IRREDUCIBLE MODULES OF GENERAL LINEAR GROUPS AND

COMPOSITION FACTORS IN THE POLYNOMIAL ALGEBRA

Pham Anh Minh

Abstract. Let p be a prime number and let GL_{n} be the group of all invertible n × n-matrices over the prime field 𝔽_{p}. It is known that every irreducible GL_{n}-module can be occurred as a composition factor of P, the polynomial algebra with n variables over 𝔽_{p}. Given an irreducible GL_{n}-module ρ corresponding to the regular partition ((p − 1)α_{1}, …, (p − 1)α_{n}), we determine the first value of the degree d of which ρ occurs as a composition factor of P_{d}, the subset of P consisting of homogeneous polynomials of degree d. In some sense, this generalizes Carlisle-Kuhn’s result to the case of any prime p.

1. Introduction

Let p be a prime number and let GL_{n} = GL(n, 𝔽_{p}) be the group of all invertible n × n-matrices over the prime field 𝔽_{p}. Denote by P = P [x_{1}, …, x_{n}] the polynomial algebra with variables x_{1}, …, x_{n} over 𝔽_{p}. There is then an action of GL_{n} on P by linear substitutions. It was shown in [9] that every irreducible 𝔽_{p}GL_{n}-module occurs as a composition factor in P. For p = 2, given an irreducible GL_{n}-module ρ, in [1], Carlisle and Kuhn have computed the first value d for which ρ occurs as a composition factor in P_{d}, the subset of P consisting of polynomials of degree d. We briefly sketch their result as follows.

Let us recall (see e.g. [5, Chapter 8]) that the irreducible GL_{n}-modules are indexed, up to isomorphism, by column regular partitions, i.e., by sequences of non-negative integers

α_{1}, …, α_{n}

satisfying 0 ≤ α_{i} − α_{i+1} ≤ p − 1 for 1 ≤ i ≤ n − 1 and 0 ≤ α_{n} ≤ p − 1. Namely, the irreducible GL_{n}-modules are indexed by column regular partitions α_{1}, …, α_{n} satisfying α_{n} ≤ p − 2. Irreducible GL_{n}-modules were originally constructed in [5] (see also [6]) using the standard ‘Weyl module’. Another construction using modular invariants was given by Tri in [15]. He proved that the irreducible GL_{n}-module corresponding to the column regular partition α = (α_{1}, …, α_{n−1}, α_{n}) is isomorphic to the GL_{n}-module, says T_{α}, generated by

L^{{α}} = L_{1}^{α_{1}−α_{2}}L_{2}^{α_{2}−α_{3}} … L_{n−1}^{α_{n−1}−α_{n}}L_{n}^{α_{n}}

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 20C20.

with

\begin{aligned}L_{k} & = ∏_{i=1}^{k} ∏_{λ_{j}∈𝔽_{p}}(λ_{1}x_{1} + ⋯ + λ_{i−1}x_{i−1} + x_{i}) \\ & = | \begin{matrix}x_{1} & … & x_{k} \\ x_{1}^{p} & … & x_{k}^{p} \\ ⋯ & … & ⋯ \\ x_{1}^{p^{k−1}} & … & x_{k}^{p^{k−1}} \end{matrix} | , 1 ≤ k ≤ n. \end{aligned}

Given a column partition α = (α_{1}, …, α_{n−1}, α_{n}), the conjugate partition α′ of α is defined to be α′ = (α′_{1}, …, α′_{n}), with α′_{i} the number of rows k such that α_{k} ≥ i. We have

Theorem A (Carlisle and Kuhn). For p = 2, the irreducible GL_{n}-module corresponding to the column regular partition α occurs for the first time as a composition factor in P_{d} when

d = ∑_{j≥1} α′_{j}2^{j−1},

where α′ is the conjugate partition of α. □

Note that, as shown in [4], every irreducible GL_{n}-module ρ also occurs as a submodule of P (this fact also follows, of course, from Tri’s construction). Hence, we also have an analogous problem for the first occurrence of ρ in P. For the prime 2, this problem was solved by Schwartz ([13]) and, indepedently, by Tri ([16]); recently, for the case of any prime p, it was completely solved in [8]. As observed in [19], these first occurrence problems were partly motivated by topology, as they were closely related to the description of stable decomposition of topological spaces.

Let ρ be an irreducible GL_{n}-module corresponding to the column regular partition α. Let d(α) be the smallest integer such that there exists an element f_{ρ} of P_{d(α)} and ρ is isomorphic to a quotient of ⟨f_{ρ}⟩, the GL_{n}-module generated of f_{ρ}. (For p = 2, it follows from Theorem A that d(α) = ∑_{j≥1} α′_{j}2^{j−1}. However, the proof of Theorem A given in [1] does not provide any information on f_{ρ}.) The purpose of this paper is to investigate d(α) and f_{ρ} in a few cases, and, in some sense, to generalize Theorem A to the case of any prime p. We shall consider, unless otherwise specified, column regular partitions α = (α_{1}, …, α_{n}) satisfying the following condition:

(∗) every α_{i} is a multiple of p − 1, 1 ≤ i ≤ n − 1, and α_{n} = 0.

(So, for p = 2, as p − 1 = 1, every column regular partition α with α_{n} = 0 satifies this condition.) With such an α, we shall construct an f_{ρ} and an GL_{n}-epimorphism ζ : ⟨f_{ρ}⟩ → ρ. Our main result is stated as follows. For convenience, we shall use the following notation. Given 0 ≤ k ≤ n, define X_{0} = 1, X_{k} = (x_{1} … x_{k})^{p−1}, α̃_{k} =

#{j|α_{j} ≥ k(p − 1)}, Q_{0} = 1 and Q_{k} = L_{k}^{p−1}. Set X_{(α)} = Xp^{n−2} ; α̃_{n−1}Xp^{n−3} ; α̃_{n−2} … Xp ; α̃_{2}Xα̃_{1}. It

follows that X_{(α)} is of degree

d{α} = ∑_{j≥1} α̃_{j}(p − 1)p^{j−1}.

The main result of this paper is the following theorem, of which the proof for p = 2 was (first) given by Walker ([17, 18]).

Theorem B. For every column regular partition α satisfying (∗), there exists a Steenrod operation S_{α} such that

S_{α}(X_{(α)}) = Qp^{n−2} ; α̃_{n−1}Qp^{n−3} ; α̃_{n−2} … Qp ; α̃_{2}Qα̃_{1}.

Hence, as GL_{n}-modules, T_{α} is isomorphic to a quotient of ⟨X_{(α)}⟩.

Note that, for α satisfying (∗), d{α} is nothing but the weight w(α) of α (see e.g. [2]) for the definition of w(α)). So, by [2, Prop. 2.13], d(α) ≥ d{α}. Combining with Theorem B, we have then

Corollary. If α satisfies (∗), then d(α) = d{α}; in other words, the irreducible GL_{n}- module corresponding to α occurs for the first time as a composition factor in P_{d} when d = d{α}. □

Especially, for p = 2, this gives an alternative proof of Theorem A.

2. Proof of Theorem B

We first recall some facts concerning the Milnor basis of the Steenrod algebra A(p). For details, the reader can refer to [11,12,14].

Consider P as the polynomial subring of the mod-p cohomology ring of an elementary abelian p-group of rank n. There is then an action of A(p) on P. Recall that A(p), as a vector space, has a basis consisting of the admissible monomials in Sq^{i} (resp. P^{i} and β) for p = 2 (resp. p > 2). Let A(p)^{∗} be the dual of A(p), and let ξ_{k} (resp. τ_{s} and ξ_{k}) be the duals of Sq^{2^{k−1}} … Sq^{2}Sq^{1} (resp. βP^{p^{s−1}} … P^{p}P^{1} and P^{p^{k−1}} … P^{p}P^{1}), with repect to that basis. In [7], Milnor showed that A(p)^{∗} is a commutative Hopf algebra, and, as an algebra

A(p)^{∗} = \{ \begin{matrix}P (ξ_{1}, ξ_{2}, …) & for p = 2, \\ E(τ_{0}, τ_{1}, …) ⊗ P (ξ_{1}, ξ_{2}, …) & for p > 2. \end{matrix}

It follows that A(p)^{∗} has the basis consisting of the monomials

\{ \begin{matrix}ξ^{R} & = ξ_{1}^{r_{1}} … ξ_{n}^{r_{k}} for p = 2, \\ τ_{S}ξ^{R} & = τ_{s_{1}} … τ_{s_{ℓ}}ξ_{1}^{r_{1}} … ξ_{k}^{r_{k}} for p > 2, \end{matrix}

with S : s_{1} < ⋯ < s_{ℓ}, R = (r_{1}, …, r_{k}, 0, …). For p = 2 (resp. p > 2), denote by St(R) (resp. St(S, R), or simply St(R) if S = ∅) the dual of ξ^{R} (resp. τ_{S}ξ^{R}) with respect to this basis of A(p)^{∗}; the new basis of A(p) consisting of these duals is called the Milnor basis. In [11,12], Mui showed that elements of this basis are nothing but the cohomology operations derived from modular invariants.

We are now interested in the St(R)’s. According to [11,12], we have

4 PHAM ANH MINH

Lemma 1 (Mui). For x, y ∈ P, we have the Cartan formula

St(R)(xy) = ∑_{J+K=I}St(J)(x) · St(K)(y),

with J = (j_{1}, …, j_{m}), K = (k_{1}, …, k_{m}) and J + K = (j_{1} + k_{1}, …, j_{m} + k_{m}). □

The following notation will also be used. For i ≥ 0, set

St^{i} = \{ \begin{matrix}Sq^{i} & for p = 2, \\ P^{i} & for p > 2. \end{matrix}

Given a sequence of non-negative integers I = (i_{1}, i_{2}, …, i_{k}), write St^{i_{1}} … St^{i_{k}} = St^{I}; for s ≥ 0, set sI = (si_{1}, …, si_{k}). For k ≥ 1, denote by (p − 1)_{k} the sequence (\overbrace{p − 1, …, p − 1}^{k−1 times}) (resp. (0)) if k > 1 (resp. k = 1). If X ∈ A(p) having X = ∑_{I}St^{I} as the decomposition of X in terms of admissible monomials, set

X^{[s]} = ∑_{I} St^{p^{s}I},

for s ≥ 0.

Here are some further properties of the St(I)’s. For 1 ≤ i ≤ k, define

L_{k,i} = | \begin{matrix}x_{1} & … & x_{k} \\ x_{1}^{p} & … & x_{k}^{p} \\ ⋯ & … & ⋯ \\ x_{1}^{p^{i−2}} & … & x_{k}^{p^{i−2}} \\ x_{1}^{p^{i}} & … & x_{k}^{p^{i}} \\ ⋯ & … & ⋯ \\ x_{1}^{p^{k}} & … & x_{k}^{p^{k}} \end{matrix} | ,

(so L_{k,1} = L_{k}^{p}) and

L_{k,(i)} = | \begin{matrix}x_{1}^{p} & … & x_{i−1}^{p} & x_{i+1}^{p} & … & x_{k}^{p} \\ x_{1}^{p^{2}} & … & x_{i−1}^{p^{2}} & x_{i+1}^{p^{2}} & … & x_{k}^{p^{2}} \\ ⋯ & … & ⋯ & ⋯ & … & ⋯ \\ x_{1}^{p^{k−1}} & … & x_{i−1}^{p^{k−1}} & x_{i+1}^{p^{k−1}} & … & x_{k}^{p^{k−1}} \end{matrix} | .

Note that, for x ∈ P_{1}, the action of St^{i} on x^{k} is given by

St^{i}(x^{k}) = \binom{k}{i}x^{k+i(p−1)}, i ≤ k.

The following is then obvious by Cartan formula.

Lemma 2. For k ≥ 1,

St^{i}(L_{k}) = \{ \begin{matrix}L_{k} & for i = 0, \\ L_{k,j} & for i = p^{j−1} + ⋯ + p^{k−1}, \\ 0 & otherwise, \end{matrix}
□

Lemma 3. If Sq(I)(x^{p}) ≠ 0, with I = (i_{1}, …, i_{k}) and x ∈ P, there exists J such that I = pJ; furthermore, St(I)(x^{p}) = [St(J)(x)]^{p};

Proof. By Cartan formula, we have St(I)(x^{p}) = ∑_{I_{1}+···+I_{p}=I}St(I_{1})(x) … St(I_{p})(x). Set

B = {(I_{1}, …, I_{p})|I_{1} + ⋯ + I_{p} = I}. Two elements (I_{1}, …, I_{p}), (J_{1}, …, J_{p}) of B are said to be equivalent if there exists a permutation σ on p letters such that I_{σ(s)} = J_{s}, for all 1 ≤ s ≤ p. Let C be an equivalence class and let I be a representative of C.

Write I = (\overbrace{I_{1}, …, I_{1}}^{m_{1} times}, …, \overbrace{I_{j}, …, I_{j}}^{m_{j} times}) with m_{1} + ⋯ + m_{j} = p and I_{s} ≠ I_{t} for s ≠ t.

This implies that C consists of \binom{p}{m_{1}}\binom{p−m_{1}}{m_{2}} … \binom{p−m_{1}−···−m_{j−1}}{m_{j}} elements. If I is not a multiple of p, it follows that j > 1, so the number of elements of C is a multiple of p; therefore ∑_{(J_{1},...,J_{p})∈C}St(J_{1})(x) … St(J_{p})(x) = 0, which implies that St(I)(x^{p}) = 0. So

I should be a multiple of p, says I = pJ. In this case, the unique equivalence class C for which ∑_{(J_{1},...,J_{p})∈C}St(J_{1})(x) … St(J_{p})(x) ≠ 0 only consists of the element (J, …, J).

Hence St(I)(x^{p}) = [St(J)(x)]^{p}. □

The proof of the above lemma also gives the following

Lemma 4. If St^{i}(x^{p}) ≠ 0 for x ∈ P, then i is a multiple of p, says i = pj, and St^{i}(x^{p}) = [St^{j}(x)]^{p}. □

The first part of the following lemma was due to Monks ([10] for p = 2) and Carlisle-Walker-Wood ([3] for p > 2).

Lemma 5. For k ≥ 1 and for every x, y ∈ P, we have

(i) St((p − 1)_{k}) = St^{p^{k−1}−1} … St^{p^{2}−1}St^{p−1}; (ii) St((p − 1)_{k})(x^{p} · y) = x^{p} · St((p − 1)_{k})(y); (iii) St((p − 1)_{k})^{[s]}(x · L_{m}^{i}) = St((p − 1)_{k})^{[s]}(x) · L_{m}^{i} if m ≥ k + s, i ≥ 0; (iv) St((p − 1)_{k})(X_{k}) = Q_{k}.

Proof. (ii) follows directly from Cartan formula and Lemma 3. We now prove (iii). By (i), St((p − 1)_{k})^{[s]} = St^{p^{k+s−1}−p^{s}} … St^{p^{s+2}−p^{s}}St^{p^{s+1}−p^{s}}. As p^{s+1} − p^{s} < ⋯ < p^{k+s−1} − p^{s} < p^{m−1}, it follows from Lemma 2 and from Cartan formula that St((p − 1)_{k})^{[s]}(x · L_{m}^{i}) = St((p − 1)_{k})^{[s]}(x) · L_{m}^{i}. (iv) is proved by induction on k. It is clear that (iv) holds for k = 1. Assume that it holds for k = i. Set

Y = | \begin{matrix}x_{1}^{p} & … & x_{i+1}^{p} \\ x_{1}^{p+p−1} & … & x_{i+1}^{p+p−1} \\ ⋯ & … & ⋯ \\ x_{1}^{p^{i}+p−1} & … & x_{i+1}^{p^{i}+p−1} \end{matrix} | .

It follows that Y = X_{i+1} · L_{i+1}. So (iii) implies that

St((p − 1)_{i+1})(Y) = St((p − 1)_{i+1})(X_{i+1} · L_{i+1})
= St((p − 1)_{i+1})(X_{i+1}) · L_{i+1}.  (∗∗)

By Laplace development, we also have

Y = ∑_{j=1}^{i+1}(−1)^{j+1}x_{j}^{p} · \frac{X_{i+1}}{x_{j}^{p−1}} · L_{i+1,(j)}.

Since

St((p − 1)_{i+1})(x_{i+1}^{p}X_{i}L_{i}^{p}) = x_{i+1}^{p}St((p − 1)_{i+1})(X_{i})L_{i}^{p}  by (ii)
= x_{i+1}^{p}St^{p^{i}−1}St((p − 1)_{i})(X_{i})L_{i}^{p} by (i)
= x_{i+1}^{p}St^{p^{i}−1}(Q_{i})L_{i}^{p} by the inductive hypothesis
= x_{i+1}^{p}Q_{i}^{p}L_{i}^{p}
= x_{i+1}^{p}L_{i}^{p^{2}},

we have

\begin{aligned}St((p − 1)_{i+1})(Y) & = ∑_{j=1}^{i+1}(−1)^{j+1}x_{j}^{p}L_{i+1,(j)}^{p} \\ & = L_{i+1}^{p}. \end{aligned}

Combining with (∗∗), we have St((p − 1)_{i+1})(X_{i+1}) = Q_{i+1}. The lemma follows. □

Lemma 6. For 0 ≤ i < k ≤ k_{1} < ⋯ < k_{r} and x ∈ P,

St^{p^{k}−p^{i}}(x^{p} · Q_{k_{1}}Q_{k_{2}} … Q_{k_{r}}) = \{ \begin{matrix}x^{p} · Q_{k_{1}}^{p}Q_{k_{2}} … Q_{k_{r}} & if i = 0 and k = k_{1}, \\ St^{p^{k}−p^{i}}(x^{p}) · Q_{k_{1}}Q_{k_{2}} … Q_{k_{r}} & if i > 0 and k < k_{1}. \end{matrix}

Especially, for every regular column partition α = (α_{1}, …, α_{n}) satisfying (∗), there exists a suitable I such that

St^{I}(L^{{α}}) = Qp^{n−2} ; α̃_{n−1}Qp^{n−3} ; α̃_{n−2} … Qp ; α̃_{2}Qα̃_{1}.

Proof. Set Q = Q_{k_{1}} … Q_{k_{r}}. Note that

St^{p^{k}−p^{i}}(x^{p} · Q) = ∑_{0≤j≤p^{k}−p^{i}} St^{p^{k}−p^{i}−j}(x^{p}) · St^{j}(Q).

By Lemma 4, St^{p^{m}−1−j}(x^{p}) ≠ 0 implies that j +1 is a multiple of p. By Cartan formula and by Lemma 1, for such a j, St^{j}(Q) ≠ 0 also implies that j = p^{k_{1}} − 1, so i = 0, k = k_{1} and St^{p^{m}−1}(x^{p} · Q) = x^{p} · Q_{k_{1}}^{p}Q_{k_{2}} … Q_{k_{r}}. If i > 0 and k < k_{1}, then j ≤ p^{k} − p^{i} implies j < p^{k} ≤ p^{k_{1}−1}, so St^{j}(Q) = 0; hence St^{p^{k}−p^{i}}(x^{p} · Q) = St^{p^{k}−p^{i}}(x^{p}) · Q.

Let ℓ be the largest integer such that α̃_{ℓ} ≠ 0 and set m_{i} = α̃_{ℓ+1−i}. It follows that m_{1} ≤ m_{2} ≤ ⋯ ≤ m_{ℓ−1} ≤ m_{ℓ} and L^{{α}} = Q_{m_{1}}Q_{m_{2}} … Q_{m_{ℓ−1}}Q_{m_{ℓ}}. For convenience, given a sequence (i_{1}, …, i_{ℓ}), write Q^{i_{1},...,i_{ℓ}} = Q_{m_{1}}^{p^{i_{1}}} … Q_{m_{ℓ}}^{p^{i_{ℓ}}}. So L^{{α}} = Q^{0,...,0}. Our task is now to prove that there exists I such that St^{I}(L^{{α}}) = Q^{ℓ−1,ℓ−2,...,1,0}. In so doing, it is sufficient to prove the following assertions:

(a) there exists t such that St^{t}(Q^{r,...,r,r−1,...,1,0}) = Q^{r+1,r...,r,r−1,...,1,0};

(b) there exists q such that

St^{q}(Q^{r+1,...,r+1,\overbrace{r, …, r}^{s times},r−1,...,1,0}) = Q^{r+1,...,r+1,\overbrace{r, …, r}^{s−1 times},r−1,...,1,0}.

Set Y = Q^{r,...,r,r−1,...,1,0}. (a) follows from

St^{p^{r}(p^{m_{1}} −1)}(Y) = St^{p^{r}(p^{m_{1}} −1)}(Q_{m_{1}}^{p^{r}} … Q_{m_{j−r}}^{p^{r}}Q_{m_{j−r+1}}^{p^{r−1}} … Q_{m_{j−2}}^{p^{2}}Q_{m_{j−1}}^{p})Q_{m_{j}}
= [St^{p^{r−1}(p^{m_{1}} −1)}(Q_{m_{1}}^{p^{r−1}} … Q_{m_{j−r}}^{p^{r−1}}Q_{m_{j−r+1}}^{p^{r−2}} … Q_{m_{j−2}}^{p}Q_{m_{j−1}})]^{p}Q_{m_{j}}

by Lemma 4

= [St^{p^{r−1}(p^{m_{1}} −1)}(Q_{m_{1}}^{p^{r−1}} … Q_{m_{j−r}}^{p^{r−1}}Q_{m_{j−r+1}}^{p^{r−2}} … Q_{m_{j−2}}^{p})]^{p}Q_{m_{j−1}}^{p}Q_{m_{j}}

= [St^{p^{r−2}(p^{m_{1}} −1)}(Q_{m_{1}}^{p^{r−2}} … Q_{m_{j−r}}^{p^{r−2}}Q_{m_{j−r+1}}^{p^{r−3}} … Q_{m_{j−2}})]^{p^{2}}Q_{m_{j−1}}^{p}Q_{m_{j}}

= …
= [St^{(p^{m_{1}} −1)}(Q_{m_{1}})]^{p^{r}} … Q_{m_{j−r}}^{p^{r}}Q_{m_{j−r+1}}^{p^{r−1}} … Q_{m_{j−2}}^{p^{2}}Q_{m_{j−1}}^{p}Q_{m_{j}}
= Q^{r+1,r,...,r,r−1,...,0}.

We now prove (b). By setting Z = Q^{r+1,...,r+1,\overbrace{r, …, r}^{s times},r−1,...,1,0} and j = ℓ − r − s + 1,

we have

Z = Q_{m_{1}}^{p^{r+1}} … Q_{m_{j−1}}^{p^{r+1}}Q_{m_{j}}^{p^{r}} … Q_{m_{j+s−1}}^{p^{r}}Q_{m_{j+s}}^{p^{r−1}} … Q_{m_{ℓ−1}}^{p}Q_{m_{ℓ}}.

From the first part of this lemma and from Lemma 4, applying St^{p^{r}(p^{m_{j}} −1)} yields

St^{p^{r}(p^{m_{j}} −1)}(Z) = [St^{p^{m_{j}} −1}(Q_{m_{1}}^{p} … Q_{m_{j−1}}^{p}Q_{m_{j}} … Q_{m_{j+s−1}})]^{p^{r}}Q_{m_{j+s}}^{p^{r−1}} … Q_{m_{ℓ−1}}^{p}Q_{m_{ℓ}}

= [Q_{m_{1}}^{p} … Q_{m_{j−1}}^{p}Q_{m_{j}}^{p} … Q_{m_{j+s−1}}]^{p^{r}}Q_{m_{j+s}}^{p^{r−1}} … Q_{m_{ℓ−1}}^{p}Q_{m_{ℓ}}

= Q^{r+1,...,r+1,\overbrace{r, …, r}^{s−1 times},r−1,...,0}.

The lemma follows. □

Proof of Theorem B. Let k be the largest integer such that α̃_{k} ≠ 0. Therefore

X_{(α)} = Xp^{k−1} ; α̃_{k}Xp^{k−2} ; α̃_{k−1} … Xp ; α̃_{2}Xα̃_{1}. Set τ_{i} = St((p − 1)α̃_{i}), 1 ≤ i ≤ k, and S_{α} =

τ_{k}^{[k−1]}τ_{k−1}^{[k−2]} … τ_{2}^{[1]}τ_{1}. By Lemma 5, τ_{i}(Xα̃_{i}) = Qα̃_{i}. So, by Lemmas 3 and 5, we have

S_{α}(X_{(α)}) = τ_{k}^{[k−1]}τ_{k−1}^{[k−2]} … τ_{2}^{[1]}τ_{1}(X_{(α)})

= τ_{k}^{[k−1]}τ_{k−1}^{[k−2]} … τ_{2}^{[1]}(Xp^{k−1} ; α̃_{k}Xp^{k−2} ; α̃_{k−1} … Xp ; α̃_{2})Qα̃_{1}

= [τ_{k}^{[k−2]}τ_{k−1}^{[k−3]} … τ_{2}(Xp^{k−2} ; α̃_{k}Xp^{k−3} ; α̃_{k−1} … Xα̃_{2})]^{p}Qα̃_{1}

= [τ_{k}^{[k−2]}τ_{k−1}^{[k−3]} … τ_{3}^{[1]}(Xp^{k−2} ; α̃_{k}Xp^{k−3} ; α̃_{k−1} … Xp ; α̃_{3})Qα̃_{2}]^{p}Qα̃_{1}

= …

= Qp^{k−1} ; α̃_{k}Qp^{k−2} ; α̃_{k−1} … Qp ; α̃_{2}Qα̃_{1}.

Set X = Qp^{k−1} ; α̃_{k}Qp^{k−2} ; α̃_{k−1} … Qp ; α̃_{2}Qα̃_{1}. Note that the Steenrod operation commutes with the action of GL_{n}. Hence, if x, y ∈ P and St^{i}(x) = y, St^{i} can be viewed as a GL_{n}- homomorphism from ⟨x⟩ to ⟨y⟩. So, as GL_{n}-modules, ⟨X ⟩ is isomorphic to a quotient of ⟨X_{(α)}⟩. Besides, Lemma 6 implies that ⟨X ⟩ is isomorphic, as GL_{n}-modules, to a quotient of the irreducible module T_{α}, hence to T_{α}. The theorem follows.

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Department of Mathematics, College of Sciences, University of Hue, Hue, Vietnam E-mail address: paminh@bdvn.vnd.net