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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 Math Proc Camb Phil Soc (2007), 142, 63 c 2007 Cambridge Philosophical Society 63 doi:10.1017/S0305004106009637 Printed in the United Kingdom The smallest subgroup whose invariants are hit by the Steenrod algebra ˜ˆ N H V HU.NG B Y NGUYE Department of Mathematics, Vietnam National University, Hanoi 334 Nguyˆe˜n Tr˜ai Street, Hanoi, Vietnam e-mail: nhvhung@vnu.edu.vn AND TRAN DINH LUONG† Department of Mathematics, Quynhon University, 170 An Duong Vuong Street, Quynhon, Vietnam e-mail: tdinhluong@yahoo.com (Received 18 May 2005; revised December 2005) Abstract Let V be a k-dimensional F2 -vector space and let W be an n-dimensional vector subspace of V Denote by G L(n, F2 ) • 1k−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 ) • 1k−3 is, in some sense, the smallest subgroup of G L(V ) % G L(k, F2 ), whose invariants are hit by the Steenrod algebra acting on the polynomial algebra, H ∗ (BV ; F2 ) % F2 [x1 , , xk ] The result is some aspect of an algebraic version of the classical conjecture that the only spherical classes in Q S are the elements of Hopf invariant one and those of Kervaire invariant one Introduction and statement of results Let Pk F2 [x1 , , xk ] be the polynomial algebra over the field of two elements, F2 , in k variables x1 , , xk , each of degree It is equipped with the usual structure of (right) module over G L k G L(k, F2 ) by means of substitution of variables as follows For every ω = (ωi j )k×k ∈ G L k and any f ∈ Pk , one defines ( f ω)(x1 , , xk ) = f (x1 ω, , xk ω), where k xjω = ωi j xi (1 j k) i=1 The mod Steenrod algebra, A, acts upon Pk by use of the formula ⎧ ⎨xi , j = 0, Sq j (xi ) = xi2 , j = 1, ⎩ 0, otherwise, † Partially supported by the National Research Program, grant no 140804 N GUY Eˆ˜ N H V H U.NG AND T RAN D INH L UONG 64 and subject to the Cartan formula n Sq ( f g) = Sq j ( f )Sq n− j (g), n j=0 for f, g ∈ F2 [x1 , , xk ] Let G be a subgroup of G L k Then Pk inherits the structure of G-module Denote by PkG the subalgebra of all G-invariants in Pk Since the action of G L k and that of A on Pk commute with each other, PkG is also an A-module In [2], the first author was interested in the homomorphism jG : F2 ⊗ PkG −→ (F2 ⊗ Pk )G A A induced by the identity map on Pk He also set up the following conjecture for G = G L k and showed that it is equivalent to a weak algebraic version of the long-standing conjecture stating that the only spherical classes in Q S are the elements of Hopf invariant one and those of Kervaire invariant one C ONJECTURE 1·1 ([2]) jG L k = in positive degrees for k > This conjecture has been proved in [3] (See Corollary 1·3 below.) That the conjecture is no longer valid for k = and k = 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 jG = in positive degrees? It should be noted that, as observed in the introduction of [2], jG = in positive degrees ⇐⇒ PkG + ⊆ A+ · Pk , where (PkG )+ and A+ denote respectively the submodules of PkG and A 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 ⊗ Pk This problem was first studied by F Peterson [6], R Wood [10], W Singer [8] A 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 product F2 ⊗ Pk has explicitly been computed for k (see [5]) It seems unlikely that A an explicit description of F2 ⊗ Pk for general k will appear in the near future There is also A another approach, the qualitative one, to the problem By this we mean giving conditions on elements of Pk to show that they go to zero in F2 ⊗ Pk , i.e belong to A+ · Pk Peterson’s A conjecture [6], which has been established by Wood [10], claims that F2 ⊗ Pk = in certain A degrees The main result of [4] (see Theorem 1·2 below) gives a huge family of elements, which are hit by A in Pk The smallest subgroup whose invariants are hit Let G i be a subgroup of G L ki for i following subgroup of G L k : ⎧⎛ A1 ⎪ ⎪ ⎪ ⎨⎜ A2 ⎜ G1 • G2 • · · · • Gm ⎜ ⎪ ⎝ ⎪ ⎪ ⎩ 65 m with k1 + · · · + km = k We consider the ∗ ⎫ ⎪ ⎪ ⎪ ⎬ ⎞ ⎟ ⎟ ⎟ Ai ∈ G i with ⎠ i m Am ⎪ ⎪ ⎪ ⎭ ⊆ G Lk In particular, G L k1 , ,km = G L k1 • · · · • G L km is a parabolic subgroup of G L k The main result of [4] is the following n T HEOREM 1·2 ([4]) Let 1k−n be the unit subgroup of G L k−n and G = G L n • 1k−n for k Then G jG : F2 ⊗ PkG −→ F2 ⊗ Pk A is zero in positive degrees if and only if n A C OROLLARY 1·3 ([3, 4]) Let G i be a subgroup of G L ki for km = k Then, for G = G L k1 • G · · · • G m , i m with k1 + · · · + G jG : F2 ⊗ PkG −→ F2 ⊗ Pk A A is zero in positive degrees if and only if k1 In particular, jG L k1 , ,km = in positive degrees if and only if k1 3, and jG L k = in positive degrees if and only if k In this paper, we are interested in the following problem: what is the smallest subgroup G of G L k with jG = in positive degrees, or equivalently with (PkG )+ ⊆ A+ · Pk ? It should be noted that giving a smaller subgroup G with jG = in positive degrees means giving a bigger family (PkG )+ of elements, which are hit by the Steenrod algebra The following is the main result of the present paper T HEOREM 1·4 Let G be a proper subgroup of G L and G i a subgroup of G L ki for i m with + k2 + · · · + km = k Then G jG : F2 ⊗(PkG ) −→ F2 ⊗ Pk A A is non trivial in positive degrees for G = G • G · · · • G m By combining the above two theorems, G L • 1k−3 turns out to be, in some sense, the smallest subgroup of G L k with jG = in positive degrees It should be noted that G L(k, F2 ) is a simple group if and only if k (see e.g Suzuki [9, page 80]) So, G L is the smallest group in this simple group family In order to prove Theorem 1·4, it suffices to show that jG in positive degrees for G = G • G · · · • G m , where G is any maximal subgroup of G L On the other hand, −1 it is easily seen that Pkω Gω = PkG ω, for ω ∈ G L k As the actions of G L k and A on Pk −1 commute with each other, (Pkω Gω )+ is contained in A+ Pk if and only if (PkG )+ is So, it is sufficient to prove the theorem for G being one of representatives of the conjugate classes of maximal subgroups in G L Therefore, we need the following classification of all the maximal subgroups of G L up to conjugacy N GUY Eˆ˜ N H V H U.NG AND T RAN D INH L UONG 66 P ROPOSITION 1·5 (See e.g [1]) There are exactly three conjugate classes of maximal subgroups of G L , whose representatives are the following three subgroups: (i) M0 = a, b , where ⎛ ⎞ ⎛ ⎞ 0 1 a = ⎝1 0⎠ , b = ⎝1 0⎠; 1 0 A ∗ A ∈ G L2 ; 1 ∗ (iii) L = A ∈ G L2 A Here M0 is of order 21, while L and L are of order 24 (ii) L = As there is no reference for a proof of Proposition 1·5 in the Atlas of finite simple groups [1, p 3], we give a proof of this proposition in the appendix Proof of Theorem 1·4 We start this section with an obvious observation, whose proof is omitted L EMMA 2·1 If G is a subgroup of G L k and ω ∈ G L k , then Pkω So, (Pkω −1 −1 Gω = PkG ω Gω + ) is contained in A+ Pk if and only if (PkG )+ is We will see that Theorem 1·4 can be reduced to the following special case L EMMA 2·2 If G is a proper subgroup of G L , then G jG : F2 ⊗ P3G −→ F2 ⊗ P3 A A is non trivial in positive degrees, or equivalently P3G + ; A+ P3 Proof Since P3G ⊆ P3H when H ⊆ G, it suffices to consider G to be a maximal subgroup of G L Further, by Lemma 2·1 and Proposition 1·5, we need only to investigate G to be one of the three groups M0 , L , L in Proposition 1·5, which are representatives of the three conjugate classes of all maximal subgroups in G L Case 1: G = L The Dickson invariant Q 2,1 x12 + x1 x2 + x22 is an L -invariant, however Q 2,1 ^ A+ P3 Indeed, it is easy to see that x12 , x22 ∈ A+ P3 , and x1 x2 ^ A+ P3 Case 2: G = L The Dickson invariant Q 1,0 x1 is an L -invariant, however x1 ^ A+ P3 Case 3: G = M0 = a, b Since b ∈ G L ⊆ G L , we first want to find polynomials in two variables x1 , x2 , which are invariant under the action of b: x1 → x2 , x2 → x1 + x2 Clearly, there is no degree one polynomial in two variables x1 , x2 , which is b-invariant The smallest subgroup whose invariants are hit 67 Every degree two polynomial in two variables x1 , x2 is of the form r x12 + sx1 x2 + t x22 with r, s, t ∈ F2 We have r x12 + sx1 x2 + t x22 b = r x22 + sx2 (x1 + x2 ) + t (x1 + x2 )2 = t x12 + sx1 x2 + (r + s + t)x22 So, this polynomial is b-invariant if and only if r = t, t = r + s + t, or equivalently r = s = t ∈ F2 Thus, the only degree two b-invariant polynomial in two variables x1 , x2 is Q 2,1 x12 + x1 x2 + x22 Similarly, there are exactly degree three polynomials in two variables x1 , x2 , which are b-invariant Namely P = x13 + x12 x2 + x23 , Q = x13 + x1 x22 + x23 , Q 2,0 = P + Q = x12 x2 + x1 x22 Note that (x3 )b = x3 So, by the above arguments, any degree three b-invariant polynomial in three variables x1 , x2 , x3 is of the form f = r P + s Q + t Q 2,1 x3 + (0x1 + 0x2 )x32 + vx33 , 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 + x1 x22 + x23 + x12 + x1 x2 + x22 x3 + x33 is the only degree three M0 -invariant polynomial in three variables x1 , x2 , x3 Now we show that f ^ A+ P3 Let (x2 , x3 ) be the ideal generated by x2 , x3 in P3 This is an A-ideal Suppose to the contrary that f ∈ A+ P3 Then [ f ] = x13 ∈ P1 = P3 /(x2 , x3 ) also belongs to A+ P1 The conclusion is impossible, because Sq (x1 ) = and Sq (x12 ) = This contradiction shows that f ^ A+ P3 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 (PkG )+ ; A+ Pk From the assumption G = G • G · · · • G m , we have obviously P3G ⊆ PkG So, it suffices to show that P3G + ; A+ Pk Suppose to the contrary that (P3G )+ ⊆ A+ Pk Since the ideal (x4 , x5 , , xk ) is an A-ideal and Pk /(x4 , , xk ) % P3 , it implies that (P3G )+ ⊆ A+ P3 This contradicts Lemma 2·2 The theorem is completely proved 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 L is a simple group of order 168 = 8·3·7 (see e.g Suzuki [9, page 80]) 68 N GUY Eˆ˜ N H V H U.NG AND T RAN D INH L UONG In order to prove Proposition 1·5, we need some results on the normalizers of Sylow subgroups of G L The following three lemmas investigate the Sylow p-subgroups of G L with p = 2, and respectively L EMMA 3·1 If H is a Sylow 7-subgroup of G L , then its normalizer NG L (H ) is a maximal subgroup of order 21 in G L Further, if M is a proper subgroup of G L and M ⊇ H , M H , then M = NG L (H ) Proof Denote σ7 = [G L : NG L (H )] As σ7 is a divisor of [G L : H ] = 24 and σ7 ≡ (mod 7), we have either σ7 = or σ7 = If σ7 = 1, then G L = NG L (H ), and thus H G L This contradicts the fact that G L is a simple group Therefore σ7 = 8, and NG L (H ) is of order 21 = 168/8 Suppose M is a proper subgroup of G L with M ⊇ H , M H Set η7 = [M : N M (H )] As η7 is a divisor of [G L : H ] = 24 and η7 ≡ (mod 7), we have either η7 = or η7 = Suppose to the contrary that η7 = Then from the fact [G L : NG L (H )] = [M : N M (H )] = it implies that all Sylow 7-subgroups of G L are contained in M Thus, the subgroup generated by all Sylow 7-subgroups of G L is a proper normal subgroup of G L This contradicts again the fact that G L is simple Hence η7 = and M = N M (H ) ⊆ NG L (H ) Combining this with the facts that M H and that NG L (H ) is of order 21, we get M = NG L (H ) As a consequence, NG L (H ) is a maximal subgroup of G L The lemma is proved L EMMA 3·2 If S is a Sylow 3-subgroup of G L , then NG L (S) is a non-abelian subgroup of order Further, if K is a subgroup of order in G L with K ⊇ S, then K = NG L (S) = NG L (K ) Proof Obviously, there is a subgroup M ⊇ S of order 21 of G L If N M (S) = M, then S M, and so M ⊆ NG L (S) As M is a maximal subgroup of G L (by Lemma 3·1), we get either NG L (S) = G L or NG L (S) = M The first equality is impossible, since G L is simple The second equality is also impossible, as in this case we have [G L : NG L (S)] = 168/21 = ) (mod 3) So N M (S) M, and we have N M (S) = S, since [M : S] = is a prime number As a consequence, the number of Sylow 3-subgroups of M is [M : N M (S)] = 21/3 = Denoting σ3 = [G L : NG L (S)], we get σ3 If σ3 = 7, then all the Sylow 3subgroups of G L are subgroups of M So the subgroup of G L generated by these Sylow 3-subgroups is a proper normal subgroup of G L This contradicts the fact that G L is a simple group Thus σ3 > Since σ3 is a divisor of [G L : S] = 168/3 = 56 and σ3 ≡ (mod 3), we have σ3 = 28 Hence, NG L (S) is of order 168/28 = If K is a subgroup of order in G L with K ⊇ S, then S K , and so K ⊆ NG L (S) As shown above, NG L (S) is also of order 6, hence K = NG L (S) Since S is the unique Sylow 3-subgroup of K , we get S NG L (K ) ⊆ NG L (S) = K Thus, NG L (K ) = K Suppose to the contrary that K is an abelian subgroup of order Then K is cyclic and it contains exactly two elements of order As the number of subgroups, which are conjugate to K in G L , is [G L : NG L (K )] = [G L : K ] = 28, there are at least 28 · = 56 elements of order in G L On the other hand, there are exactly · = 48 elements of order and · 28 = 56 elements of order in G L The equality 168 − (56 + 48 + 56) = shows that there is exactly one Sylow 2-subgroup of order in G L This contradicts the fact that G L is simple Therefore, K is non-abelian The smallest subgroup whose invariants are hit 69 The lemma is proved Let T be the subgroup of G L consisting of all upper triangular matrices with on the main diagonals Recall that T is a Sylow 2-subgroup of G L with T % D8 , the dihedral group of elements L EMMA 3·3 If U is a Sylow 2-subgroup of G L , then NG L (U ) = U Proof Let Z (U ) denote the center of U From U % D8 it implies that Z (U ) is of order Suppose u ∈ Z (U ), u We will show that C G L (u) = U Obviously, U ⊆ C G L (u) If C G L (u) is of an order divisible by 3, then it contains a Sylow 3-subgroup A, and so B = A, u is an abelian subgroup of order in G L This contradicts Lemma 3·2 If C G L (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 L (u) is of order 21 or C G L (u) = G L The first case is impossible, since C G L (u) ⊇ U The second case is also impossible, as G L is simple Therefore, C G L (u) is of order Hence C G L (u) = U Notice that Z (U ) char U , that is every automorphism of U maps Z (U ) to itself Thus, we get Z (U ) NG L (U ) (see e.g (6·14) in Suzuki [9, page 51]) So NG L (U ) ⊆ C G L (u) = U , and NG L (U ) = U The lemma is proved Proof of Proposition 1·5 Let M be a maximal subgroup of G L If M is of an order divisible by 7, then it contains a Sylow 7-subgroup H of G L Then, by Lemma 3·1, M = NG L (H ) is a maximal subgroup of order 21 in G L On the other hand, a direct M0 , and computation shows that a is of order 7, b is of order 3, and b−1 ab = a So a hence M0 is of order 21 As the Sylow 7-subgroups of G L are conjugate in G L , M is also conjugate to M0 in G L 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 L So, M is not maximal because T ⊆ L and T L If M is of order 3, then it is not maximal by Lemma 3·2 If M is of order 6, then by using Lemma 3·2 we can show that M is conjugate to a subgroup of L Hence M is also not maximal Suppose M is of order 12 Let H be a Sylow 3-subgroup of G L with H ⊆ M From Lemma 3·2, NG L (H ) is of order If NG L (H ) ⊆ M, then it is an index subgroup of M, thus NG L (H ) M This contradicts Lemma 3·2 asserting that NG L (NG L (H )) = NG L (H ) Hence, N M (H ) = NG L (H ) M = H So, M contains exactly [M : H ] = Sylow 3subgroups, and elements of order As a consequence, M contains exactly one Sylow 2-subgroup A of order 4, and therefore A M Let B be a Sylow 2-subgroup of G L with B ⊇ A As [B : A] = 2, we get A B Note that M is a proper subgroup of L M, B Further, combining A L and the fact that G L is simple, we have L G L So, M is not a maximal subgroup of G L This is a contradiction Finally, suppose M is of order 24 We first show that M % , the symmetric group on letters Let H be a Sylow 3-subgroup of G L with H ⊆ M We have H ⊆ N M (H ) ⊆ NG L (H ) If N M (H ) = H , then M contains exactly 24/3 = Sylow 3-subgroups, and · = 16 elements of order Thus, M contains exactly one Sylow 2-subgroup U of order 70 N GUY Eˆ˜ N H V H U.NG AND T RAN D INH L UONG It implies that U M This contradicts Lemma 3·3 asserting that NG L (U ) = U Therefore N M (H ) H By Lemma 3·2, N M (H ) = NG L (H ) is a non-abelian subgroup of order So M contains exactly 24/6 = Sylow 3-subgroups, denoted by H1 , H2 , H3 , H4 Then M acts on the set {H1 , H2 , H3 , H4 } by conjugacy Let ϕ: M → be the corresponding group homomorphism We have Kerϕ = t ∈ M| Hit = Hi for i = 1, 2, 3, = N M (Hi ) i=1 As N M (Hi ) = NG L (Hi ) is of order 6, Hi is the unique subgroup of order in N M (Hi ) for every i Since the subgroups H1 , H2 , H3 , H4 are distinct, Kerϕ is not of order or If Kerϕ is of order 2, let z be the order element in Kerϕ, then for any u ∈ M we have z ∈ N M (Hiu ) u i=1 = N M (Hi ) = Kerϕ i=1 It implies that z u = z for every u ∈ M, or equivalently z ∈ Z (M), the center of M So, H, z is an abelian subgroup of order This contradicts Lemma 3·2 As a consequence, Kerϕ = {1} and we get M % Since M % , there is uniquely an elementary abelian subgroup A of order such that A M (see e.g (2·9) in Suzuki [9, page 294]) Thus M ⊆ NG L (A) As G L is simple, NG L (A) G L Since M is maximal, NG L (A) = M Because T % D8 , there are exactly two elementary abelian subgroups of order 4, denoted by A1 , A2 , in T Indeed, as D8 has the presentation D8 = x, y | x = y = [x, y]2 = , a direct computation shows that all the elements of order in D8 are x, y, [x, y], x[x, y], y[x, y] Note that any elementary abelian subgroup of order in D8 must contain Z (D8 ) = {1, [x, y]} So, D8 has exactly two elementary abelian subgroups of order 4, namely A1 = Z (D8 ), x and A2 = Z (D8 ), y As A is of order 4, it is contained in a Sylow 2-subgroup of G L So A is conjugate to one of the two subgroups A1 and A2 Hence M = NG L (A) is conjugate to either NG L (A1 ) or NG L (A2 ) So, there are at most conjugate classes of subgroups of order 24 in G L To complete the proof of the proposition, we need only to show that L and L are not conjugate subgroups of order 24 in G L Obviously, L and L both are subgroups of order 24 We have clearly L L = T Suppose to the contrary that L and L are conjugate in G L , that is L u1 = L for some u ∈ G L From T ⊆ L we get T u ⊆ L u1 = L As T and T u are Sylow 2-subgroups of L , there exists v ∈ L such that T uv = T , or equivalently uv ∈ NG L (T ) On the other hand, NG L (T ) = T (by Lemma 3·3), so uv ∈ T This implies −1 that u ∈ L and hence L = L u2 = L This is a contradiction Consequently, L , L are not conjugate in G L The proposition is completely proved REFERENCES [1] J H C ONWAY, R T C URTIS , S P N ORTON , R A PARKER and R A W ILSON Atlas of Finite Groups Maximal Subgroups and Ordinary Characters for Simple Groups With computational assistance from J G Thackray (Oxford University Press, 1985), xxxiv + 252 pp The smallest subgroup whose invariants are hit 71 [2] N GUY Eˆ˜ N H V H U.NG Spherical classes and the algebraic transfer Trans Amer Math Soc 349 (1997), 3893–3910 `ˆ N N GO C NAM The hit problem for the Dickson algebra Trans Amer [3] N GUY Eˆ˜ N H V H U.NG and T R A Math Soc 353 (2001), 5029–5040 `ˆ N N GO C NAM The hit problem for the 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contradicts the fact that G L is simple Therefore, K is non-abelian The smallest subgroup whose invariants are hit 69 The lemma... [4] (see Theorem 1·2 below) gives a huge family of elements, which are hit by A in Pk The smallest subgroup whose invariants are hit Let G i be a subgroup of G L ki for i following subgroup