_{j} where b_{ij} ∈ K and for some k, l the coefficient b_{kl} ≠ 0 and y_{l} ∉ N_{2}. Let

Ω_{t} = {ν ∈ X(V_{2}) | μ_{2} − ν is a sum of t roots of Π(S_{2})},

J_{t} = {j | ω(y_{j}) ∈ Ω_{t}}. Choose maximal t such that for some l ∈ J_{t} and some k the coefficient b_{kl} ≠ 0 and y_{l} ∉ N_{2}. Fix these k and l. Observe that N contains weight components of its vectors. In particular, if a linear combination x ⊗ v_{2} + ∑_{j} u_{j} ⊗ w_{j} ∈ N, x, u_{j} ∈ V_{1}, w_{j} ∈ V_{2} are weight vectors and ω(w_{j}) < μ_{2}, then x ⊗ v_{2} ∈ N. Hence t > 0 since V′_{1} ∩ N = 0. Set Θ = \{q | ω(y_{q}) = ω(y_{l}), b_{kq} ≠ 0\}, y = ∑_{q∈Θ} b_{kq}y_{q}. Let ȳ and \overline{v_{2}} be the images of y and v_{2} under the canonical homomorphism V_{2} → M. Then ȳ ≠ 0. Since only highest weight vectors of M are fixed by the group U^{+}(S_{2}), there exists an operator X = X_{β_{1},d_{1}} … X_{β_{s},d_{s}} where β_{1}, …, β_{l} ∈ R^{+}(S_{2}) such that Xȳ = cv_{2} where c ∈ K^{∗}. As the operator X sends distinct weight subspaces to distinct ones and for each m > t either J_{m} = ∅, or y_{j} ∈ N_{2} for all j ∈ J_{m}, it is not difficult to deduce that Xz has a nonzero weight component of the form x ⊗ v_{2}. This yields a contradiction since Xz ∈ N. Hence N ⊂ V_{1} ⊗ N_{2} as required. □

Lemma 2.24 ([19, Lemma 2.6]) Let Γ = SL_{2}(K), μ = aω_{1} ∈ X^{+}(Γ), M = M(μ), and V = V (μ).

i) We have V = ⟨X_{−α,d}v | 0 ≤ d ≤ a⟩ where v is a nonzero highest weight vector and dim V (μ) = a + 1.

ii) Let a < p, x ∈ Γ be a nonidentity positive root element and m ∈ M be a nonzero highest weight vector. Then dim M = a + 1 and (x − 1)^{a}M = ⟨m⟩.

iii) Let a = ∑_{j=0}^{k} a_{j}p^{j} with all a_{j} < p. Then dim M = ∏_{j=0}^{k}(a_{j} + 1) and

k_{M}(x) = min{p, 1 + ∑_{j=0}^{k} a_{j}}.

Proof. The formula for k_{M}(x) in iii) is the unique assertion of the lemma which was not fornulated explicitly in [19, Lemma 2.6]. This formula follows from Steinberg’s tensor product theorem [16, Theorem 1.1], ii) and Lemma 2.17. □

Lemma 2.25 Let G = B_{r}(K) and n = 2r +1 = p^{s} −2 with s > 1. Let x ∈ G be a regular unipotent element. Assume that χ ∈ Irr G, ω(χ) ∈ {2ω_{1}, ω_{i} | 2 ≤ i ≤ r} and s > 2 if p = 3 and ω(χ) = ω_{r}. Then k_{χ}(x) = p^{s}.

Proof. One can assume that x = ∏_{i=1}^{r} x_{−i}(1). Let M be the standard G-module and v ∈ M be a nonzero highest weight vector. It is well known that x has a single Jordan block on M and (x − 1)^{p^{s}−3}v ≠ 0. Set ω = ω(χ).

First assume that ω = 2ω_{1}. Apply Proposition 2.22 with λ_{1} = λ_{2} = ω_{1}, k_{1} = p^{s} − 3, k_{2} = 2. Put V = V (ω). By [3, ch. VIII, § 7, Proposition 1], X(V) = {±ε_{i} ± ε_{j} | 1 ≤ i, j ≤ r}. Hence X(V) ∩ X^{+} = {ω, ω_{2}, 0}. As ω_{2} = ω − α_{1}, [17, Theorem 39] and Lemma 2.19 yield that dim U_{ω_{2}} = dim V_{ω_{2}} = 1 for each G-module U with highest weight ω_{2}. Now it is clear that dim(M(ω_{2})_{μ}) = dim V_{μ} for all μ ∈ X^{+}(V)\{0}. In particular, this holds for μ ∈ X_{k} in the notation of 2.22. Since \binom{p^{s}−1}{2} ≢ 0(mod p), we obtain that k_{χ}(x) = p^{s}.

Next, let ω = ω_{i} with 2 ≤ i < r. It is well known that χ can be regarded as a representation of the group Ḡ = SO_{n}(K) and is equivalent to the restriction τ|Ḡ where τ = φ(ω_{i}) ∈ Irr SL_{n}(K) (see, for instance, [11, (8.1)]). The image of x under the standard homomorphism G → Ḡ is a regular unipotent element in SL_{n}(K). Now 2.11 yields the required claim.

Now let ω = ω_{r}. Set Γ = H(i | i < r). Then Γ ≅ A_{r−1}(K). It is well known that X(χ) coincides with the orbit of ω under W (see, for instance, [9, Lemma 5.4.10 ]). So one easily observes that the weight ν = (∑_{i=1}^{r−4} ε_{i} − ∑_{j=r−3}^{r} ε_{j})/2 ∈ X(χ) and ν + lα_{i} ∉ X(χ) for i < r and l ∈ Z^{+}\{0}. Then it is clear that the module M(ω) has a nonzero vector m of weight ν fixed by the groups X_{i} for 1 ≤ i < r. The vector m generates an indecomposable Γ-module with highest weight δ = ω_{r−4}. Hence ρ = φ(δ) is a composition factor of χ|Γ. Set y = ∏_{i=1}^{r−1} x_{i}(1). Then y is a regular unipotent element of Γ. By Lemma 2.12, k_{ρ}(y) = p^{s} (observe that r > 10 in our case). Now Lemma 2.16 implies that k_{χ}(x) = k_{χ}(y) = k_{ρ}(y) = p^{s}. □

Lemma 2.26 Let G = B_{3}(K), p = 3, ω = ω_{1} + ω_{3} ∈ X(G), φ = φ(ω). Let x ∈ G be a regular unipotent element. Then k_{φ}(x) = 9.

Proof. One can assume that x = ∏_{i=1}^{3} x_{−i}(1). Apply Proposition 2.22 with λ_{1} = ω_{1}, λ_{2} = ω_{3}, k_{1} = 6 and k_{2} = 2. Put M_{i} = M(λ_{i}), and let v_{i} ∈ M_{i} be nonzero highest weight vectors for i = 1, 2. Then M_{1} is the standard G-module and (x − 1)^{6}v_{1} ≠ 0. Applying 2.19 several times, we get that X_{−2}X_{−1}v_{2} ≠ 0. Formula (1) yields that X_{−2}X_{−1}v_{2} is the weight component of weight ω − α_{1} − α_{2} of the vector (x − 1)^{2}v_{2}. Hence (x − 1)^{2}v_{2} ≠ 0. Using Jantzen’s filtration formula [8, Part II, Proposition 8.19] and the fact that the Euler characteristic χ(λ) = 0 if λ ∈ X and ⟨λ, α_{i}⟩ = −1 for some i with 1 ≤ i ≤ r which follows, for instance, from [8, Part II, Proposition 5.4a)], one can check that the module V (ω) is irreducible. Since \binom{8}{2} ≢ 0(mod 3), Proposition 2.22 yields that k_{φ}(x) > 8. Hence k_{φ}(x) = 9 as |x| = 9. □

3 The proof of Theorem 1.1

In this section φ ∈ Irr is a p-large representation, ω = ∑_{i=1}^{r} a_{i}ω_{i} = ω(φ), M = M(ω), v ∈ M is a nonzero highest weight vector. For a classical algebraic group H over K we shall say that (H, s)-conjecture holds if k_{φ}(g) = p^{s} for every p-large φ ∈ Irr H and each element g ∈ H of order p^{s}. In this context we write B_{l}, C_{l}, D_{l} instead of B_{l}(K), C_{l}(K) and D_{l}(K).

3.1 The general scheme of the proof of Theorem 1.1. As we have mentioned in the Introduction, for G = A_{r}(K) Theorem 1.1 has been obtained in [20, Theorem 4]. So it suffices to assume that G ∈ {B_{r}(K), C_{r}(K), D_{r}(K)}. It seems convenient to consider the cases G = C_{r}(K), B_{r}(K) and D_{r}(K) separately. Proposition 2.18 shows that it suffices to handle p-restricted representations. By 2.5, Theorem 1.1 holds for elements of order p. This creates the base for induction on the order of elements. So we are done if we prove Theorem 1.1 for elements of order p^{s+1}, s ≥ 1 assuming that (H, s)-conjecture holds for some fixed classical sg H ⊂ G. Recall that the maximal order of a p-element in SL_{n}(K) is equal to the smallest power of p which ≥ n. For both symplectic and orthogonal cases we also want to use induction on r. So we first consider the cases where G = B_{r}(K) or C_{r}(K) and 2r = p^{s} + 1 which yield the groups of the minimal rank containing elements of the given order. These cases are handled separately by arguments of similar nature. In what follows we call them the SM and OM cases. Let one of these cases hold. Set x = ∏_{i=1}^{r} x_{i}(1) ∈ G, x^{p^{s}} = y, M_{y} = M_{1}. Using the description of conjugacy classes in [15, ch. IV, Exercise 2.15], one easily concludes that all elements of G of order p^{s+1} are conjugate to x. In both SM and OM cases y is a root element of G associated with the maximal root β. We construct a nonzero weight vector m ∈ M such that ⟨ω(m), β⟩ = p−1 and the group X_{β} fixes m. By Lemma 2.24, m ∈ M_{1}. Put C = C_{G}(y). Then KCm ⊂ M_{1}. Set I = {2, …, r} for G = C_{r}(K) and I = {1, 3, …, r} for G = B_{r}(K), R′ = R^{+}\I and H = H(i | i ∈ I) for both groups. Looking at the explicit formula for β, one easily observes that H ⊂ C. Hence the KH-module N generated by m is contained in M_{1}. Let z = ∏_{i∈I} x_{i}(1). Then z is a regular unipotent element of H. We claim that it suffices to prove that k_{N}(z) = p^{s}. Indeed, in this case there exists a weight vector u ∈ N ⊂ M_{1} with (z − 1)^{p^{s}−1}u ≠ 0. One can write x = x′z where x′ ∈ ⟨X_{α} | α ∈ R′⟩. Now 2.16 implies that (x − 1)^{p^{s}−1}u ≠ 0. Hence k_{M_{1}}(x) = p^{s} as this value cannot be greater. Then Proposition 2.10 yields that k_{φ}(x) = p^{s+1}.

So to handle the cases SM and OM it is crucial to show that k_{N}(z) = p^{s}. This equality is proved separately for each case with the use of Results 2.11-2.26 and (H, s)-conjecture.

Now let G = C_{r}(K) and 2r > p^{s} + 1. We assume that (C_{r−1}, s + 1)-conjecture holds. It follows from 2.8 that for any x of order p^{s+1} lying in G the closure \overline{cl(x)} contains an element from the subgroup H = C_{r−1}(K) naturally embedded into G. We construct a p-large composition factor of the restriction φ|H. Then 2.9 implies our assertion for G = C_{r}(K).

Finally, let G = B_{r}(K) or D_{r}(K) and 2r > p^{s} + 1. We assume that (D_{r}, s + 1)- conjecture holds for G = B_{r}(K) and (B_{r−1}, s + 1)-conjecture holds for G = D_{r}(K). Let x ∈ G be a unipotent element of order p^{s+1}. Lemma 2.8 implies that for G = B_{r}(K) the set \overline{cl(x)} contains an element of order p^{s+1} lying in the naturally embedded subgroup D_{r}(K) and that for G = D_{r}(K) the closure \overline{cl(x)} contains an element of this order lying in a naturally embedded subgroup of type B_{r−1} or A_{r−1}. We show that the restrictions of φ to the relevant subgroups have p-large composition factors. Now Lemma 2.9 yields that k_{φ}(x) = p^{s+1}. □

Remark. Unfortunately, the proof of Theorem 1.1 for G = A_{r}(K) requires a more strong inductive hypothesis when we apply induction on r. Here, when dealing with the case r = p^{s}, we need to show that a regular unipotent element of A_{r−2}(K) (≅ SL_{p^{s}−1}(K)) has the minimal polynomial of degree p^{s} in an irreducible representation φ if ω(φ) ∉ {0, p^{l}ω_{1}, p^{l}ω_{r−3}}. We cannot prove this without assuming that the whole Theorem 2.2 holds for groups of type A_{j}, j < r, Item I which handles regular elements seems not enough. And the proof of the latter theorem is quite involved and could occupy a hundred of pages, [20] contains only a scheme of the proof with the assertions of key tool lemmas. Naturally, for the special linear group we would also like to have a self-contained proof of Theorem 1.1 with an inductive hypothesis concerning p-large representations only, but such proof is not available.

From now on until the end of the section φ ∈ I_{p}. We proceed to prove Theorem 1.1 for G = C_{r}(K). Recall that for this group n = 2r. According to [2, Table III], β = 2(α_{1} + … + α_{r−1}) + α_{r}.

Proposition 3.2 Let G = C_{r}(K), n = p^{s} + 1, s ≥ 1, and let (C_{r−1}, s)-conjecture hold. Then k_{φ}(x) = p^{s+1} for each x ∈ G of order p^{s+1}.

Proof. We use the scheme of 3.1. One can take x = ∏_{i=1}^{r} x_{i}(1). Set y = x^{p^{s}}, M_{y} = M_{1}. Considering the action of x and y on the standard G-module, one easily concludes that y = x_{β}(1). Observe that ⟨α_{1}, β⟩ = 1 and ⟨α_{i}, β⟩ = 0 for i > 1. Since φ is p-large and p-restricted, there exist j < r and d ∈ Z^{+} with 0 < d ≤ a_{j} such that

a_{j} − d + ∑_{i=j+1}^{r} a_{i} = p − 1.  (2)

Put H = H(2, …, r), m = v(1, j, d), μ = ω_{H}(m). Lemma 2.20 shows that the vector m ≠ 0 and generates an indecomposable H-module N with highest weight μ. One easily deduces that

μ = (∑_{i=1}^{j−2} a_{i}ω_{i}) + (a_{j−1} + a_{j} − d)ω_{j−1} + (d + a_{j+1})ω_{j} + ∑_{l=j+1}^{r−1} a_{l+1}ω_{l}

if j > 1 (here for j = 2 the first summand is (a_{1} + a_{2} − d)ω_{1} and for j = r − 1 the last summand is (d + a_{r})ω_{r−1}),

μ = (a_{2} + d)ω_{1} + ∑_{i=2}^{r−1} a_{i+1}ω_{i}

if j = 1, r > 2 and

μ = (a_{2} + d)ω_{1}

for r = 2. It is clear that X_{β} preserves m, and one can directly verify that ⟨ω(m), β⟩ = p − 1. Set N = KHm, z = ∏_{i=2}^{r} x_{i}(1), ψ = φ(μ) ∈ Irr H. The arguments in 3.1 yield that it suffices to show that K_{N}(z) = p^{s}. Obviously, N has a composition factor ψ. Hence k_{φ}(x) = p^{s+1} if k_{ψ}(z) = p^{s}.

Observe that μ ≠ 0. Put t = z^{p^{s−1}}. Then |t| = p. Lemma 2.7 forces that t has p^{s−1} − 1 Jordan blocks of size p and one block of size p − 1. Now Results 2.1, 2.3 and 2.4 yield that for a representation π ∈ Irr H with ω(π) ≠ 0 the degree k_{π}(t) = p if s > 1 and for s = 1 this degree is at least p − 1. Hence in all cases k_{π}(z) ≥ (p − 1)p^{s−1}. Since p > 2, now it follows from Lemma 2.17 that k_{ψ}(z) = p^{s} as required if ψ is tensor decomposable. So we can and shall exclude cases with tensor decomposable ψ. We shall exclude cases with p-large ψ as well since then k_{ψ}(z) = p^{s} by our assumptions.

For r > 2 put Γ = H(2, …, r − 1), u = ∏_{i=2}^{r−1} x_{i}(1). Then Γ ≅ A_{r−2}(K) and u is a regular unipotent element of Γ. Lemma 2.16 implies that k_{ψ}(z) ≥ k_{ρ}(u) for every ρ ∈ Irr Γ such that the restriction ψ|Γ has a composition factor ρ. Write μ = ∑_{i=1}^{r−1} b_{i}ω_{i}. If 0 < b_{r−1} < p, it follows from Lemma 2.19 that X_{−r}^{b}m ≠ 0 for all nonnegative integers b ≤ b_{r−1}. This implies that ψ|Γ has composition factors with highest weights

(∑_{i=1}^{r−3} b_{i}ω_{i}) + (b_{r−2} + 2b)ω_{r−2}

(for r = 3 the first sum is zero) for all integer b with 0 ≤ b ≤ b_{r−1}. The same Lemma 2.19 forces that ψ|Γ always has such factor with b = b_{r−1}. Equality (2) yields that d + ∑_{i=j+1}^{r} a_{i} ≠ 1 since d > 0 and a_{j} ≤ p − 1. Hence d + a_{r} ≥ 2 if j = r − 1. It is clear that a_{j−1} + a_{j} − d, a_{j+1} + d < 2p. Since we have excluded the situations where ψ is tensor decomposable, we can assume that a_{j−1} + a_{j} − d, a_{j+1} + d ≤ p. If j = r − 1, r > 2 and a_{r} +d = p, it suffices to consider the cases where a_{i} = 0 for i ≤ r −3 and a_{r−2} +a_{r−1} −d ∈ {0, p}. Observe that the case p = 3, j = r − 1, and a_{r−2} + a_{r−1} − d = a_{r} + d = p is impossible since it contradicts (2). Now, using the arguments above on composition factors of ψ|Γ, one can deduce from Lemmas 2.14 and 2.15 that either k_{ρ}(u) = p^{s} for some of these factors ρ and hence k_{ψ}(z) = p^{s}, or ψ is p-large, or one of the following holds:

1) j = r − 1 > 1, d = a_{r−1}, a_{r−1} + a_{r} = p and a_{i} = 0 for i ≤ r − 2;

2) j = r − 2 > 2, d = a_{r−2}, a_{r−1} + a_{r−2} ∈ {1, 2, p} and a_{i} = 0 for i ≠ 1, r − 2, r − 1;

3) j = r − 3, r > 5, d = a_{r−3}, a_{r−3} + a_{r−2} ∈ {1, p} and a_{i} = 0 for i ≠ r − 3, r − 2;

4) j = 2, r > 2 and ω = ω_{2} + (p − 1)ω_{3};

5) j = 1, r > 2, r ≠ 5, a_{2} + d ∈ {1, 2, p}, a_{i} = 0 for 3 ≤ i ≤ r − 2, a_{r−1} ≤ 1 if r > 3, a_{r} = 0 for p > 3 and a_{r} ≤ 1 for p = 3; moreover, a_{i} = 0 for i > 2 if a_{2} + d > 1;

6) r = 5;

7) r = 2.

Let Case 1) hold. By Formula (2), we have ω = ω_{r−1} + (p − 1)ω_{r}, m = v(1, r − 1, 1). Lemmas 2.20 and 2.21 yield that m is fixed by X_{i} for i > 1 and m_{1} = X_{−r}m ≠ 0. Then the commutator relations in L yield that the vector m_{1} is fixed by U^{+}(H). Now it is clear that m_{1} generates an indecomposable H-module with highest weight μ_{1} = 2ω_{r−2} +(p−2)ω_{r−1}. As μ_{1} is p-large, the assumptions of the proposition yield that k_{N}(z) = p^{s} as required.

Now let Case 2) hold. If a_{r−1}+a_{r−2} = 1 or 2, then a_{r−1} ≤ 1. Since a_{r} = 0 and d = a_{r−2}, Formula (2) yields a contradiction. Hence a_{r−1} + a_{r−2} = p. We can assume that a_{1} = 0 since otherwise ψ is p-large. Take m_{1} = X_{−(r−1)}m. It follows from Lemmas 2.20 and 2.21 that m_{1} ≠ 0 and is fixed by the groups X_{i} with i > 1 and i ≠ r − 1. The commutator relations in L yield that X_{r−1}m_{1} = pm = 0. So m_{1} is fixed by X_{r−1} as well. Hence m_{1} generates an indecomposable H-module with highest weight ε = ω_{r−3}+(p−2)ω_{r−2}+ω_{r−1}. As ε is p-large, our assumptions yield that k_{N}(z) = p^{s}.

Next, let Case 3) hold. Using Formula (2) as above, we come to a contradiction if a_{r−2} + a_{r−3} = 1. Hence a_{r−2} + a_{r−3} = p. Applying Lemmas 2.20 and 2.21 and arguing as in Cases 1) and 2), we deduce that m_{1} = X_{−(r−2)}m ≠ 0, X_{r−2}m_{1} = pm = 0, the vector m_{1} is fixed by U^{+}(H) and generates an indecomposable H-module with highest weight δ = ω_{r−4} + (p − 2)ω_{r−3} + ω_{r−2}. As δ is p-large, our assumptions force k_{N}(z) = p^{s}.

Now consider Case 4). Here m = X_{−1}X_{−2}v. Put m_{1} = X_{−3}m. Using Lemmas 2.20 and 2.21 as in Cases 1)-3), we conclude that m_{1} generates an indecomposable H-module with p-large highest weight ω_{H}(m_{1}) = ω_{1} + (p − 2)ω_{2} + ω_{3}. As before, this implies that k_{N}(z) = p^{s}.

Next consider Case 5). First assume that a_{2} +d = 1. Then a_{2} = 0, d = 1. Formula( 2) yields that a_{1} − 1 + a_{r−1} + a_{r} = p − 1. As a_{1} ≤ p − 1, this forces a_{r−1} + a_{r} > 0, hence r > 3 and a_{r−1} = 1 for p > 3. One easily deduces that we have the following possibilities: r > 3 and ω = (p − 1)ω_{1} + ω_{r−1}; p = 3 and ω = ω_{1} + ω_{r−1} + ω_{r}; and p = 3 and ω = 2ω_{1} + ω_{r}.

Let r > 3 and ω = (p − 1)ω_{1} + ω_{r−1}. Then μ = ω_{1} + ω_{r−2}. If s = 1, Lemma 2.15 forces k_{ψ}(z) = p^{s} as required. Let s > 1. By our assumptions, r ≥ 13. Put m_{1} = X_{−r}X_{−(r−1)}m. Applying 2.19 several times, we get that m_{1} ≠ 0. Since ω(m) − α_{r} ∉ X(M), one easily concludes that X_{r−1}m_{1} = 0 and the group X_{r−1} fixes m_{1}. This implies that U^{+}(Γ) fixes m_{1} and m_{1} generates an indecomposable Γ-module with highest weight λ = ω_{Γ}(m_{1}) = ω_{1} + ω_{r−3} + ω_{r−2}. Put ρ = φ(λ). Lemma 2.14 yields that k_{ρ}(u) = p^{s}. Hence k_{N}(z) = p^{s} as required.

If p = 3 and ω = ω_{1} + ω_{r−1} + ω_{r}, the representation ψ is p-large.

Let p = 3 and ω = 2ω_{1} + ω_{r}. Then r ≥ 14 since s > 2. We have μ = ω_{1} + ω_{r−1}, m = X_{−1}v. Put c = X_{−r}m. Then c ≠ 0 by Lemma 2.19. One easily observes that c generates an indecomposable Γ-module N_{1} with highest weight ν = ω_{1} + 2ω_{r−2}. As ν is p-large, this forces k_{N}(z) = k_{N_{1}}(u) = p^{s}. Thus the subcase a_{2}+d = 1 has been considered.

Now let a_{2} + d = 2. Then ω = a_{1}ω_{1} + a_{2}ω_{2} and a_{2} > 0 since φ is p-large. Hence a_{2} = d = 1 and a_{1} = p − 1. It is well known (see, for instance, [11, (8.1)]) that ψ is equivalent to the restriction to H of the irreducible representation θ = φ(2ω_{1}) of the group H_{1} = A_{p^{s}−2}(K). Observe that z is a regular element of H_{1}. By Lemma 2.7, for 0 < l < s an element z^{p^{l}} has p^{l} −1 Jordan blocks of size p^{s−l} and one block of size p^{s−l} −1. Results 2.2-2.4 force that k_{ψ}(z) = k_{θ}(z) = p^{s} as required. Observe that the arguments of this paragraph hold for r = 5 as well.

Next, let a_{2} + d = p. Then ω = a_{1}ω_{1} + a_{2}ω_{2}, m = X_{−1,d}v. Set m_{1} = X_{−2}m. Applying Lemmas 2.19 and 2.21 and arguing as in Cases 1)-4), we conclude that m_{1} ≠ 0, X_{2}m_{1} = 0, the vector m_{1} is fixed by U^{+}(H) and generates an indecomposable H-module with highest weight (p − 2)ω_{1} + ω_{2}. Hence N and N|Γ have composition factors σ and τ, respectively, with highest weight (p − 2)ω_{1} + ω_{2}. If s = 1, apply 2.15 and get k_{σ}(z) = p^{s}. For s > 1 use 2.14 to obtain k_{τ}(u) = p^{s}. Thus in both situations k_{N}(z) = p^{s}, and we are done. This completes the analysis of Case 5).

Let Case 6) hold. Then r = 5, p = 3. Considering the subcases j = 1, 2, 3, and 4 separately and applying Formula (2), one can check that either ψ is tensor decomposable or p-large, or

ω ∈ {ω_{4} + 2ω_{5}, ω_{3} + 2ω_{4}, ω_{2} + 2ω_{3}, 2ω_{1} + ω_{2}, 2ω_{1} + ω_{3}, 2ω_{1} + ω_{4}, 2ω_{1} + ω_{5}, ω_{1} + 2ω_{2}}.

The situations where ω = ω_{4} + 2ω_{5}, ω_{3} + 2ω_{4}, ω_{2} + 2ω_{3} and 2ω_{1} + ω_{2} were handled during the analysis of Cases 1)-3) and 5), respectively.

Let ω = 2ω_{1} + ω_{a} with 3 ≤ a ≤ 5. Then μ = ω_{1} + ω_{a−1}. Take m_{1} = X_{−5}X_{−4} … X_{−a}m. It is clear that the weights ω(m) − α_{4} − α_{5} and ω(m) − α_{3} − α_{5} ∉ X(φ) for a = 3 and the weight ω(m) − α_{5} ∉ X(φ) for a = 4. Applying Lemma 2.19 several times, one can show that in all cases m_{1} ≠ 0. Now it is evident that m_{1} is fixed by U^{+}(Γ) and generates an indecomposable Γ-module with highest weight σ = ω_{Γ}(m_{1}). We have σ = 2ω_{1} + ω_{3}, ω_{1} + ω_{2} + ω_{3} and ω_{1} + 2ω_{3} for a = 3, 4, and 5, respectively. Let χ = φ(σ). Since in all cases χ is p-large, we get k_{ψ}(z) = k_{χ}(u) = p^{s} as required.

Now let ω = ω_{1} + 2ω_{2}. Then m = X_{−1}v. Aplying Lemmas 2.19 and 2.21 and arguing as in Cases 1), 2) and 3)-5), we deduce that m_{1} = X_{−2}m ≠ 0, X_{2}m_{1} = 0, the vector m_{1} is fixed by U^{+}(H) and generates an indecomposable H-module with highest weight ξ = ω_{1} + ω_{2}. Let π = φ(ξ) ∈ Irr H. It has been actually proven in the previous paragraph that k_{π}(z) = p^{s}. Hence k_{N}(z) = p^{s}, and we are done. This completes the analysis of Case 6).

Finally, consider Case 7). Then p = 3 and ω ∈ {ω_{1} + 2ω_{2}, 2ω_{1} + ω_{2}, 2ω_{1} + 2ω_{2}}. We have m = X_{−1}v if ω = ω_{1} + 2ω_{2} or 2ω_{1} + ω_{2} and m = X_{−1,2}v if ω = 2ω_{1} + 2ω_{2}; H ≅ SL_{2}(K), μ = 3ω_{1} for ω = ω_{1} + 2ω_{2}, 2ω_{1} for ω = 2ω_{1} + ω_{2} and 4ω_{1} for ω = 2ω_{1} + 2ω_{2}. Theorem 2.1 and Proposition 2.3 yield that k_{ψ}(z) = 3 = p^{s} for ω = 2ω_{1} + ω_{2} or 2ω_{1} + 2ω_{2}. Let ω = ω_{1} + 2ω_{2}. It follows from 2.21 that m_{1} = X_{−2}m ≠ 0. Now [19, Lemmas 2.12 and 2.11] imply that m generates an H-module isomorphic to the Weyl module V (3ω_{1}) and k_{N}(z) = 3 as desired. So all the possibilities have been considered. □

Proposition 3.3 Let s > 0, G = C_{r}(K), n > p^{s} + 1, and let φ ∈ Irr be p-large. Assume that (C_{r−1}, s + 1)-conjecture holds. Then k_{φ}(x) = |x| for each element x ∈ G of order p^{s+1}.

Proof. Let x be as in the assertion of the proposition. Lemma 2.8 and the description of unipotent conjugacy classes in [15, ch. IV, Exercise 2.15] imply that \overline{cl(x)} contains an element y of order p^{s+1} lying in the group H = H(2, …, r). By Lemma 2.9, k_{φ}(x) = p^{s+1} if k_{φ}(y) = p^{s+1}. Now our assumptions yield that it suffices to show that φ|H has a p-large composition factor.

We have n > 4. First assume that ∑_{i=3}^{r} a_{i} ≠ 0. By Smith’s theorem [13], φ|H has a composition factor ψ = φ(∑_{i=1}^{r−1} a_{i+1}ω_{i}). If ∑_{i=2}^{r} a_{i} ≥ p, the representation ψ is p-large, and we are done.

Let ∑_{i=2}^{r} a_{i} < p. Since φ is p-large, we have b+∑_{i=2}^{r} a_{i} = p for some b ∈ Z^{+}, 0 < b ≤ a_{1}. As ∑_{i=3}^{r} a_{i} ≠ 0, the sum a_{2}+b < p. Set v_{1} = X_{−1,b}v, λ = (a_{2}+b)ω_{1}+∑_{i=2}^{r−1} a_{i+1}ω_{i} ∈ X(H), ρ = φ(λ). By 2.19, v_{1} ≠ 0. It is clear that v_{1} generates an indecomposable H-module with highest weight λ which has a p-large composition factor ρ.

Finally, let ω = a_{1}ω_{1} + a_{2}ω_{2}. Put v_{2} = v(1, 2, a_{2}). Lemma 2.20 yields that v_{2} ≠ 0 and the groups X_{i} fix v_{2} for i > 1. So one easily observes that v_{2} generates an indecomposable H-module with highest weight τ = a_{1}ω_{1} + a_{2}ω_{2}. Hence φ|H has a p-large composition factor π = φ(τ).

Now the required factor is constructed in all cases which completes the proof. □

Using induction on r and s, we see that Theorem 1.1 for G = C_{r}(K) follows from Results 2.5, 2.18, 3.2, 3.3 and the general scheme in 3.1.

Now let G = B_{r}(K) or D_{r}(K). Recall that n = 2r + 1 and r > 2 for G = B_{r}(K), n = 2r and r > 3 for G = D_{r}(K).

Notation block B and some comments. For G = B_{r}(K) set H = H(1, 3, …, r), H_{1} = H(1), H_{2} = H(3, …, r). In this case we shall write a weight λ ∈ X(H) in the form (a, λ_{2}) where a = ⟨λ, α_{1}⟩ and λ_{2} is the restriction of λ to H_{2}. By [17, Corollary a) of Lemma 68], a representation ψ ∈ Irr H with highest weight μ = (a, μ_{2}) is equivalent to the tensor product of the representations φ(aω_{1}) ∈ Irr H_{1} and φ(μ_{2}) ∈ Irr H_{2}. Put ψ_{1} = φ(aω_{1}) ∈ Irr H_{1}, ψ_{2} = φ(μ_{2}) ∈ Irr H_{2}. For a weight vector m ∈ M set ω_{β}(m) = ⟨ω(m), β⟩. Recall that β = α_{1} + 2 ∑_{i=2}^{r} α_{i} and l(φ) = a_{1} + 2(∑_{i=2}^{r−1} a_{i}) + a_{r}. We also need some special comments for the case where n = p^{s} + 2. In this situation put x = ∏_{i=1}^{r} x_{i}(1), y = x^{p^{s}}, z_{1} = x_{1}(1), z_{2} = ∏_{i=3}^{r} x_{i}(1), z = z_{1}z_{2}. One easily observes that y ∈ ⟨X_{β}⟩ and H ⊂ C_{G}(y). Lemma 2.16 implies that if ψ and ψ_{2} are such as above, then k_{ψ}(z) ≥ k_{ψ_{2}}(z_{2}). This notation will be used until the end of the section.

Proposition 3.4 Let G = B_{r}(K), n = p^{s} + 2 and ω = (p − 1)ω_{1} + a_{2}ω_{2} with a_{2} ≠ 0. Then k_{φ}(x) = p^{s+1}.

Proof. For a_{2} > 1 put t_{1} = v(r, 2, a_{2} − 1), t = X_{−2}^{a_{2}}X_{−3}^{a_{2}−1} … X_{−(r−1)}^{a_{2}−1}t_{1}. If a_{2} = 1, set t = X_{−2}v. Lemmas 2.19 and 2.20 and [19, Lemma 4.2] yield that t ≠ 0, the groups X_{i} with i > 0 and i ≠ 2 fix t, ω_{β}(t) = p, ω_{H}(t) = (p − 2 + 2a_{2}, ω_{1}) for r > 3 and ω_{H}(t) = (p − 2 + 2a_{2}, 2ω_{1}) if r = 3. We claim that X_{−β}t ≠ 0.

Until the end of the proof for vectors x, y ∈ M or endomorphisms x, y of M we write x = sc · y if x = cy for some c ∈ K^{∗}. We need this notation to avoid introducing many different constants whose exact values are not needed.

Let T ⊂ G be the maximal torus with respect to which the root system is considered. For 1 ≤ i ≤ r let s_{i} be an involution in the group ⟨X_{i}, X_{−i}⟩ which normalizes T and maps onto the reflection associated with the root α_{i} under the canonical homomorphism N_{G}(T) → W. Write w = s_{2} … s_{r−1}s_{r}s_{r−1} … s_{2}, and let w′ ∈ W be the image of w under the homomorphism mentioned above. One can directly check that w′α_{1} = β. Using [17, Lemma 19], one gets X_{−β} = sc·wX_{−1}w^{−1} and s_{2}X_{−1} = sc·X_{−(α_{1}+α_{2})}s_{2} (as endomorphisms of M). We shall need the following assumption.

(*) Let m ∈ M\{0}, α ∈ R, X_{α} fixes m and ⟨ω(m), α⟩ = a > 0. Then s_{α}m = sc · X_{−α,a}m.

Indeed, by the universal property of the Weyl module [8, Part II, Lemma 2.13b)], the ⟨X_{α}, X_{−α}⟩-module generated by m is a quotient of the Weyl module V (aω_{1}). One may assume that m is the image of a highest weight vector m′ ∈ V (aω_{1}) under the canonical homomorphism onto this module. Using the construction of the Chevalley groups in [17, § 3] and [17, Theorem 39], we get the required equality for m′ and then transfer it to m.

Observe that our vector t is obtained from t′ = X_{−2,a_{2}−1}v in the result of a series of the following steps: one has a vector l such that X_{i} fixes l for some i and ⟨ω(l), α_{i}⟩ = a and then takes l_{1} = X_{−i,a}l. Using this observation and assumptions (*), we deduce that w^{−1}t = sc · s_{2}^{−1}X_{−2,a_{2}−1}v. By [17, Lemma 19], ω(s_{2}^{−1}X_{−2,a_{2}−1}v) = ω − α_{2}. Since dim M_{ω−α_{2}} = 1, we get s_{2}^{−1}X_{−2,a_{2}−1}v = sc · X_{−2}v. Now it is clear that X_{−β}t = sc · wX_{−1}X_{−2}v ≠ 0 by Lemma 2.21.

Set X_{−β}t = m. As ⟨β, α_{i}⟩ = 0 for i ≠ 2, the weight ω_{H}(m) = ω_{H}(t) and X_{−β} commutes with H. Hence the groups X_{i}, i ≠ 2, fix m. It is clear that X_{β} fixes t. Now Lemmas 2.12 and 2.11 of [19] imply that the vector t generates the Weyl module V (pω_{1}) with respect to the group ⟨X_{β}, X_{−β}⟩ and m ∈ M_{y}. Put M_{y} = M_{1} and N = KHm. Then N ⊂ M_{1} as H ⊂ C_{G}(y). Set μ = ω_{H}(t) and ψ = φ(μ) ∈ Irr H. It is clear that k_{ψ_{2}}(z_{2}) = p^{s} − 2. If k_{ψ_{1}}(z_{1}) > 2, Lemma 2.17 forces k_{ψ}(z) = p^{s}. Since p−2+2a_{2} < 3p ≤ p^{2}, Lemma 2.24 yields that k_{ψ_{1}}(z_{1}) > 2 if a_{2} > 1. Let a_{2} = 1. Then t = X_{−2}v. Put q = X_{−1}m = X_{−1}X_{−β}t. We claim that q ≠ 0. Indeed, put v_{1} = s_{2}X_{1}X_{2}v. The arguments above show that v_{1} = sc · X_{−(α_{1}+α_{2})}v (observe that s_{2}v = sc · X_{2}v as a_{2} = 1). Then

q = sc · X_{−1}wX_{−1}X_{−2}v = sc · X_{−1}X_{−2}^{2}X_{−3} … X_{−(r−1)}X_{−r}^{2}X_{−(r−1)} … X_{−3}v_{1}.

Using commutator relations in L, we first get

X_{2}q = sc · X_{−1}X_{−2}X_{−3} … X_{−(r−1)}X_{−r}^{2}X_{−(r−1)} … X_{−3}v_{1},

then conclude that

X_{3} … X_{r−1}X_{r}^{2}X_{r−1} … X_{3}X_{2}q = sc · X_{−1}X_{−2}v_{1}

for r > 3 and X_{3}^{2}X_{2}q = X_{−1}X_{−2}v_{1} for r = 3. Hence it suffices to show that X_{−1}X_{−2}v_{1} ≠ 0. Put v_{2} = X_{−2}v_{1}. We claim that v_{2} ≠ 0. Indeed, using commutator relations in L and taking into account that ⟨ω(v_{1}), α_{2}⟩ = 0, we get that X_{2}v_{2} = sc · X_{−2}X_{−1}v ≠ 0 by Lemma 2.19. Therefore v_{2} ≠ 0. Set S = H(1, 2). By Smith’s theorem [13], v_{2} generates an irreducible S-module M_{S} with highest weight (p−1)ω_{1}+ω_{2}. It is clear that v_{2} ∈ M_{S} and is not a lowest weight vector in M_{S}. As M_{S} is p-restricted, [1, Lemma 6.2 and Theorem 6.4] yield that X_{−α}v_{2} ≠ 0 for some α ∈ R^{+}(S) ([1, Lemma 6.2] considers R^{+}(S) and highest weight vectors, but one can replace R^{+}(S) by R^{−}(S) and the highest weight by the lowest one). Now it follows from the commutator relations in L(S) that X_{−i}v_{2} ≠ 0 for i = 1 or 2. One easily observes that ω(X_{−2}v_{2}) ∉ X(φ). Hence X_{−2}v_{2} = 0 and X_{−1}v_{2} ≠ 0. Thus we have deduced the desired inequality X_{−1}X_{−2}v_{1} ≠ 0 which forces q ≠ 0. By [19, Lemma 2.12], KH_{1}m ≅ V (pω_{1}) (as an H_{1}-module). Now Proposition 2.23 yields that N has a quotient module N_{1} isomorphic to V (pω_{1}) ⊗ M(λ) where λ ∈ X(H_{1}), λ = ω_{1} if r > 3 and λ = 2ω_{1} for r = 3. It follows from [19, Lemma 2.11] that z_{1} has the minimal polynomial of degree p in the module V (pω_{1}). Then Lemma 2.17 forces k_{N_{1}}(z) = p^{s} as z = z_{1}z_{2}. Hence in all cases k_{N}(z) = p^{s}. Thus k_{M_{1}}(z) = p^{s}. Using Lemma 2.16 and arguing as in 3.1, we deduce that k_{M_{1}}(x) = p^{s}. By Corollary 2.5, k_{φ}(y) = p. Now Proposition 2.10 yields that k_{φ}(x) = p^{s+1}. □

Lemma 3.5 Theorem 1.1 holds for G = B_{3}(K).

Proof. Corollary 2.5 yields that it suffices to consider the cases p = 3 and p = 5 only since all unipotent elements of G have order p for p > 5.

Let p = 3. Then the order of a unipotent element in G is at most 9. It follows from 2.8 that \overline{cl(u)} contains an element w with J(w) = diag(J_{5}, E_{2}) for all elements u of order 9 in G. Put G_{1} = H(2, 3), G_{2} = H(α_{2}, α_{1}, α_{2} + 2α_{3}). Then G_{1} ≅ B_{2}(K) ≅ C_{2}(K), G_{2} ≅ D_{3}(K) ≅ A_{3}(K). Using [15, ch. IV, Exercise 2.15], one can take for w elements of both groups G_{1} and G_{2}. By 2.9, since Theorem 1.1 holds for groups of type A_{l} and C_{l}, it suffices to show that φ|G_{1} or φ|G_{2} has a p-large composition factor.

As φ is p-large, we have a_{1} + 2a_{2} + a_{3} ≥ 3. Then a_{2} + a_{3} > 0 since a_{1} ≤ 2. By Smith’s theorem [13], φ|G_{1} has a composition factor with highest weight a_{2}ω_{1} + a_{3}ω_{2}. So we are done if a_{2} + a_{3} ≥ 3.

Let a_{2} + a_{3} ≤ 2. If a_{1}, a_{2} ≠ 0 and a_{3} = 0, set v_{1} = v(1, 2, a_{2}). By Lemma 2.20, v_{1} ≠ 0 and X_{i} fixes v_{1} for i > 1. One can easily check that v_{1} generates an indecomposable G_{1}-module with highest weight λ ∈ {ω_{1} + 4ω_{2}, 2ω_{1} + 4ω_{2}, ω_{1} + 2ω_{2}, 2ω_{1} + 2ω_{2}}. In all cases φ|G_{1} contains a p-large composition factor φ(λ).

If a_{2} = 0, put v_{2} = X_{−1,a_{1}}v. By 2.19, v_{2} ≠ 0. It is clear that v_{2} generates a G_{1}-module which contains a composition factor τ with highest weight a_{1}ω_{1} +a_{3}ω_{2}. Since φ is p-large, one has a_{1} + a_{3} ≥ 3 and τ is p-large.

If ω = a_{1}ω_{1} + ω_{2} + ω_{3}, one can easily deduce that the vector v generates a G_{2}-module with a p-large composition factor.

So in all the cases above a desired p-large factor is constructed.

It remains to consider the case ω = 2ω_{2}. In this situation φ|G_{1} has a composition factor with highest weight 2ω_{1} and φ(ω_{1}) is the standard 5-dimensional representation of B_{2}(K). Applying Proposition 2.22 with λ_{1} = λ_{2} = ω_{1}, k_{1} = k_{2} = 4 to the group G_{1} and taking w ∈ U^{−}(G_{1}), we get k_{φ}(w) = 9 and complete the proof for p = 3.

Now let p = 5. Here we use the general scheme from 3.1, but this particular case has some specific features since r is small, so we handle it separately. Recall that β = α_{1} + 2α_{2} + 2α_{3}. Since n = p + 2, only regular unipotent elements have order > p. Below we use the notation and comments from block B. For a weight vector l ∈ M we shall write ω_{H}(l) in the form (c_{1}, c_{2}) where c_{1} = ⟨ω(l), α_{1}⟩, c_{2} = ⟨ω(l), α_{3}⟩. By Proposition 3.4, k_{φ}(x) = 25 for ω = 4ω_{1} + a_{2}ω_{2}, so we exclude this case. We shall find a vector m ∈ M such that ω_{β}(m) = 4 and the groups X_{β}, X_{1} and X_{3} stabilize m. Then the arguments in 3.1 imply that m ∈ M_{y}, the H-module N = KHm ⊂ M_{y} and it suffices to show that k_{N}(z) = 5. Until the end of the proof set μ = ω_{H}(m), ψ = φ(μ) ∈ Irr H. The construction of m can be naturally divided into 2 subcases.

I. Let a_{1} + a_{2} > 4. Then a_{1} + a_{2} − a = 4 for some positive integer a ≤ a_{2}. Put m = X_{−2,a_{2}+a_{3}}X_{−3,a_{3}+2a}X_{−2,a}v. It follows from 2.19 and [19, Lemma 4.2] that m ≠ 0 and the groups X_{β}, X_{1} and X_{3} fix m. One easily observes that ω_{β}(m) = 4 and ω_{H}(m) = (a_{1}+a_{2}+a_{3}+a, a_{3}+2a_{2}−2a). Set b_{1} = a_{1}+a_{2}+a_{3}+a, b_{2} = a_{3}+2a_{2}−2a. By Lemma 2.24, for a p-restricted representation ρ ∈ Irr(SL_{2}(K)) with highest weight lω_{1} and a nontrivial unipotent element t we have k_{ρ}(t) = l + 1. Write b_{1} = 5c_{1} + c_{2}, b_{2} = 5d_{1} + d_{2} where 0 ≤ c_{1}, c_{2}, d_{1}, d_{2} < 5 (this is possible since b_{1}, b_{2} < 4p < p^{2}). Now standard facts on representations of semisimple groups [17, Corollary a) of Lemma 68] and Lemmas 2.24 and 2.17 imply that k_{ψ}(z) = min{5, c_{1} + c_{2} + d_{1} + d_{2} + 1}. Set c = c_{1} + c_{2} + d_{1} + d_{2}. We are done if c ≥ 4. Observe that b_{1} ≥ 6. So c_{1} + c_{2} ≥ 2 and c ≥ 4 if d_{1} + d_{2} ≥ 2. One can easily deduce that d_{1} + d_{2} ≠ 0 since otherwise we would get the excluded case ω = 4ω_{1} + a_{2}ω_{2}.

Let d_{1} + d_{2} = 1. Then either a = a_{2} and a_{3} = 1, or b_{2} = 5. In the first case we get b_{1} = 5 + 2a_{2} ≠ 10 and b_{1} ≥ p + 2. So c_{1} + c_{2} ≥ 3 and k_{ψ}(z) = 5. In the second one a_{3} is odd and b_{1} = 4 + a_{3} + 2a. Hence b_{1} ≠ 10 and b_{1} ≥ 7. As above, k_{ψ}(z) = 5 in this case also.

II. Now let a_{1} + a_{2} ≤ 4. First assume that a_{1} + a_{2} + a_{3} > 4. Then a_{1} + a_{2} + a_{3} − a = 4 for some a ∈ Z, 0 < a ≤ a_{3}. Put m = v(2, 3, a). It is obvious that X_{β} and X_{1} fix m. By Lemma 2.20, m ≠ 0 and the group X_{3} also fixes m. One easily computes that ω_{H}(m) = (a_{1} + a_{2} + a, 2a_{2} + a_{3}) and ω_{β}(m) = 4. Put b_{1} = a_{1} + a_{2} + a, b_{2} = 2a_{2} + a_{3}. It is clear that 0 < b_{1}, b_{2} < 25. Moreover, b_{1} > 1. Arguing as in Case I, one can show that k_{ψ}(z) = 5 if both b_{1}, b_{2} ∉ {1, 5}.

Let b_{1} = 5. By Lemma 2.21, X_{−1}m ≠ 0. Now [19, Lemmas 2.11 and 2.12] show that the H_{1}-module generated by m is isomorphic to the Weyl module V (pω_{1}) and k_{N}(z_{1}) = 5. Hence, by 2.16, k_{N}(z) = 5 as required.

If b_{2} = 1, we have a_{2} = 0, a = a_{3} = 1, a_{1} = 4 and b_{1} = 5, so we are in the case just handled.

Next, let b_{2} = 5. Then a_{3} is odd. Hence b_{1} is odd as a_{1} + a_{2} + a_{3} − a is even. Thus b_{1} ≥ 3. The case b_{1} = 5 has been settled, so let b_{1} ≠ 5. Then b_{1} = 5c_{1} + c_{2} where c_{1}, c_{2} ∈ Z^{+}, c_{1}, c_{2} < 5 and c_{1} + c_{2} ≥ 3. Arguing as in Case I, we get k_{N}(z) = 5.

Now all the possibilities with a_{1} + a_{2} + a_{3} > 4 have been handled.

Finally, let a_{1} + a_{2} + a_{3} ≤ 4. Then one can write 4 = a_{1} + a_{2} + a_{3} + a_{2} − a with a ∈ Z^{+}, 0 < a ≤ a_{2}. Put m = X_{−2}^{a}v. It is clear that the groups X_{β}, X_{1} and X_{3} fix m. We have ω_{H}(m) = (a_{1} + a, a_{3} + 2a), ω_{β}(m) = 4. Set b_{1} = a_{1} + a, b_{2} = a_{3} + 2a. Obviously, b_{1} ≥ 1, b_{2} ≥ 2. Arguing as above, one can show that k_{N}(z_{1}) = 5 if b_{1} = 5.

Let b_{2} = 5. Then the commutator relations in L and Lemma 2.19 show that X_{2}X_{−3}m = cX_{−3}X_{−2}^{a−1}v ≠ 0 (c ∈ K^{∗}). Hence X_{−3}m ≠ 0. Applying [19, Lemmas 2.11 and 2.12], as in the case where b_{1} = 5, we get that k_{N}(z_{2}) = 5. By Lemma 2.16, k_{N}(z) = 5 in this case.

Now let both b_{1}, b_{2} ≠ 5. Arguing as before in this proof, one easily deduces that k_{ψ}(z) = 5 if b_{1} ≠ 1. If b_{1} = 1, one can directly verify that a_{1} = 0, a = 1 and b_{2} = 5 which yields a contradiction. This completes the proof. □

Proposition 3.6 Let G = B_{r}(K), n = p^{s} + 2 > 7, and let φ ∈ I_{p} be p-large. Assume that (B_{r−2}, s)-conjecture holds. Then k_{φ}(g) = |g| for each g ∈ G of order p^{s+1}.

Proof. Apply the general scheme of 3.1. Using [15, ch. IV, Exercise 2.15], one easily concludes that it suffices to consider regular unipotent elements since only they have order p^{s+1}. Below the notation from block B is used. Our goal is to prove that k_{φ}(x) = p^{s+1}. We shall construct a nonzero vector m ∈ M such that the groups X_{i} with i > 0 and i ≠ 2 and X_{β} fix m and ω_{β}(m) = p − 1. Set N = KHm, N_{2} = KH_{2}m. According to the arguments in 3.1, it suffices to show that k_{N}(z) = p^{s}. It is clear that N is an indecomposable H-module with highest weight μ = ω_{H}(m). Set ψ = φ(μ) ∈ Irr H. For the weight μ and the representation ψ define a, μ_{2}, ψ_{1} and ψ_{2} as in block B. Obviously, k_{N}(z) ≥ k_{ψ}(z). Recall that k_{ψ}(z) ≥ k_{ψ_{2}}(z_{2}). As n > 7, we have r > 3. If ψ_{2} is p-large, then k_{ψ_{2}}(z_{2}) = p^{s} by our assumptions, so we can exclude these cases.

Put Γ = H(3, …, r − 1), u = ∏_{i=3}^{r−1} x_{i}(1). The group Γ ≅ A_{r−3}(K), and u is a regular unipotent element in Γ. Lemma 2.16 implies that k_{ψ_{2}}(z_{2}) = p^{s} if k_{ρ}(u) = p^{s} for some composition factor ρ of the restriction ψ_{2}|Γ. If μ_{2} = ∑_{i=1}^{r−2} b_{i}ω_{i}, Smith’s theorem [13] implies that ψ_{2}|Γ has a composition factor ρ_{0} = φ(∑_{i=1}^{r−3} b_{i}ω_{i}). Set τ_{0} = ∑_{i=1}^{r−3} b_{i}ω_{i} ∈ X(Γ). One easily observes that the vectors of the form X_{−r,c}m are either zero, or generate indecomposable Γ-modules with highest weight τ_{c} = τ_{0} + cω_{r−3}. Put ρ_{c} = φ(τ_{c}). By Lemma 2.19, X_{−r,c}m ≠ 0 if b_{r−2} < p and 0 ≤ c ≤ b_{r−2}. Since X_{r}m = 0, the commutator relations in L yield that X_{−r}m ≠ 0 if b_{r−2} ≢ 0(mod p). Hence in all these situations the relevant representations ρ_{c} are composition factors of ψ_{2}|Γ. Throughout this proof we refer to the arguments of the present paragraph on the existence of such factors as to the ρ_{c}-arguments.

Now we exclude the cases where ψ_{2} is tensor decomposable. Our immediate goal is to prove the following claim: (*) If ψ_{2} is tensor decomposable, then k_{ψ_{2}}(z_{2}) = p^{s}.

Indeed, first suppose that s > 1. Set h = z_{2}^{p^{s−1}}. Using Lemma 2.7, we get that h has p^{s−1} − 2 Jordan blocks of size p and two blocks of size p − 1. Then results 2.1, 2.3 and 2.4 yield that k_{θ}(h) = p for each nontrivial θ ∈ Irr H_{2}. This forces k_{θ}(z_{2}) > (p − 1)p^{s−1}. If s = 1, the element z_{2} has the single Jordan block of size p − 2 and it follows from 2.1, 2.3 and 2.4 that k_{θ}(z_{2}) ≥ p − 2 for p > 7 and ≥ 4 for p = 7 if θ is such as above. Observe that p > 5 and 2p − 4 > p if s = 1 since n > 7. Now Lemma 2.17 yields that in all cases k_{ψ_{2}}(z_{2}) = p^{s} if ψ_{2} is tensor decomposable. So (*) holds.

Thus we can and shall exclude tensor decomposable cases in further analysis. Obviously, ψ_{2} is tensor decomposable if ρ_{0} is tensor decomposable or ρ_{0} is not p-restricted and a_{r} ≠ 0.

Our construction of the vector m can be naturally divided into 3 subcases.

I. Let ∑_{i=1}^{r−1} a_{i} > p − 1. Then there exist j, b ∈ Z^{+}, 1 < j < r, 0 < b ≤ a_{j} such that a_{j} − b + ∑_{i=1}^{j−1} a_{i} = p − 1. Put w = v(r, j, b), c_{r−1} = ⟨ω(w), α_{r−1}⟩, c_{l} = ⟨ω(w), α_{l}⟩ + c_{l+1} for 2 ≤ l < r − 1, m = X_{−2,c_{2}} … X_{−(r−1),c_{r−1}}w. Applying [19, Lemma 4.2], one can conclude that the groups X_{β} and X_{i} with i > 0 and i ≠ 2 fix m, verify that ω_{β}(m) = p − 1 and find ω_{H}(m). We have

a = b + a_{r} + ∑_{i=1}^{j} a_{i} + 2 ∑_{i=j+1}^{r−1} a_{i}  (3)

(here the second sum is zero if j = r − 1);

μ_{2} = (∑_{i=1}^{j−3} a_{i+1}ω_{i}) + (a_{j−1} + b)ω_{j−2} + (a_{j} − b + a_{j+1})ω_{j−1} + ∑_{i=j}^{r−2} a_{i+2}ω_{i}  (4)

for 2 < j < r − 1 (here the first sum is zero if j = 3);

μ_{2} = (∑_{i=1}^{r−4} a_{i+1}ω_{i}) + (a_{r−2} + b)ω_{r−3} + (2a_{r−1} − 2b + a_{r})ω_{r−2}  (5)

for j = r − 1 (here the first sum is zero for r = 4); and

μ_{2} = (a_{2} − b + a_{3})ω_{1} + ∑_{i=2}^{r−2} a_{i+2}ω_{i}  (6)

for j = 2. Recall that r ≥ 4. It is clear that a_{j−1} + b < 2p and a_{j} − b + a_{j+1} < 2p. If μ_{2} = 0, Formulae (4)-(5) yield that j = 2 and ω = (p − 1)ω_{1} + a_{2}ω_{2}. This case has been handled in Proposition 3.4. So we assume that μ_{2} ≠ 0.

First assume that s > 1 and s > 2 for p = 3. Then p^{s} − 2 ≥ 23, r ≥ 13. It follows from Lemma 2.12 that k_{ρ_{0}}(u) = p^{s}, or ψ_{2} is tensor decomposable or p-large, or one of the following holds:

1) j = 3, b = a_{3}, a_{2} + a_{3} ∈ {1, 2, p}, a_{i} = 0 for 4 ≤ i ≤ r − 1 and a_{r} = 0 if a_{2} + a_{3} = p;

2) j = 3, b = a_{3}, a_{2} + a_{3} = 1, a_{r−1} = 1 and a_{i} = 0 for 4 ≤ i ≤ r − 2;

3) j = 4, b = a_{4}, a_{3} + a_{4} ∈ {1, p}, a_{2} = 0, a_{i} = 0 for 5 ≤ i ≤ r − 1 and a_{r} = 0 if a_{3} + a_{4} = p;

4) p = 3, r = 14, j = 5, b = a_{5}, a_{4} + a_{5} ∈ {1, 3}, a_{i} = 0 for i = 2, 3 and for 6 ≤ i ≤ 14;

5) p = 3, r = 14, j = 11, b = a_{11}, a_{10} + a_{11} ∈ {1, 3}, a_{i} = 0 for 2 ≤ i ≤ 10 and for 12 ≤ i ≤ 14;

6) j = r − 2, b = a_{r−2}, a_{r−3} + a_{r−2} ∈ {1, p}, a_{i} = 0 for 2 ≤ i ≤ r − 4, a_{r−1} = 0 and a_{r} = 0 if a_{r−3} + a_{r−2} = p;

7) j = r − 1 and τ_{0} ∈ {ω_{1} + ω_{r−3}, ω_{r−3}, pω_{r−3}, 2ω_{r−3}};

8) j = 2, and either τ_{0} ∈ {0, ω_{1}, 2ω_{1}, pω_{1}, ω_{1} + ω_{r−3}, ω_{2}, ω_{r−4}, ω_{r−3}, 2ω_{r−3}} and a_{r} = 0 if τ_{0} = pω_{1}, or r = 14, p = 3, τ_{0} ∈ {ω_{3}, ω_{9}} and a_{r} = 0.

We shall consider Cases 1)-8) in turn taking into account the ρ_{c}-arguments.

Let 1) hold. First assume that a_{2} + a_{3} = 1. This forces a_{2} = 0, a_{3} = b = 1 and a_{1} = p − 1. If a_{r} ≥ 2, the ρ_{c}-arguments yield that ψ_{2}|Γ has the composition factor ρ_{2}. By Lemma 2.12, k_{ρ_{2}}(u) = p^{s} which implies k_{ψ_{2}}(z_{2}) = p^{s}.

Let a_{r} = 1. One easily observes that the weight τ = μ_{2} − α_{r−1} − 2α_{r} ∈ X(H_{2}) lies in the W (H_{2})-orbit of μ_{2}. So τ ∈ X(ψ_{2}). Obviously, τ + α_{i} ∉ X(ψ_{2}) if 3 ≤ i < r. Hence the H_{2}-module M(μ_{2}) has a nonzero vector of H_{2}-weight τ fixed by X_{i} for 3 ≤ i < r. This forces that ψ_{2}|Γ has a composition factor π with highest weight ω_{1}+ω_{r−4}. By Lemma 2.12, k_{π}(u) = p^{s} as desired.

Now let a_{r} = 0. Then a = p + 1, and it follows from 2.24 that k_{ψ_{1}}(z_{1}) = 3. Since in our case ψ_{2} is the standard realization of H_{2}, we have k_{ψ_{2}}(z_{2}) = p^{s} − 2. Then Lemma 2.17 yields that k_{ψ}(z) = p^{s} as required. This completes the analysis of the subcase a_{2} +a_{3} = 1.

Next, suppose a_{2} + a_{3} = 2. If a_{r} ≠ 0, the ρ_{c}-arguments imply that ψ_{2}|Γ has a composition factor ρ_{1}. Lemma 2.12 yields that k_{ρ_{1}}(u) = p^{s}, and we are done.

Let a_{r} = 0. Then μ_{2} = 2ω_{1}, and k_{ψ_{2}}(z_{2}) = p^{s} by Lemma 2.25.

Finally, let a_{2}+a_{3} = p and a_{r} = 0. Then a_{1}+a_{2} = p−1, a_{2}, a_{3} ≠ 0, a = a_{1}+a_{2}+2a_{3} ≥ p + 1. One easily observes that a < 3p ≤ p^{2}. Arguing as in the case a_{2} + a_{3} = 1, a_{r} = 0 and using Lemma 2.24, we get that k_{ψ_{1}}(z_{1}) ≥ 3 and k_{ψ_{2}}(z_{2}) = p^{s} − 2. Then k_{N}(z) = p^{s} by Lemma 2.17. Now Case 1) has been settled.

Let Case 2) take place. If a_{r} ≠ 0, we apply the ρ_{c}-arguments and Lemma 2.12 and conclude that ψ_{2}|Γ has a composition factor ρ_{1} and k_{ρ_{1}}(u) = p^{s} as desired.

Let a_{r} = 0. Then μ_{2} = ω_{1} + ω_{r−2}. By Lemma 2.19, m_{1} = X_{−r}^{2}X_{−(r−1)}m ≠ 0. The commutator relations in L show that X_{r−1}m_{1} = 0. Now it is clear that m_{1} generates an indecomposable Γ-module with highest weight λ = ω_{1}+ω_{r−4}+ω_{r−3}. Put ρ = φ(λ) ∈ Irr Γ. By Lemma 2.12, k_{ρ}(u) = p^{s}, and Case 2) is done.

Next, let Case 3) hold. First assume that a_{r} = 0. Then k_{ψ_{2}}(z_{2}) = k_{χ}(z_{2}) where χ = φ(ω_{2}) ∈ Irr H_{2}. By Lemma 2.25, k_{χ}(z_{2}) = p^{s}, and we are done.

If a_{r} ≠ 0, we have a_{3} + a_{4} = 1. In this situation by the ρ_{c}-arguments, ψ_{2}|Γ has a composition factor ρ_{1} = φ(ω_{2} + ω_{r−3}) and by Lemma 2.12, k_{ρ_{1}}(u) = p^{s}. This completes the analysis of Case 3).

It seems convenient to consider Cases 4) and 5) together. We have k_{ψ_{2}}(z_{2}) = k_{θ}(z_{2}) where θ ∈ IrrH_{2} and ω(θ) = ω_{3} or ω_{9}. By Lemma 2.25, k_{θ}(z_{2}) = p^{s}.

Let Case 6) take place. The arguments are similar to those of Case 3). If a_{r} = 0, we reduce the question to the subcase μ_{2} = ω_{r−4}. Then Lemma 2.25 yields that k_{ψ_{2}}(z_{2}) = p^{s}. If a_{r} ≠ 0, the ρ_{c}-arguments and Lemma 2.12 imply that ψ_{2}|Γ has a composition factor ρ_{1} = ω_{r−4} + ω_{r−3} and k_{ρ_{1}}(u) = p^{s}. This settles Case 6).

Consider Case 7). Set c = ⟨μ_{2}, α_{r}⟩. Using claim (*), one can assume that either τ_{0} ∈ {ω_{1} + ω_{r−3}, ω_{r−3}, 2ω_{r−3}} and c < p, or τ_{0} = pω_{r−3} and c ∈ {0, p, 2p}. Observe that c = 2a_{r−1} − 2b + a_{r} < 3p ≤ p^{2}.

Let ψ_{2} ∈ I_{p}(H_{2}). First assume that c = 0. Take m_{1} = X_{−r}^{2}X_{−(r−1)}m. The analysis of the weight system X(ψ_{2}) and Lemma 2.19 yield that m_{1} ≠ 0 and the groups X_{i} fix m_{1} for 3 ≤ i < r. Hence m_{1} generates an indecomposable Γ-module with highest weight σ ∈ {ω_{1} + ω_{r−4} + ω_{r−3}, ω_{r−4} + ω_{r−3}, ω_{r−4} + 2ω_{r−3}}. Put π = φ(σ). Then ψ_{2}|Γ has a composition factor π and by Lemma 2.12, k_{π}(u) = p^{s}.

Now let c ≠ 0. One can assume p > 3 as ψ_{2} is p-large for p = 3. If τ_{0} ≠ ω_{r−3}, by the ρ_{c}-arguments, the restriction ψ_{2}|Γ has a composition factor ρ_{1} with highest weight τ_{1} ∈ {ω_{1} + 2ω_{r−3}, 3ω_{r−3}} and by Lemma 2.12, k_{ρ_{1}}(u) = p^{s}.

Let τ_{0} = ω_{r−3}. If c > 1, then by the ρ_{c}-arguments, ψ_{2}|Γ has a composition factor ρ_{2} = 3ω_{r−3} and we have just seen that k_{ρ_{2}}(u) = p^{s}.

Let c = 1. Set m_{1} = X_{−r}^{3}X_{−(r−1)}m. Lemma 2.19 and the analysis of the system X(ψ_{2}) yield that the vector m_{1} ≠ 0 and is fixed by the groups X_{i}, 3 ≤ i < r. Hence m_{1} generates an indecomposable Γ-module with highest weight ν = ω_{r−4} + 2ω_{r−3}. Set θ = φ(ν). Then θ is a composition factor of ψ_{2}|Γ and by 2.12, k_{θ}(u) = p^{s}. Hence in all the cases with p-restricted τ_{0} we have shown that k_{ψ_{2}}(z_{2}) = p^{s}.

Now let τ_{0} = pω_{r−3}. Then μ_{2} = pδ where δ ∈ X^{+}. Put λ = φ(δ). It is clear that k_{ψ_{2}}(z_{2}) = k_{λ}(z_{2}). But λ is one of the representations just considered. This completes the analysis of Case 7).

Finally, turn to Case 8). The subcase where p = 3, r = 14 and τ_{0} ∈ {ω_{3}, ω_{9}} can be handled just like Cases 4) and 5). Assume that this is not the case. Arguing as in Cases 1)-3) and 6)-7), we can exclude the situations where

τ_{0} ∈ {2ω_{1}, ω_{1} + ω_{r−3}, ω_{2}, ω_{r−4}, ω_{r−3}, 2ω_{r−3}}.

Then either τ_{0} ∈ {0, ω_{1}}, or τ_{0} = pω_{1} and a_{r} = 0.

First let τ_{0} = 0 or ω_{1}. Let a_{r} ≥ 3 if τ_{0} = 0 and a_{r} ≥ 2 if τ_{0} = ω_{1}. Set ρ = ρ_{3} for τ_{0} = 0 and ρ = ρ_{2} for τ_{0} = ω_{1}. Then by the ρ_{c}-arguments and Lemma 2.12, k_{ρ}(u) = p^{s}.

Let τ_{0} = ω_{1} and a_{r} ≤ 1. If a_{r} = 1, the arguments used in the similar situation in Case 1) yield that k_{ψ_{2}}(z_{2}) = p^{s}.

Let a_{r} = 0. We have a = a_{1} + a_{2} + b + 2a_{3}, a_{2} − b + a_{3} = 1 and a_{1} + a_{2} − b = p − 1. Hence either a_{2} = b, a_{1} = p − 1 and a_{3} = 1, or a_{2} = b + 1, a_{3} = 0 and a_{1} = p − 2. Obviously, a = p + 1 + 2a_{2} > p + 1 in the first case and a = p − 3 + 2a_{2} ≥ p + 1 in the second one (here a_{2} > 1). In both cases a ≠ p^{l} since a is even. Then k_{ψ_{1}}(z_{1}) ≥ 3 by 2.24. It is well known that k_{ψ_{2}}(z_{2}) = p^{s}. Now Lemma 2.17 forces k_{ψ}(z) = p^{s}.

Let τ_{0} = 0 and a_{r} ≤ 2. Since μ_{2} ≠ 0, we have a_{r} > 0.

If a_{r} = 1, we get k_{ψ_{2}}(z_{2}) = p^{s} by Lemma 2.25.

Next, let a_{r} = 2. Several applications of 2.19 yield that m_{1} = X_{−r}^{2}X_{−(r−1)}X_{−r}m ≠ 0. Analyzing the weight system X(ψ), we conclude that the groups X_{i} fix m_{1} for 3 ≤ i < r. Now it is clear that m_{1} generates an indecomposable Γ-module with highest weight λ = ω_{r−4} + ω_{r−3}. So ψ_{2}|Γ has a composition factor τ = φ(λ). By Lemma 2.12, k_{τ}(u) = p^{s}. This completes the analysis of the case τ_{0} = 0.

Finally, let τ_{0} = pω_{1} and a_{r} = 0. We have a = a_{1} + a_{2} + b + 2a_{3}, a_{2} − b + a_{3} = p and a_{1} + a_{2} − b = p − 1. Hence a_{3} > a_{1}, a = p − 1 + 2(a_{3} + b) ≥ p + 3 and a is even and is not a power of p. Then we apply Lemmas 2.24 and 2.17 as for τ_{0} = ω_{1} and conclude that k_{ψ}(z) = p^{s}.

Thus we have settled Cases 1)-8), and all possibilities where s > 1 and p^{s} > 9 have been considered.

Now let p^{s} = 9. Then n = 11, r = 5. If ψ_{2} is p-large, the assumptions of the proposition yield that k_{ψ_{2}}(z_{2}) = p^{s}. The arguments in the proof of (*) imply that k_{ψ_{2}}(z_{2}) > 6 as μ_{2} ≠ 0. By 2.24, k_{ψ_{1}}(z_{1}) ≥ 3 if a ≠ 3^{l} with l ≥ 0 (Formula (3) forces a ≠ 0). Since ψ ≅ ψ_{1}⊗ψ_{2}, Lemma 2.17 implies that k_{ψ_{2}}(z_{2}) = 9 if a ≠ 3^{l}. Recall that a_{j}−b+∑_{i=1}^{j−1} a_{i} = 2. Now Formulae (3)-(6) show that there is the unique possibility with a = 3^{l} and ψ_{2} not p-large and not tensor decomposable: j = 2, ω = 2ω_{1} + 2ω_{2} + ω_{3} + ω_{5} and μ_{2} = ω_{1} + ω_{3}. By Lemma 2.26, k_{ψ_{2}}(z_{2}) = 9 for this μ_{2}. This settles the case r = 5.

Next, let s = 1. Observe that p > 5 as n > 7. First assume that p = 7. Then H_{2} ≅ B_{2}(K) ≅ C_{2}(K). It will be more convenient for us to regard the group H_{2} as B_{2}(K) taking Π(H_{2}) = {α_{3}, α_{4}} (in this order). Results 2.1, 2.3 and 2.4 yield that k_{ψ_{2}}(z_{2}) = 7 if μ_{2} ∉ {7^{l}ω_{1}, 7^{l}ω_{2}}. Formulae (4)-(5) yield that it suffices to consider the following situations:

a) j = 2, a_{4} = 0 and a_{2} − b + a_{3} ∈ {1, 7};

b) j = 2, a_{2} = b and a_{3} = 0, a_{4} = 1;

c) j = 3, a_{4} = 0 and a_{3} = b, a_{2} + a_{3} ∈ {1, 7}.

By (3), we have a = a_{1} + a_{2} + b + 2a_{3}, a_{1} + a_{2} − b = 6 in Case a); a = a_{1} + 2a_{2} + 1, a_{1} = 6 in Case b); and a = a_{1} + a_{2} + 2a_{3}, a_{1} + a_{2} = 6 in Case c). Hence in all these cases a > 7 and a ≠ 7^{l}; furthermore, in Case b) a is odd and 9 ≤ a < 49. Applying Lemma 2.24 as for s > 1, we deduce that k_{ψ_{1}}(z_{1}) ≥ 3 in Cases a) and c) and k_{ψ_{1}}(z_{1}) ≥ 4 in Case b). Due to Results 2.1 and 2.4, k_{ψ_{2}}(z_{2}) = 5 in Cases a) and c) and 4 in Case b). Now Lemma 2.17 yields that k_{ψ}(z) = 7 in all Cases a)-c). This completes the analysis of the case p = 7.

Now let p > 7. Then by 2.13, k_{ψ_{2}}(z_{2}) = p if μ_{2} ≠ p^{l}ω_{1} with l ∈ Z^{+}. Formulae (4)-(6) yield that μ_{2} = p^{l}ω_{1} in the following cases only:

a) j = 3, b = a_{3}, a_{2} + a_{3} = 1 or p and a_{i} = 0 for 4 ≤ i ≤ r;

b) j = 2, a_{2} − b + a_{3} = 1 or p and a_{i} = 0 for 4 ≤ i ≤ r.

Observe that r > 5 since p > 7. Formula (3) implies that in Case a) a = a_{1} + a_{2} + 2a_{3} and a_{1} + a_{2} = p − 1, in Case b) a = a_{1} + a_{2} + a_{3} + b and a_{1} + a_{2} + a_{3} − b = p − 1. In both cases k_{ψ_{2}}(z_{2}) = p − 2, a is even and greater than p. We apply Lemma 2.24 as for s > 1 or p = 7 and conclude that k_{ψ_{1}}(z_{1}) ≥ 3 and k_{ψ_{2}}(z_{2}) = p as desired. Thus all the possibilities where ∑_{i=1}^{r−1} a_{i} > p − 1 have been considered.

II. Now let ∑_{i=1}^{r−1} a_{i} ≤ p − 1, but ∑_{i=1}^{r} a_{i} ≥ p. Then we have p − 1 = a_{r} − b + ∑_{i=1}^{r−1} a_{i} where 0 < b ≤ a_{r}. Take m = v(2, r, b). Lemma 2.20 implies that m ≠ 0 and the groups X_{i} fix m for i > 0 and i ≠ 2. It is clear that X_{β} fixes m. One easily deduces that ω_{β}(m) = p − 1, a = b + ∑_{i=1}^{r−1} a_{i}, and

μ_{2} = (∑_{i=1}^{r−3} a_{i+1}ω_{i}) + (2a_{r−1} + a_{r})ω_{r−2}.  (7)

Arguing as in Part I of the proof, we conclude that k_{ψ}(z) = p^{s} if k_{ψ_{1}}(x_{1})+k_{ψ_{2}}(z_{2}) > p^{s}. So we can exclude such situations. First assume that s > 1 and r > 5. Surely, 2a_{r−1} + a_{r} < 3p ≤ p^{2}. Now the ρ_{c}-arguments and Lemma 2.12 yield that either ψ_{2}|Γ has a composition factor ρ with k_{ρ}(z_{2}) = p^{s}, or ψ_{2} is p-large or tensor decomposable, or one of the following holds:

1) μ_{2} = pω_{r−2} or 2pω_{r−2};

2) μ_{2} ∈ {ω_{r−2}, 2ω_{r−2}, ω_{1} + ω_{r−2}, ω_{r−3} + ω_{r−2}}.

By 2.25, k_{ψ_{2}}(z_{2}) = p^{s} if μ_{2} = ω_{r−2}. It has been already shown in Part I when working with the case s > 1 and r > 5 that k_{ψ_{2}}(z_{2}) = p^{s} if μ_{2} is some of the other weights of Case 2). If μ_{2} = pλ where λ is p-restricted and θ = φ(λ) ∈ Irr, it is clear that k_{ψ_{2}}(z_{2}) = k_{θ}(z_{2}). So the facts proven above yield that k_{ψ_{2}}(z_{2}) = p^{s} in Case 1) as well.

Thus we have settled the case s > 1 and r > 5 for our values of the coefficients a_{i}. Now let r = 5. We have p = 3, ∑_{i=1}^{4} a_{i} ≤ 2, a_{5} ≠ 0 and a_{j} ≠ 0 for some j < 5. Taking

all this into account, it is not difficult to deduce from Formula (7) that the cases where ψ_{2} is not tensor decomposable and is not p-large are exhausted by the following ones:

1) ω = ω_{1} + ω_{4} + ω_{5},

2) ω = ω_{1} + ω_{2} + ω_{5},

3) ω = 2ω_{1} + ω_{5},

4) ω = 2ω_{4} + 2ω_{5},

5) ω = 2ω_{1} + 2ω_{5},

6) ω = ω_{1} + 2ω_{5}.

Let h = z_{2}^{3}. Then J(h) = diag(J_{3}, J_{2}, J_{2}) and [10, Theorem 1] implies that k_{ψ_{2}}(h) = 3 if μ_{2} ≠ 0. Thus (ψ_{2}(h) − 1)^{2} ≠ 0 in all such cases which forces (ψ_{2}(z_{2}) − 1)^{6} ≠ 0 and k_{ψ_{2}}(z_{2}) ≥ 7.

In Cases 1)-3) a = 3 and Lemma 2.21 yields that X_{−1}m ≠ 0. Hence m generates an H_{1}-module isomorphic to V (3ω_{1}) by [19, Lemma 2.12]. Then Proposition 2.23 implies that N contains a quotient isomorphic to V (3ω_{1}) ⊗ M(μ_{2}). [19, Lemma 2.11] yields that z_{1} has the minimal polynomial of degree 3 in V (3ω_{1}). Now it follows from 2.17 that k_{N}(z) = 9 in these situations.

We have a = 4 in Cases 4) and 5) and a = 2 in Case 6). Then Lemma 2.24 implies that k_{ψ_{1}}(z_{1}) = 3 in all these cases. Since k_{ψ_{2}}(z_{2}) ≥ 7, Lemma 2.17 forces k_{ψ}(z) = 9 completing the analysis of the case r = 5.

Finally, let s = 1. By our assumptions, p ≥ 7. For p > 7 Lemma 2.13 and Formula (7) yield that k_{ψ_{2}}(z_{2}) = p.

Let p = 7. Then Results 2.1, 2.3 and 2.4 and Formula (7) force that k_{ψ_{2}}(z_{2}) = p, except the cases where μ_{2} = ω_{2} or 7ω_{2}. Let μ_{2} = ω_{2}. Then a_{2} = a_{3} = 0, a_{4} = 1 and a_{1} = 6. Now argue as in Cases 1)-3) for r = 5. Lemma 2.21, [19, Lemmas 2.11 and 2.12] and Proposition 2.23 imply the following facts: X_{−1}m ≠ 0, the H_{1}-module generated by m is isomorphic to V (7ω_{1}), N contains a quotient isomorphic to V (7ω_{1}) ⊗ M(μ_{2}) and z_{1} has the minimal polynomial of degree 7 in V (7ω_{1}). Since ψ_{2} is the standard realization of the group C_{2}(K), we have k_{ψ_{2}}(z_{2}) = 4. Now Lemma 2.17 yields that k_{N}(z) = 7.

Next, let μ_{2} = 7ω_{2}. Then (a_{3}, a_{4}) = (3, 1), (2, 3) or (1, 5), a_{2} = 0. We have b = 1, a_{1} = 3, a = 7 in the first case, 1 ≤ b ≤ 3, a ∈ 5, 7, 9 in the second case and 1 ≤ b ≤ 5, a ∈ 3, 5, 7, 9, 11 in the third one. It is clear that k_{ψ_{2}}(z_{2}) = 4. All the situations where a = 7 are handled as in the case μ_{2} = ω_{2}. Otherwise Lemma 2.24 implies that k_{ψ_{1}}(z_{1}) ≥ 4. Then Lemma 2.17 yields that k_{N}(z) = 7 as above. This completes the arguments of Part II.

III. Finally, let ∑_{i=1}^{r} a_{i} ≤ p − 1. Then we have

p − 1 = (∑_{i=1}^{j} a_{i}) + a_{j} − b + 2( ∑_{i=j+1}^{r−1} a_{i}) + a_{r}

for some j, 2 ≤ j < r and some b ∈ Z^{+}, 0 < b ≤ a_{j} (here the second sum is 0 for j = r−1). Put m = v(2, j, b). By Lemma 2.20, m ≠ 0 and the groups X_{i} fix m for i > 0 and i ≠ 2.

It is clear that X_{β} also fixes m. One easily observes that ω_{β}(m) = p − 1, a = b + ∑_{i=1}^{j−1} a_{i},

μ_{2} = (∑_{i=1}^{j−3} a_{i+1}ω_{i}) + (a_{j−1} + a_{j} − b)ω_{j−2} + (a_{j+1} + b)ω_{j−1} + ∑_{i=j}^{r−2} a_{i+2}ω_{i}

for 3 ≤ j < r − 1 (with the first sum equal to 0 if j = 3),

μ_{2} = (a_{3} + b)ω_{1} + ∑_{i=2}^{r−2} a_{i+2}ω_{i}

for j = 2 and

μ_{2} = (∑_{i=1}^{r−4} a_{i+1}ω_{i}) + (a_{r−2} + a_{r−1} − b)ω_{r−3} + (a_{r} + 2b)ω_{r−2}

for j = r − 1 (with the first sum equal to 0 for r = 4).

Assume first that s > 1 and r > 5. Lemma 2.12 implies that either ψ_{2} is tensor decomposable or p-large, or k_{ρ_{0}}(u) = p^{s}, or one of the following holds:

1) j = 3, a_{2} = 0, a_{3} = b = 1 and a_{i} = 0 for 4 ≤ i ≤ r − 1;

2) p = 3, r = 14, j = 4, a_{4} = b = 1, a_{2} = a_{3} = 0 and a_{i} = 0 for 5 ≤ i ≤ 14;

3) p = 3, r = 14, j = 10, a_{10} = b = 1, a_{i} = 0 for 2 ≤ i ≤ 9 and for 11 ≤ i ≤ 14;

4) j = r − 3, a_{i} = 0 for 2 ≤ i ≤ r − 4, a_{r−3} = 1 and a_{r−2} = a_{r−1} = 0;

5) j = r − 2, a_{i} = 0 for 2 ≤ i ≤ r − 3, a_{r−2} = 1 and a_{r−1} = 0;

6) j = r − 2, a_{2} = a_{r−2} = 1, a_{i} = 0 for 3 ≤ i ≤ r − 3 and a_{r−1} = 0;

7) j = r − 2, b = a_{r−2}, a_{i} = 0 for 2 ≤ i ≤ r − 3 and a_{r−2} + a_{r−1} = 2;

8) j = 2, b = 1 and a_{i} = 0 for 3 ≤ i ≤ r − 1;

9) j = 2, b = a_{r−1} = 1 and a_{i} = 0 for 3 ≤ i ≤ r − 2;

10) j = 2, a_{3} + b = 2 and a_{i} = 0 for 4 ≤ i ≤ r − 1;

11) j = r − 1, τ_{0} ∈ {0, ω_{1}, 2ω_{1}, ω_{2}, ω_{r−4}, ω_{r−3}, 2ω_{r−3}, ω_{1} + ω_{r−3}} and a_{r} + 2b < p if τ_{0} ≠ 0.

If p = 3 and τ_{0} = ∑_{i=1}^{r−3} b_{i}ω_{i} with some b_{i} = 2, we may assume that ⟨μ_{2}, α_{r}⟩ = 0 since otherwise ψ_{2} is p-large. Let ⟨μ_{2}, α_{r}⟩ ≠ 0, p > 3 or all b_{i} < 2, and let some of Cases 1), 4), 6), 7), 9), 10) or Case 11) with τ_{0} ∈ {2ω_{1}, ω_{2}, ω_{r−4}, 2ω_{r−3}, ω_{1} + ω_{r−3}} hold. Then by the ρ_{c}-arguments, ψ_{2}|Γ has a composition factor ρ_{1} = φ(τ_{1}) and Lemma 2.12 implies that k_{ρ_{1}}(u) = p^{s}. Thus we may and shall exclude all these situations. By Lemma 2.25, k_{ψ_{2}}(z_{2}) = p^{s} if μ_{2} ∈ {ω_{2}, ω_{3}, ω_{r−5}, ω_{r−4}, ω_{r−3}, 2ω_{1}}. We have already shown in Part I of this proof that k_{ψ_{2}}(z_{2}) = p^{s} if μ_{2} ∈ {ω_{r−3} + ω_{r−2}, ω_{1} + ω_{r−3}, 2ω_{r−3}}. Thus k_{ψ_{2}}(z_{2}) = p^{s} in Cases 1)-4), 6), 7), 9), 10), in Case 5) with a_{r} = 0 or 1 and in Case 11) with τ_{0} ∉ {0, ω_{1}, ω_{r−3}}.

Let Case 5) take place and a_{r} > 1. We may assume that p > 3 as otherwise ψ_{2} is p-large. Then by the ρ_{c}-arguments and Lemma 2.12, ψ_{2}|Γ has a composition factor ρ_{2} = φ(3ω_{r−3}) and k_{ρ_{2}}(u) = p^{s}.

Consider Case 8). If a_{r} ≥ 2, the arguments of the previous paragraph yield that ψ_{2}|Γ has a composition factor ρ_{2} = φ(ω_{1} + 2ω_{r−3}) and k_{ρ_{2}}(u) = p^{s}.

If a_{r} = 1, then μ_{2} = ω_{1} + ω_{r−2} and we have already shown in Part I that k_{ψ_{2}}(z_{2}) = p^{s} in this case.

Let a_{r} = 0. We get a_{l} = 0 for 3 ≤ l ≤ r, b = 1, a_{1} + 2a_{2} − 1 = p − 1, a = a_{1} + 1 and k_{ψ_{2}}(z_{2}) = p^{s} − 2. Hence 0 < a_{1} < p − 1. By Lemma 2.24, k_{ψ_{1}}(z_{1}) ≥ 3. Now Lemma 2.17 implies that k_{ψ}(z) = p^{s} as desired.

It remains to consider Case 11) with τ_{0} ∈ {0, ω_{1}, ω_{r−3}} and a_{r} + 2b < p for τ_{0} ≠ 0. Assume first that τ_{0} ≠ 0. Since a_{r} + 2b ≥ 2, the ρ_{c}-arguments force that ψ_{2}|Γ has a composition factor ρ_{2} = φ(ω_{1} + 2ω_{r−3}) or φ(3ω_{r−3}). We exclude the case p = 3 since then ψ_{2} is p-large. For p > 3 Lemma 2.12 yields that k_{ρ_{2}}(u) = p^{s}.

Finally, let τ_{0} = 0. Then a_{i} = 0 for 2 ≤ i ≤ r−2, b = a_{r−1}, a_{1}+a_{r−1}+a_{r} = p−1. Hence 2a_{r−1}+a_{r} < 2p. Since we have excluded the cases where ψ_{2} is tensor decomposable, we can assume that 2a_{r−1} + a_{r} ≤ p. If 2 < 2a_{r−1} + a_{r} < p, by the ρ_{c}-arguments and Lemma 2.12, ψ_{2}|Γ has a composition factor ρ_{3} = φ(3ω_{r−3}) and k_{ρ_{3}}(u) = p^{s}. We have already seen in Part I of the proof that k_{ψ_{2}}(z_{2}) = p^{s} if ψ_{2} = 2ω_{r−2}. Since the Frobenius morphism does not change the minimal polynomials, Lemma 2.25 forces k_{ψ_{2}}(z_{2}) = p^{s} for ψ_{2} = pω_{r−2}. It is clear that 2a_{r−1} + a_{r} > 1. So all the possibilities with s > 1 and r > 5 have been considered.

Now let r = 5. Then p^{s} = 9. As we have excluded p-large and tensor decomposable representations ψ_{2}, we have to consider the following cases only:

a) j = 3 and ω = ω_{1} + ω_{3};

b) j = 2 and ω = 2ω_{2};

c) j = 2 and ω = ω_{1} + ω_{2};

d) j = 2 and ω = ω_{2} + ω_{5};

e) j = 4 and ω = ω_{1} + ω_{4};

f) j = 4, ω = ω_{4} + ω_{5}.

In Cases a) and b) μ_{2} = ω_{2} and 2ω_{1}, respectively, and k_{ψ_{2}}(z_{2}) = p^{s} by 2.25.

In Cases c) and e) a = 2 and k_{ψ_{1}}(z_{1}) = 3 by 2.24. As we have seen before, k_{ψ_{2}}(z_{2}) > 6 if μ_{2} ≠ 0. Then Lemma 2.17 yields that k_{ψ}(z) = 9 as required.

In Case d) μ_{2} = ω_{1} + ω_{3}, and k_{ψ_{2}}(z_{2}) = p^{s} by Lemma 2.26.

Consider Case f). Here m = X_{−2}X_{−3}X_{−4}v. Put m_{1} = X_{−5}m. Using commutator relations in L and 2.19, we can deduce that X_{4}X_{3}X_{2}m_{1} = X_{−5}v ≠ 0. Hence m_{1} ≠ 0. The same commutator relations show that X_{5}m_{1} = 0 and X_{5} fixes m_{1}. One easily observes that ω(m_{1}) + α_{j} ∉ X(M) for j = 3, 4. Hence the groups X_{j} also fix m_{1} for these j. Now it is clear that m_{1} generates an indecomposable H_{2}-module with highest weight χ = ω_{2} +ω_{3}. Put π = φ(χ). Since π is p-large, the assumptions of the proposition yield that k_{π}(z_{2}) = 9. Then k_{N}(z_{2}) = 9 which completes the analysis of the case r = 5.

Now let s = 1. Then according to our assumptions, p > 5. Results 2.1, 2.3, 2.4 and 2.13 imply that either k_{ψ_{2}}(z_{2}) = p, or one of the following holds: j = 2, μ_{2} = ω_{1} or r = 4, j = 3, p = 7 and μ_{2} = 7ω_{2}. In the first case we have k_{ψ_{2}}(z_{2}) = p − 2, b = 1, a_{i} = 0 for i > 2 and a_{1} + 2a_{2} = p. So a_{1} ≠ 0. Hence a = a_{1} + 1 ≥ 2 and a < p. Then k_{ψ_{1}}(z_{1}) ≥ 3 by 2.24 and k_{ψ}(z) = p by 2.17. In the second one we get a_{2} = 0 and m = v(2, 3, b). Put m_{1} = X_{−4}m. Using commutator relations in L, Lemmas 2.19 and 2.20 and Formula (1), one can deduce that X_{3}X_{2}m_{1} ≠ 0 and the groups X_{⟩} fix m_{1} for i = 3 and 4. Hence m_{1} generates an indecomposable H-module with highest weight σ = ω_{1} + 5ω_{2}. Put θ = φ(σ). Results 2.1, 2.3 and 2.4 yield that k_{θ}(z_{2}) = 7 and hence k_{N}(z_{2}) = 7. Now all the possibilities have been considered. This completes the proof. □

Proposition 3.7 Let G = B_{r}(K), n > p^{s} + 2, s > 0, and let φ ∈ I_{p} be p-large. Assume that (H, s + 1)-conjecture holds for H = B_{r−1}(K) and H = D_{r}(K). Then k_{φ}(x) = p^{s+1} for each element x ∈ G of order p^{s+1}.

Proof. Let x ∈ G and |x| = p^{s+1}. Lemma 2.8 and the description of unipotent conjugacy classes in [15, ch. IV, Exercise 2.15] imply that \overline{cl(x)} contains an element of order p^{s+1} from the subgroup H = B_{r−1}(K) or D_{r}(K) naturally embedded into G. Now Lemma 2.9 and the assumptions of our proposition yield that k_{φ}(x) = p^{s+1} provided φ|H has a p-large composition factor. We shall prove that such factor exists.

First let H = H(2, …, r) ≅ B_{r−1}(K). Let a_{2} + a_{3} ≥ p. Smith’s theorem [13] forces that φ|H has a composition factor φ_{1} with highest weight ∑_{i=1}^{r−1} a_{i+1}ω_{i}. It is clear that l(φ_{1}) > p, hence φ_{1} is p-large.

Now let a_{2} + a_{3} < p. Set v_{1} = v(1, 2, a_{2}). By Lemma 2.20, v_{1} ≠ 0 and the groups X_{i} fix v_{1} for i > 1. Hence v_{1} generates an indecomposable H-module with highest weight λ = a_{1}ω_{1} + (a_{2} + a_{3})ω_{2} + ∑_{i=3}^{r−1} a_{i+1}ω_{i}. Set ρ = φ(λ). It is clear that l(ρ) = l(φ) ≥ p and ρ is p-large.

Next, let H be the subgroup generated by all long root subgroups. It is well known that H ≅ D_{r}(K) and one can take Π(H) = {α_{1}, …, α_{r−1}, α_{r−1} + 2α_{r}}. Now one easily concludes that v generates an indecomposable H-module with highest weight

μ = (∑_{i=1}^{r−1} a_{i}ω_{i}) + (a_{r−1} + a_{r})ω_{r}.

Set ρ = φ(μ). If a_{r−1} + a_{r} < p, we get l(ρ) = l(φ) ≥ p, and ρ is p-large.

Let a_{r−1}+a_{r} ≥ p. Then a_{r−1}+a_{r} = p+s where s ∈ Z^{+} and s < a_{r}. Set k = p−1−a_{r−1} and take v_{1} = X_{−r}^{k}v. By Lemma 2.19, v_{1} ≠ 0. One easily observes that the group U^{+}(H) fixes v_{1}. Now it is clear that v_{1} generates an indecomposable H-module with highest weight

τ = (∑_{i=1}^{r−2} a_{i}ω_{i} + (a_{r−1} + k)ω_{r−1} + (a_{r−1} + a_{r} − k)ω_{r}.

As a_{r−1} + k = p − 1 and k ≤ p − 1 < a_{r−1} + a_{r}, one easily observes that l(φ(τ)) ≥ p and τ is p-large. This completes the proof. □

Proposition 3.8 Let G = D_{r}(K), n > p^{s} + 1, s > 0and let φ ∈ I_{p} be p-large. Assume that (B_{r−1}, s + 1)-conjecture holds. Then k_{φ} = p^{s+1} for each element x ∈ G of order p^{s+1}.

Proof. The proof is similar to that of 3.7. Let x ∈ G and |x| = p^{s+1}. Lemma 2.8 and [15, ch. IV, Exercise 2.15 and Result 2.27(ii)] imply that \overline{cl(x)} contains a representative lying in a subgroup H of type B_{r−1} naturally embedded into G or in one of the subgroups H = H(1, 2, …, r − 1) or H = H(1, 2, …, r − 2, r). Then Lemma 2.9, Theorem 1.1 for groups of type A and the assumptions of the proposition imply that it suffices to show that φ|H has a p-large composition factor in all these cases.

First let

H = ⟨X_{i}, X_{−i}, x_{r−1}(t)x_{r}(t), x_{−(r−1)}(t)x_{−r}(t) | 1 ≤ i ≤ r − 2, t ∈ K⟩.

It is well known (see, for instance, [11, Section 8])that H ≅ B_{r−1}(K). Let Π(H) = {β_{1}, …, β_{r−1}}. One may assume that X_{β_{i}} = X_{i} for 1 ≤ i ≤ r − 2 and X_{β_{r−1}} = X_{r−1} + X_{r}. The canonical homomorphism of weight systems X(G) → X(H) associated with the embedding H → G maps the fundamental weights ω_{i}, 1 ≤ i ≤ r − 2, of G to the weights ω_{i} ∈ X(H), the weights ω_{r−1} and ω_{r} to the weight ω_{r−1}, and the roots α_{r−1} and α_{r} to β_{r−1}. Now one easily observes that v generates an indecomposable H-module with highest weight λ = (∑_{i=1}^{r−2} a_{i}ω_{i}) + (a_{r−1} + a_{r})ω_{r−1}. If a_{r−1} + a_{r} < p, the weight λ is p-large, and we are done.

Let a_{r−1} + a_{r} ≥ p. Applying the graph automorphism if required, one can assume a_{r−1} ≥ a_{r}. It is clear that a_{r} > 0. For any h ∈ Z with 0 < h ≤ a_{r} set

S_{h} = {λ ∈ X(φ) | ω − λ = xα_{r−1} + yα_{r}, x + y = h}

and M_{h} = ⟨M_{λ} | λ ∈ S_{h}⟩. Since X_{r} and X_{r−1} commute, using Lemma 2.19, one easily concludes that dim M_{λ} = 1 for λ ∈ S_{h} and |S_{h}| = h + 1. So dim M_{h} = h + 1. It is clear that X_{β_{r−1}}M_{h} ⊂ M_{h−1}. Hence M_{h} contains a nonzero vector v_{h} which is annihilated by X_{β_{r−1}}. Now Formula (1) implies that X_{β_{r−1}} fixes v_{h}. Hence v_{h} is fixed by U^{+}(H). Now it is clear that v_{h} generates an indecomposable H-module with highest weight

ρ_{h} = (∑_{i=1}^{r−3} a_{i}ω_{i}) + (a_{r−2} + h)ω_{r−2} + (a_{r−1} + a_{r} − 2h)ω_{r−1}.

One can choose h such that a_{r−1} + a_{r} − 2h = p − 2 or p − 1. Then φ(ρ_{h}) is p-large. This completes the proof for H ≅ B_{r−1}(K).

Now let H = H(1, 2, …, r − 1). If ∑_{i=1}^{r−1} a_{i} ≥ p, Smith’s theorem [13] yields that v generates an irreducible H-module with a p-large highest weight.

Let ∑_{i=1}^{r−1} a_{i} < p ≤ ∑_{i=1}^{r} a_{i}. Take v_{1} = X_{−r,a_{r−2}+a_{r}}X_{−(r−2),a_{r−2}}v. By Lemma 2.19, v_{1} ≠ 0. Lemma 2.20 implies that X_{i} fixes v_{1} for 0 < i < r. One can easily conclude that v_{1} generates an indecomposable H-module with highest weight

μ = (∑_{i=1}^{r−4} a_{i}ω_{i}) + (a_{r−3} + a_{r−2})ω_{r−3} + a_{r}ω_{r−2} + (a_{r−2} + a_{r−1})ω_{r−1}

(the first sum is 0 if r = 4). It is clear that φ(μ) is p-large.

Finally, let ∑_{i=1}^{r} a_{i} < p. Set a = ∑_{i=1}^{r−2} a_{i}, v_{1} = X_{−r,a_{r}+a}v(r −2, 1, a_{1}). By Lemmas 2.19 and 2.20, v_{1} ≠ 0 and the groups X_{i} fix v_{1} for 0 < i < r. One easily deduces that v_{1} generates an indecomposable H-module with highest weight ν = (∑_{i=1}^{r−3} a_{i+1}ω_{i})+a_{r}ω_{r−2} + (∑_{i=1}^{r−1} a_{i})ω_{r−1}. We have l(φ(ν)) = l(φ) ≥ p, so φ(ν) is p-large. Hence φ|H always has a p-large composition factor.

Similarly we can prove the existence of such factor for H = H(1, …, r − 2, r). This completes the proof. □

Using induction on r and s, we conclude now that Theorem 1.1 for G = B_{r}(K) or D_{r}(K) follows from the general scheme in 3.1 and Results 2.18 and 3.4-3.8. Thus Theorem 1.1 holds for all the classical groups.

4 Proofs of other main results

In this section Theorems 1.2 and 1.3 and Corollary 1.4 are proved. Recall that G_{C} is the group over C of the same type as G; if φ ∈ I_{p}, then φ_{C} ∈ Irr G_{C} and ω(φ_{C}) = ω(φ).

Lemma 4.1 Let x ∈ G and y ∈ G_{C} be unipotent regular elements, and let φ ∈ I_{p}. Then k_{φ}(x) ≤ k_{φ_{C}}(y).

Proof. Set ω = ω(φ), V = V (ω) and denote by V_{C} the irreducible G_{C}-module with highest weight ω. Then φ can be realized in a quotient of V, and k_{φ}(x) ≤ k_{V} (x). The construction of the Chevalley groups in [17, §3] and the arguments in [17, Theorem 39] show that there exist such bases B in V and B_{C} in V_{C} that the matrix of x in B is obtained from that of y in B_{C} (an integer one) by reduction mod p. Hence k_{V} (x) ≤ k_{φ_{C}}(y) which yields our claim. □

Proof of Theorem 1.2. Define x and y as in 4.1. Let n and φ ∈ Irr be such as in the assertion of the theorem. Set k = (p − 1)p^{l} + 1. Proposition 2.3 and Algorithm 2.4 imply that k_{φ_{C}}(y) = k in all four cases. By Lemma 2.7, if G = A_{r}(K), then for i ≤ l the element x^{p^{i}} has one Jordan block of size p^{l−i} + 1 and (p^{i} − 1) blocks of size p^{l−i} in the standard realization of G. Now Results 2.2-2.4 yield that k_{φ}(x) = k for G = A_{r}(K).

Let G = C_{r}(K). Put G_{1} = A_{2r−1}(K). It is well known (see, for instance, [11, (8.1)]) that φ = ψ|G where ψ ∈ Irr G_{1} and ω(ψ) = (p−1)ω_{1}. Observe that x is a regular element of G_{1}. Arguing as above, we deduce that k_{φ}(x) = k.

Now let G = B_{r}(K) or D_{r}(K). By Lemma 4.1, k_{φ}(x) ≤ k_{φ_{C}}(y) = k. Let τ ∈ Irr = φ(ω_{2}), M_{1} = M(ω_{2}), and let m_{1} ∈ M_{1} be a nonzero highest weight vector. We can take x = ∏_{i=1}^{r} x_{−i}(1). We claim that (x − 1)^{2p^{l}}m_{1} is a nonzero lowest weight vector.

First let G = B_{r}(K). Then G ⊂ G_{1} = A_{2r}(K) and x is a regular element in G_{1}. The representation τ ≅ τ_{1}|G where τ_{1} ∈ IrrG_{1} and ω(τ_{1}) = ω_{2} (see, for instance, [11, (8.1)]). Now Results 2.2-2.4 imply that k_{τ}(x) = 2p^{l} + 1. One easily observes that 2ω_{2} = 2(α_{1} +2 ∑_{i=2}^{r} α_{i}). Since n = p^{l} +2, this yields that (x−1)^{2p^{l}}m_{1} is a nonzero lowest weight vector.

Next, let G = D_{r}(K), and let H ⊂ G be the naturally embedded subgroup of type B_{r−1}. It is well known that J(x) = diag(J_{n−1}, 1) and that x can be regarded as a regular element of H . By 2.3 and 2.4, k_{τ_{C}}(y) = 2p^{l} + 1. Put Π(H) = {β_{1}, …, β_{r−1}}. Using comments in [11, Section 8], we can take x_{−β_{i}}(1) = x_{−i}(1) for i ≤ r − 2 and x_{−β_{r−1}}(1) = x_{−(r−1)}(1)x_{−r}(1). Then x = ∏_{i=1}^{r−1} x_{−β_{i}}(1). One easily observes that m_{1} generates an indecomposable (in fact, irreducible ) H-module with highest weight ω_{2}. Hence τ|H has a composition factor τ′ = φ(ω_{2}). The arguments above for G = B_{r}(K) show that k_{τ′}(x) = 2p^{l} + 1. Since k_{τ}(x) ≤ k_{τ_{C}}(y) by Lemma 4.1, this implies that k_{τ}(x) = 2p^{l} + 1. Now we have ω_{2} = α_{1} + 2(∑_{i=2}^{r−2} α_{i}) + α_{r−1} + α_{r}. As 2r = p^{l} + 3, this forces that our claim holds for G = D_{r}(K) as well.

Now complete the proof for G = B_{r}(K) or D_{r}(K) using Proposition 2.22. For 1 ≤ t ≤ (p − 1)/2 set φ_{t} = φ(tω_{2}) ∈ Irr, M_{t} = M(tω_{2}), and let m_{t} ∈ M_{t} be a nonzero highest weight vector. We claim that (x − 1)^{2tp^{l}}m_{t} ≠ 0 and is a lowest weight vector in M_{t}. For t = 1 this has been just shown. Apply induction on t. Let t > 1. Assume that our claim holds for all u < t. Then use Proposition 2.22 with λ_{1} = (t−1)ω_{2}, λ_{2} = ω_{2}, k_{1} = 2(t−1)p^{l} and k_{2} = 2p^{l}. Since the highest and the lowest weights of φ_{t} are the sums of the highest and, respectively, the lowest weights of the representations φ_{t−1} and φ_{1}, we conclude that (x − 1)^{2tp^{l}}m_{t} is a lowest weight vector. According to 2.22, (x − 1)^{k_{1}+k_{2}}m_{t} ≠ 0 since \binom{2tp^{l}}{2p^{l}} ≢ 0(mod p) by [7, Lemma 22.4]. So our claim holds. Taking t = (p − 1)/2, we get k_{φ}(x) ≥ k which completes the proof as we have already seen that k_{φ}(x) ≤ k. □

In what follows G = C_{r}(K), φ ∈ I_{p} and ω = ω(φ) = ∑_{i=1}^{r} a_{i}ω_{i}. For 1 ≤ k ≤ j ≤ r set Σ_{k,j} = ∑_{i=k}^{j} a_{i}, Σ = Σ_{1,r}.

Proof of Theorem 1.3. Let a_{1} + 2Σ_{2,r} ≥ p. If Σ ≥ p, the result follows from Theorem 1.1. So assume Σ < p. Put H = H(1, 2, …, r − 1). All the elements we have to consider are conjugate to elements of H. We can also assume r > 2 since for r = 2 all p-elements of H have order p, and the result follows from Theorem 2.1. By this theorem, it suffices to consider elements of order > p. It seems convenient to subdivide the proof into the following cases.

Case 1). ω = a_{r}ω_{r}. Since p is odd, we have 2a_{r} ≥ p + 1. Set v_{1} = X_{−r,(p+1)/2}v. By Lemma 2.19, v_{1} ≠ 0. Hence v_{1} generates an indecomposable H-module with highest weight (p + 1)ω_{r−1}. Put v_{2} = X_{−(r−1),2}v_{1} and check that v_{2} ≠ 0. Indeed, by 2.19 and commutation relations in L, we get X_{r}v_{2} = cX_{−(r−1),2}X_{−r,(p−1)/2}v ≠ 0 (here c ∈ K^{∗}) and X_{r−1}v_{2} = 0. So it follows from Formula (1) that v_{2} is fixed by the groups X_{i} with i < r and hence by U^{+}(H). Now one can observe that v_{2} generates an indecomposable H-module with highest weight τ = 2ω_{r−2} + (p − 3)ω_{r−1}. Set ψ = φ(τ). Hence ψ is a composition factor of φ|H. We claim that k_{ψ}(x) = |x| for each p-element x ∈ H. Let |x| = p^{s+1}, s ≥ 0. Then r ≥ p^{s} and [14, I.2.4] implies that the Zariski closure of the conjugacy class in H containing x contains an element y ∈ H with J(y) = diag(J_{p^{s}+1}, E_{t}). (Here we mean the canonical Jordan form of y in the standard H-module.) Put H_{1} = H(i | r − p^{s} ≤ i < r). Then H_{1} ≅ A_{p^{s}}(K), and we may assume that y ∈ H_{1}. By Smith’s theorem [13], ψ|H_{1} has a composition factor ψ_{1} = φ(2ω_{p^{l}−1} +(p−3)ω_{p^{l}}). By Lemma 2.7, the element y^{p^{i}}, 0 < i ≤ s, has one Jordan block of size p^{s−i} + 1 and (p^{i} − 1) blocks of size p^{s−i}. Now Results 2.2-2.4 yield that k_{ψ}(y) = k_{ψ_{1}}(y) = p^{s+1}. Then k_{φ}(x) = k_{ψ}(x) = p^{s+1} by Lemma 2.9.

Case 2). ω = a_{r−1}ω_{r−1} + a_{r}ω_{r}, a_{r−1} ≠ 0, a_{r−1} + 2a_{r} ≥ p. Set a = a_{r−1} + 2a_{r}. First assume that a > p. Then there exists b ≤ a_{r} such that a_{r−1} + 2b = p + 2 for odd a and a_{r−1} + 2b = p + 1 for even a. Observe that p > 3 if a is odd since Σ < p. Put v_{1} = X_{−r,b}v, v_{2} = X_{−(r−1),2}v_{1} for even and v_{2} = X_{−(r−1),3}v_{1} for odd a. Arguing as in Case 1), we show that v_{2} ≠ 0 (taking the element X_{r}v_{2} for even and X_{r}^{2}v_{2} for odd a) and conclude that v_{2} generates an indecomposable H-module with highest weight τ = 2ω_{r−2} + (p − 3)ω_{r−1} for even and 3ω_{r−2} + (p − 4)ω_{r−1} for odd a. Hence φ|H has a composition factor ψ = φ(τ). Using Results 2.7, 2.9 and 2.2-2.4 as in Case 1), one can find out that k_{ψ}(x) = |x| for every p-element x ∈ H.

Now let a = p. Put v_{1} = v(r, r − 1, a_{r−1}). By Lemma 2.20, v_{1} ≠ 0 and the groups X_{i} fix v_{1} for 1 ≤ i < r. Hence v_{1} generates an indecomposable H-module with highest weight a_{r−1}ω_{r−2}+pω_{r−1}. Let H_{2} = H(r−1, r). By Smith’s theorem [13], v generates an irreducible restricted H_{2}-module M_{2}. Since ⟨ω(v_{1}), α_{r−1}⟩ > 0, v_{1} is not a lowest weight vector in M_{2}. By [1, Lemma 6.2 and Theorem 6.4], M_{2} is an irreducible L(H_{2})-module and X_{−α}v_{1} ≠ 0 for some α ∈ R^{+}(H_{2}). As p > 2, the commutator relations in L(H_{2}) yield that X_{−i}v_{1} ≠ 0 for some i ∈ {r−1, r}. Considering the weight system X(M), we conclude that X_{−r}v_{1} = 0. Hence v_{2} = X_{−(r−1)}v_{1} ≠ 0. The commutator relations in L yield that X_{r−1}v_{2} = 0. As a_{r−1} + a_{r} < p and a_{r} ≠ 0, we have a_{r−1} + 1 < p. Now Formula (1) implies that the groups X_{i} fix v_{2} for i < r. Then one easily deduces that v_{2} generates an indecomposable H-module with a p-restricted highest weight τ = (a_{r−1} + 1)ω_{r−2} + (p − 2)ω_{r−1}. Since a_{r−1} > 0, the weight τ is p-large. Now Theorem 1.1 for groups of type A_{r} implies our claim.

Case 3). Σ_{1,r−2} ≠ 0 and Σ + a_{r} ≥ p. Set s = Σ + a_{r}. There exists b ≤ a_{r} such that Σ_{1,r−1} + 2b = p for odd and p + 1 for even s. Put v_{1} = X_{−r,b}v. By Lemma 2.19, the vector v_{1} ≠ 0 and generates an indecomposable H-module with highest weight τ = (∑_{i=1}^{r−2} a_{i}ω_{i})+(a_{r−1} +2b)ω_{r−1}. If a_{r−1} +2b < p, we get that l(φ(τ)) ≥ p and φ(τ) is p-large.

Let a_{r−1} + 2b ≥ p. Since Σ_{1,r−2} ≠ 0 and Σ_{1,r−1} + 2b ≤ p + 1, there is the unique possibility Σ_{1,r−2} = 1 and a_{r−1}+2b = p. Put v_{2} = X_{−(r−1)}v_{1}. By 2.19 and the commutator relations in L, we have X_{r}v_{2} = cX_{−(r−1)}X_{−r,b−1}v ≠ 0 (here c ∈ K^{∗}) and X_{r−1}v_{2} = 0. Hence the vector v_{2} ≠ 0 is fixed by the groups X_{i} for 1 ≤ i < r and generates an indecomposable H-module with highest weight τ_{1} = (∑_{i=1}^{r−3} a_{i}ω_{i}) + (a_{r−2} + 1)ω_{r−2} + (p − 2)ω_{r−1}. Since Σ_{1,r−2} = 1 and p > 2, the representation φ(τ_{1}) is p-restricted and p-large. Hence in all situations φ|H has a p-large composition factor, and our claim follows from Theorem 1.1 for G = A_{r}(K).

Case 4). Σ + a_{r} < p. Put v_{1} = v(r, 1, a_{1}). By Lemma 2.20, the vector v_{1} ≠ 0 and is fixed by the groups X_{i} for 1 ≤ i < r. Now one can easily conclude that v_{1} generates an indecomposable H-module with highest weight τ = (∑_{i=1}^{r−2} a_{i+1}ω_{i})+(Σ+a_{r})ω_{r−1}. It is not difficult to see that l(φ(τ)) = a_{1} + 2Σ_{2,r} ≥ p. Hence φ|H contains a p-large composition factor, and the proof is completed as in Case 3). Now all the possibilities have been considered, and the theorem is proved. □

Proof of Corollary 1.4. Define H as in the proof of 1.3. Lemma 2.8 implies that for each element x from the assertion of the corollary \overline{cl(x)} contains an element y ∈ H with |y| = |x|. Then apply Lemma 2.9 and Theorem 1.3. □

Acknowledgements

This research has been done in the framework of the joint project ”Representations of finite groups” performed by scientists of the Institute of Experimental Mathematics, University of Essen, and the Institute of Mathematics, Academy of Sciences of Belarus. A part of the paper has been written during the author’s visit to the IEM. The author is grateful to the Volkswagen Foundation for financial support of the project and to Prof.Dr. G.O. Michler and his colleagues at the IEM for their generous hospitality and helpful discussions.

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