DSpace at VNU: The smallest subgroup whose invariants are hit by the Steenrod algebra

DSpace at VNU: The smallest subgroup whose invariants are hit by the Steenrod algebra tài liệu, giáo án, bài giảng , luậ...

Mathematical Proceedings of the Cambridge

Philosophical Society

http://journals.cambridge.org/PSP

Additional services for Mathematical Proceedings of the

Cambridge Philosophical Society:

Email alerts: Click here

Subscriptions: Click here

Commercial reprints: Click here

Terms of use : Click here

The smallest subgroup whose invariants are hit by the Steenrod algebra

NGUYỄN H V HƯNG and TRAN DINH LUONG

Mathematical Proceedings of the Cambridge Philosophical Society / Volume 142 / Issue 01 / January 2007, pp

63 - 71

DOI: 10.1017/S0305004106009637, Published online: 12 February 2007

Link to this article: http://journals.cambridge.org/abstract_S0305004106009637

How to cite this article:

NGUYỄN H V HƯNG and TRAN DINH LUONG (2007) The smallest subgroup whose invariants are hit by the Steenrod algebra Mathematical Proceedings of the Cambridge Philosophical Society, 142, pp 63-71 doi:10.1017/S0305004106009637

Request Permissions : Click here

Downloaded from http://journals.cambridge.org/PSP, IP address: 138.37.21.81 on 16 Mar 2015

doi:10.1017/S0305004106009637 Printed in the United Kingdom

The smallest subgroup whose invariants are hit

by the Steenrod algebra

BYNGUY ˆE˜ N H V HU.NG

Department of Mathematics, Vietnam National University, Hanoi 334 Nguyˆe˜n Tr˜ai Street, Hanoi, Vietnam.

e-mail: nhvhung@vnu.edu.vn

Department of Mathematics, Quynhon University,

170 An Duong Vuong Street, Quynhon, Vietnam.

e-mail: tdinhluong@yahoo.com

Abstract

Let V be a k-dimensionalF2-vector space and let W be an n-dimensional vector subspace

of V Denote by G L (n, F2) • 1 k −n the subgroup of G L (V ) consisting of all isomorphisms

ϕ : V → V with ϕ(W) = W and ϕ(v) ≡ v (mod W) for every v ∈ V We show that

G L (3, F2) • 1 k−3 is, in some sense, the smallest subgroup of G L (V )
GL(k, F2), whose

invariants are hit by the Steenrod algebra acting on the polynomial algebra, H∗(BV ; F2)

F2[x1, , x k] The result is some aspect of an algebraic version of the classical conjecture

that the only spherical classes in Q0S0are the elements of Hopf invariant one and those of Kervaire invariant one.

1 Introduction and statement of results Let P kF2[x1, , x k] be the polynomial algebra over the field of two elements, F2, in

k variables x1, , x k, each of degree 1 It is equipped with the usual structure of (right)

module over G L k GL(k, F2) by means of substitution of variables as follows For every

ω = (ω i j ) k ×k ∈ GL k and any f ∈ P k, one defines( f ω)(x1, , x k ) = f (x1ω, , x k ω),

where

x j ω =

k

i=1

ω i j x i (1
j
k).

Sq j (x i ) =

⎧

⎨

⎩

x i , j = 0,

x2

i , j = 1,

† Partially supported by the National Research Program, grant no 140804.

64 NGUYEˆ ˜NH V HU.NG ANDTRAN DINHLUONG

and subject to the Cartan formula

Sq n ( f g) =

n

j=0

Sq j ( f )Sq n − j (g),

for f , g ∈ F2[x1, , x k]

P G

k the subalgebra of all G-invariants in P k Since the action of G L k and that ofA on P k

In [2], the first author was interested in the homomorphism

j G: F2⊗

A



P k G

A P k ) G

and showed that it is equivalent to a weak algebraic version of the long-standing conjecture

stating that the only spherical classes in Q0S0 are the elements of Hopf invariant one and those of Kervaire invariant one.

CONJECTURE1·1 ([2]) j G L k = 0 in positive degrees for k > 2.

This conjecture has been proved in [3] (See Corollary 1·3 below.) That the conjecture is

no longer valid for k = 1 and k = 2 is respectively shown in [2] to be an exposition of the

existence of the Hopf invariant one and the Kervaire invariant one classes

In [4], the following problem has been investigated: which subgroup G of G L k possesses

j G = 0 in positive degrees? It should be noted that, as observed in the introduction of [2],

⊆ A+· P k ,

k andA consisting of all

elements of positive degree Therefore, the smaller the group G is the harder the problem

turns out to be

The problem we are discussing is closely related to the hit problem of determining

F2⊗

and S Priddy [7], who show its relationships to several classical problems in cobordism

theory, modular representation theory, the Adams spectral sequence for the stable homotopy

of spheres, and the stable homotopy type of classifying spaces of finite groups The tensor productF2⊗

an explicit description ofF2⊗

A P k for general k will appear in the near future There is also

another approach, the qualitative one, to the problem By this we mean giving conditions on

elements of P k to show that they go to zero inF2⊗

A P k, i.e belong toA+· P k Peterson’s

conjecture [6], which has been established by Wood [10], claims thatF2⊗

A P k = 0 in certain

which are hit byA in P

Let G i be a subgroup of G L k i for 1
i
m with k1+ · · · + k m = k We consider the following subgroup of G L k:

G1• G2• · · · • G m

⎧

⎪

⎨

⎪

⎪

⎛

⎜

⎜

⎝

A2

⎞

⎟

⎟

⎠











⎫

⎪

⎬

⎪

k

In particular, G L k1, ,k m = GL k1• · · · • GL k m is a parabolic subgroup of G L k

The main result of [4] is the following.

THEOREM1·2 ([4]) Let 1 k −n be the unit subgroup of G L k −n and G = GL n• 1k −n for

j G: F2⊗

A



P k G

−→



F2⊗

A P k

G

is zero in positive degrees if and only if n  3.

COROLLARY1·3 ([3, 4]) Let G i be a subgroup of G L k i for 2
i
m with k1+ · · · +

k m = k Then, for G = GL k1• G2· · · • G m ,

j G: F2⊗

A



P k G

−→



F2⊗

A P k

G

is zero in positive degrees if and only if k1  3 In particular, j G L k1, ,km = 0 in positive

degrees if and only if k1 3, and j G L k = 0 in positive degrees if and only if k  3.

In this paper, we are interested in the following problem: what is the smallest subgroup G

of G L k with j G = 0 in positive degrees, or equivalently with (P G

k )+ ⊆ A+· P k ? It should

bigger family(P G

k )+of elements, which are hit by the Steenrod algebra.

The following is the main result of the present paper

THEOREM1·4 Let G1 be a proper subgroup of G L3 and G i a subgroup of G L k i for

2
i
m with 3 + k2+ · · · + k m = k Then

j G: F2⊗

A (P G

k ) −→



F2⊗

A P k

G

is non trivial in positive degrees for G = G1• G2· · · • G m

smallest subgroup of G L k with j G = 0 in positive degrees It should be noted that GL(k, F2)

is a simple group if and only if k  3 (see e.g Suzuki [9, page 80]) So, GL3is the smallest group in this simple group family

In order to prove Theorem 1·4, it suffices to show that j G  0 in positive degrees for

G = G1• G2· · · • G m , where G1 is any maximal subgroup of G L3 On the other hand,

it is easily seen that P ω−1G ω

k ω, for ω ∈ GL k As the actions of G L k andA on P k

commute with each other,(P ω−1G ω

kif and only if(P G

k )+is So, it is

sufficient to prove the theorem for G1being one of representatives of the conjugate classes

of maximal subgroups in G L3 Therefore, we need the following classification of all the

66 NGUYEˆ ˜NH V HU.NG ANDTRAN DINHLUONG

PROPOSITION1·5 (See e.g [1]) There are exactly three conjugate classes of maximal

subgroups of G L3, whose representatives are the following three subgroups:

(i) M0

a=

⎛

⎝01 00 10

⎞

⎠ , b =

⎛

⎝01 11 00

⎞

⎠;

(ii) L1 =



 

 A ∈ GL2



;

(iii) L2 =



 

 A ∈ GL2



Here M0is of order 21, while L1and L2are of order 24.

groups [1, p 3], we give a proof of this proposition in the appendix.

We start this section with an obvious observation, whose proof is omitted

P k ω−1G ω = P G

k ω.

So, (P ω−1G ω

k )+is contained in A+P

k if and only if (P G

k )+is.

We will see that Theorem 1·4 can be reduced to the following special case

LEMMA2·2 If G is a proper subgroup of GL3, then

j G: F2⊗

A



P3G

−→



F2⊗

A P3

G

is non trivial in positive degrees, or equivalently



P G

3

+

3 Proof Since P G

3 ⊆ P H

of G L3 Further, by Lemma 2·1 and Proposition 1·5, we need only to investigate G to be one of the three groups M0, L1, L2in Proposition 1·5, which are representatives of the three

conjugate classes of all maximal subgroups in G L3

Case 1: G = L1 The Dickson invariant Q2,1 x2

1+ x1x2+ x2

2 is an L1-invariant, however

Q2,1 ^ A+P

3 Indeed, it is easy to see that x2

1, x2

2 ∈ A+P

3, and x1x2^ A+P

3

Case 2: G = L2 The Dickson invariant Q1,0 x1is an L2-invariant, however x1 ^ A+P

3

variables x1, x2, which are invariant under the action of b : x1 → x2, x2→ x1+ x2

b-invariant.

Every degree two polynomial in two variables x1, x2is of the form r x12+ sx1x2+ tx2

2with

r , s, t ∈ F2 We have



r x12+ sx1x2+ tx2

2



b = rx2

2+ sx2(x1+ x2) + t(x1+ x2)2

= tx2

1 + sx1x2+ (r + s + t)x2

2.

r = s = t ∈ F2 Thus, the only degree two b-invariant polynomial in two variables x1, x2is

Q2,1 x2

1 + x1x2+ x2

2

Similarly, there are exactly 3 degree three polynomials in two variables x1, x2, which are

b-invariant Namely

P = x3

1 + x2

1x2+ x3

2,

Q = x3

1 + x1x22+ x3

2,

Q2,0 = P + Q = x2

1x2+ x1x22.

Note that(x3)b = x3 So, by the above arguments, any degree three b-invariant polyno-mial in three variables x1, x2, x3is of the form

f = r P + s Q + t Q2,1 x3+ (0x1+ 0x2)x2

3+ vx3

3,

where r , s, t, v ∈ F2 This polynomial is M0-invariant if and only if it is invariant under

the action of a: x1 → x2, x2 → x3, x3 → x1+ x3 A routine computation shows that the equation( f )a = f is equivalent to r = 0, s = t = v So

f =x13+ x1x22+ x3

2

 +x12+ x1x2+ x2

2



x3+ x3

3.

is the only degree three M0-invariant polynomial in three variables x1, x2, x3

3 Let(x2, x3) be the ideal generated by x2, x3in P3 This is

anA-ideal Suppose to the contrary that f ∈ A+P

3 Then[ f ] = x3

1 ∈ P1= P3/(x2, x3) also

1 The conclusion is impossible, because Sq2(x1) = 0 and Sq1(x2

1) = 0.

This contradiction shows that f ^ A+P

3 The lemma is proved

We are now ready to prove the main theorem of the paper

Proof of Theorem 1·4 Recall that, as shown in the introduction, the theorem is equivalent

to(P G

k )+ A+P k.

From the assumption G = G1• G2· · ·• G m , we have obviously P G1

3 ⊆ P G

k So, it suffices

to show that



P G1

3

+

k

3 )+ ⊆ A+P

k Since the ideal (x4, x5, , x k ) is an A-ideal and P k /(x4, , x k )
P3, it implies that (P G1

3 )+ ⊆ A+P

3 This contradicts

The theorem is completely proved

3 Appendix: Maximal subgroups of G L (3, F2)

In order to make the paper self-contained, we give here a direct proof for Proposition 1·5

Our proof will basically be based on the fact that G L3is a simple group of order 168= 8·3·7

(see e.g Suzuki [9, page 80]).

68 NGUYEˆ ˜NH V HU.NG ANDTRAN DINHLUONG

subgroups of G L3 The following three lemmas investigate the Sylow p-subgroups of G L3 with p = 2, 3 and 7 respectively.

LEMMA3·1 If H is a Sylow 7-subgroup of GL3, then its normalizer N G L3(H) is a

max-imal subgroup of order 21 in G L3 Further, if M is a proper subgroup of G L3and M ⊇ H,

Proof Denote σ7 = [GL3 : N G L3(H)] As σ7 is a divisor of [GL3 : H] = 24 and

and thus H  GL3 This contradicts the fact that G L3is a simple group Thereforeσ7 = 8,

and N G L3(H) is of order 21 = 168/8.

η7 = 8 Suppose to the contrary that η7 = 8 Then from the fact [GL3 : N G L3(H)] = [M :

N M (H)] = 8 it implies that all Sylow 7-subgroups of GL3 are contained in M Thus, the subgroup generated by all Sylow 7-subgroups of G L3is a proper normal subgroup of G L3

N G L3(H) Combining this with the facts that M  H and that NG L3(H) is of order 21, we

get M = N G L3(H) As a consequence, NG L3(H) is a maximal subgroup of GL3

The lemma is proved

LEMMA3·2 If S is a Sylow 3-subgroup of GL3, then N G L3(S) is a non-abelian subgroup

of order 6 Further, if K is a subgroup of order 6 in G L3with K ⊇ S, then K = N G L3(S) =

N G L3(K ).

Proof Obviously, there is a subgroup M ⊇ S of order 21 of GL3 If N M (S) = M,

we get either N G L3(S) = GL3 or N G L3(S) = M The first equality is impossible, since

N G L3(S)] = 168/21 = 8  1 (mod 3) So NM (S)  M, and we have N M (S) = S, since

[M : S] = 7 is a prime number As a consequence, the number of Sylow 3-subgroups of M

3-subgroups is a proper normal subgroup of G L3 This contradicts the fact that G L3 is

σ3≡ 1 (mod 3), we have σ3= 28 Hence, N G L3(S) is of order 168/28 = 6.

If K is a subgroup of order 6 in G L3with K ⊇ S, then S  K , and so K ⊆ N G L3(S) As

shown above, N G L3(S) is also of order 6, hence K = NG L3(S) Since S is the unique Sylow

3-subgroup of K , we get S  N G L3(K ) ⊆ NG L3(S) = K Thus, NG L3(K ) = K

Suppose to the contrary that K is an abelian subgroup of order 6 Then K is cyclic and it

contains exactly two elements of order 6 As the number of subgroups, which are conjugate

to K in G L3, is[GL3: N G L3(K )] = [GL 3: K ] = 28, there are at least 28·2 = 56 elements

of order 6 in G L3 On the other hand, there are exactly 6· 8 = 48 elements of order 7 and

2· 28 = 56 elements of order 3 in GL3 The equality 168− (56 + 48 + 56) = 8 shows that there is exactly one Sylow 2-subgroup of order 8 in G L3 This contradicts the fact that G L3

is simple Therefore, K is non-abelian.

The lemma is proved.

Let T be the subgroup of G L3consisting of all upper triangular matrices with 1 on the

group of 8 elements

LEMMA3·3 If U is a Sylow 2-subgroup of GL3, then N G L3(U) = U.

Proof Let Z (U) denote the center of U From U
D8it implies that Z (U) is of order

2 Suppose u ∈ Z(U), u  1 We will show that C G L3(u) = U.

Obviously, U ⊆ C G L3(u) If CG L3(u) is of an order divisible by 3, then it contains a Sylow

Lemma 3·2 If C G L3(u) is of an order divisible by 7, then it contains a Sylow 7-subgroup.

So, by Lemma 3·1, we conclude that either C G L3(u) is of order 21 or C G L3(u) = GL 3 The

first case is impossible, since C G L3(u) ⊇ U The second case is also impossible, as GL 3 is simple

Therefore, C G L3(u) is of order 8 Hence CG L3(u) = U.

get Z (U)  N G L3(U) (see e.g (6·14) in Suzuki [9, page 51]) So NG L3(U) ⊆ CG L3(u) = U,

and N G L3(U) = U.

The lemma is proved

Proof of Proposition 1 ·5 Let M be a maximal subgroup of GL3 If M is of an order

M = N G L3(H) is a maximal subgroup of order 21 in GL 3 On the other hand, a direct

hence M0is of order 21 As the Sylow 7-subgroups of G L3are conjugate in G L3, M is also conjugate to M0in G L3

In the remaining case, when the order of M is not divisible by 7, this order is a divisor

of 24

If the order of M is a divisor of 8, then M is conjugate to a subgroup of T , the subgroup

of all upper triangular matrices in G L3 So, M is not maximal because T ⊆ L1and T  L1

If M is of order 3, then it is not maximal by Lemma 3·2

Lemma 3·2, N G L3(H) is of order 6 If NG L3(H) ⊆ M, then it is an index 2 subgroup of M,

thus N G L3(H) M This contradicts Lemma 3·2 asserting that N G L3(NG L3(H)) = NG L3(H).

3-subgroups, and 8 elements of order 3 As a consequence, M contains exactly one Sylow

B

Further, combining A  L and the fact that GL3is simple, we have L  GL3 So, M is not

a maximal subgroup of G L3 This is a contradiction

N G L3(H) If NM (H) = H, then M contains exactly 24/3 = 8 Sylow 3-subgroups, and

70 NGUYEˆ ˜NH V HU.NG ANDTRAN DINHLUONG

8 It implies that U  M This contradicts Lemma 3·3 asserting that N G L3(U) = U Therefore

N M (H)  H By Lemma 3·2, N M (H) = N G L3(H) is a non-abelian subgroup of order 6.

So M contains exactly 24 /6 = 4 Sylow 3-subgroups, denoted by H1, H2, H3, H4 Then M

acts on the set{H1, H2, H3, H4} by conjugacy Let ϕ: M → 4be the corresponding group homomorphism We have

Kerϕ =t ∈ M| H t

i = H i for i = 1, 2, 3, 4=

4



i=1

N M (H i ).

As N M (H i ) = N G L3(Hi ) is of order 6, H i is the unique subgroup of order 3 in N M (H i ) for

every i Since the subgroups H1, H2, H3, H4are distinct, Kerϕ is not of order 6 or 3 If Kerϕ

is of order 2, let z be the order 2 element in Ker ϕ, then for any u ∈ M we have

z u ∈

4



i=1

N M (H u

i ) =

4



i=1

N M (H i ) = Kerϕ.

Kerϕ = {1} and we get M
4

N G L3(A)  GL 3 Since M is maximal, N G L3(A) = M.

Because T
D8, there are exactly two elementary abelian subgroups of order 4, denoted

by A1, A2, in T Indeed, as D8has the presentation

a direct computation shows that all the elements of order 2 in D8are

x , y, [x, y], x[x, y], y[x, y].

{1, [x, y]} So, D8has exactly two elementary abelian subgroups of order 4, namely

As A is of order 4, it is contained in a Sylow 2-subgroup of G L3 So A is conjugate to one of the two subgroups A1and A2 Hence M = N G L3(A) is conjugate to either NG L3(A 1)

or N G L3(A 2) So, there are at most 2 conjugate classes of subgroups of order 24 in GL3

conjugate subgroups of order 24 in G L3 Obviously, L1and L2both are subgroups of order

24 We have clearly L1 L2 = T Suppose to the contrary that L1and L2are conjugate in

G L3, that is L u

1 = L2 As T and

T u are Sylow 2-subgroups of L2, there existsv ∈ L2 such that T u v = T , or equivalently

u v ∈ N G L3(T ) On the other hand, NG L3(T ) = T (by Lemma 3·3), so uv ∈ T This implies

that u ∈ L2 and hence L1 = L u−1

2 = L2 This is a contradiction Consequently, L1, L2are

not conjugate in G L3

The proposition is completely proved

REFERENCES

[1] J H CONWAY , R T C URTIS , S P N ORTON , R A P ARKER and R A W ILSON Atlas of Fi-nite Groups Maximal Subgroups and Ordinary Characters for Simple Groups With computational

assistance from J G Thackray (Oxford University Press, 1985), xxxiv + 252 pp.

[2] NGUY E ˆ ˜N H V HU.NG Spherical classes and the algebraic transfer Trans Amer Math Soc 349 (1997), 3893–3910.

[3] NGUY E ˆ ˜NH V HU.NGand TRA`ˆNNGO

C N AM The hit problem for the Dickson algebra Trans Amer.

Math Soc 353 (2001), 5029–5040.

[4] NGUY E ˆ ˜NH V HU.NGand TRA`ˆNNGO

C N AM The hit problem for the modular invariants of linear

groups Jour Algebra 246 (2001), 367–384.

[5] M KAMEKO Products of Projective Spaces as Steenrod Modules Thesis (Johns Hopkins University,

1990).

[6] F P PETERSON Generators of H∗(RP∞∧ RP∞) as a module over the Steenrod algebra Abstracts

Amer Math Soc., No 833 (1987).

[7] S PRIDDY On characterizing summands in the classifying space of a group, I Amer Jour Math 112

(1990), 737–748.

[8] W M SINGER The transfer in homological algebra Math Zeit 202 (1989), 493–523.

[9] M SUZUKI Group Theory (Springer-Verlag, 1982).

[10] R M W WOOD Modular representations of G L (n, F p ) and homotopy theory Lecture Notes in

Math 1172 (Springer-Verlag, 1985), 188–203.