2-Cohomologies of the groups SL (n,q)

Algebra and Logic, Vol. 47, No. 6, 2008

2-COHOMOLOGIES OF THE GROUPS SL(n, q) V. P. Burichenko

UDC 512.542.5

Keywords: cohomologies of groups, finite simple group. Let G = SL(n, q), where q is odd, V be a natural module over G, and L = S 2 (V ) be its symmetric square. We construct a 2-cohomology group H 2 (G, L). The group is one-dimensional over Fq if n = 2 and q = 3, and also if (n, q) = (4, 3). In all other cases H 2 (G, L) = 0. Previously, such groups H 2 (G, L) were known for the cases where n = 2 or q = p is prime. We state that H 2 (G, L) are trivial for n  3 and q = pm , m  2. In proofs, use is made of rather elementary (noncohomological) methods.

INTRODUCTION Let G be a group, M a G-module, and H 2 (G, M ) a group of 2-cohomologies of G with coefficients in M (see [1-3] for definitions). It is known that H 2 (G, M ) is nonzero iff there exists a nonsplit extension of M by G. Describing H 2 (G, M ) is therefore important to many issues in finite group theory. Of particular interest is the case where M is irreducible. In the present paper, we handle the case where G = SL(n, q) and M is the symmetric square of a natural G-module. Let q = pm , p be an odd prime, F = Fq be a field with q elements, and V be an n-dimensional space over F , n  2. Suppose G = SL(V ) = SL(n, q) and L = S 2 (V ). We claim that the following theorem is true. THEOREM 1. The group H 2 (G, L) is nonzero if and only if either n = 2 and q = 3, or n = 4 and q = 3. In these cases H 2 (G, L) is one-dimensional as a space over F . Previously, this theorem was known for some particular cases. For example, let n = 2. PROPOSITION 2 [4, Prop. 4.2]. Let q = pd and V be a space of column vectors of length 2. Then H (SL(2, q), Hom0k (V, V )) ∼ = Fq , for q = 3, and is zero for q = 3. Here we use the terminology created in [4]. Assume Hom0k (V, V ) is the space of k-endomorphisms of trace 0 on V , which can be identified with a space of (2 × 2)-matrices of trace 0, and the action of G = SL(2, q) is given by the rule g(m) = gmg −1 . There is a G-invariant nondegenerate skewsymmetric bilinear form on V , and so V ∗ ∼ = V as G-modules. Hence, End(V ) ∼ = V ⊗ V as G-modules. An explicit way of identifying End(V ) with V ⊗ V is, = V∗⊗V ∼ for example, the following:   −c a ae1 ⊗ e1 + be1 ⊗ e2 + ce2 ⊗ e1 + de2 ⊗ e2 ↔ . −d b 2

Institute of Mathematics, National Academy of Sciences, ul. Kirova 32a, Gomel, 246000 Belarus; [email protected]. Translated from Algebra i Logika, Vol. 47, No. 6, pp. 687-704, November-December, 2008. Original article submitted April 24, 2008. 384

c 2008 Springer Science+Business Media, Inc. 0002-5232/08/4706-0384


If p = 2, then we have a G-module decomposition End(V ) = I ⊕ Hom0k (V, V ), where I is the space of scalar endomorphisms. Furthermore, for p = 2, we also have V ⊗ V = Λ2 (V ) ⊕ S 2 (V ). It is clear that under the above identification of End(V ) and V ⊗ V , the subspaces I and Hom0k (V, V ) are identified with Λ2 (V ) and S 2 (V ), respectively. Thus, Proposition 2 implies that H 2 (G, S 2 (V )) is equal to 0 for q = 3 and is isomorphic to F for q = 3—this is exactly what Theorem 1 states for n = 2. The case q = p is also known. PROPOSITION 3 [5, Thm. 1.4]. Let p > 2, n > 2, and 0
i
p − 1. Then H 2 (SL(n, p), S i (V )) ∼ = H 2 (SL(n, p), S i (V ∗ )) = 0 but for the following cases: (1) p > 3, n = 3, and i = p − 4; (2) p = 3, n = 4, and i = 2. In the exceptional cases, the dimension of a cohomology group is equal to 1. (In [5], this theorem was formulated in terms of groups of Lie type and modules of highest weight.) Thus, to prove Theorem 1, it is sufficient to establish the following vanishing theorem for cohomologies (which can be viewed as the main result of the present paper). THEOREM 4. If q = pm , m > 1, and n  3, then H 2 (G, L) = 0. It should be noted that although the statement of the last theorem speaks of cohomology groups, cohomologies practically are not involved in proving. For our goals to be met, we can confine ourselves to classical results on extensions, namely, the G¨ aschutz and Schur–Zassenhaus theorems. Nor do we even need to couch a definition for cohomologies. It is easy to see that Theorem 4 is equivalent to the following: PROPOSITION 5. Let q = pm , m > 1, n  3, G, L, . . . be as above. Assume E:

 −→ G −→ 1 1 −→ L −→ G i

ρ

⊂G  that is isomorphically mapped is an extension of L by G. Then E is split, i.e., there is a subgroup G by ρ onto G. Recall that an extension of a G-module M by G is a short exact sequence of groups ρ i  −→ 1 −→ M −→ G G −→ 1,

 and which agrees with a G-module structure on M , i.e., is such that gi(m)g −1 = i(ρ(g)m) for any g ∈ G any m ∈ M .  for The proposition will be proved later. Namely, in Sec. 1 we choose some special preimages in G standard generators of G, i.e., for diagonal and monomial matrices and elements xij (t). In Secs. 1 and 2, for n  4 and n = 3, respectively, we show that preimages of elements xij (t) with i < j satisfy the same relations as are satisfied by the elements themselves. This implies the splittability of an extension. We make yet other remarks. (1) In the present paper, Theorem 1 is proved only for the case where n  3 and q = p; yet, we also have a proof for the general case, which is independent of previous contributions. This seems useful, since works dealing in calculation of small-dimensional cohomologies are often not free of computational errors which affect the final result. (2) We prove that H 2 (G, L) ∼ = Z3 for G = SL(4, 3). The method that we use is based on generators and relations and is interesting in its own right. However, it is not demonstrated here, for we are preparing a paper in which this method will be employed for finding 2-cohomology groups of Ω(7, 3) and Ω− (8, 3) with coefficients in their natural modules. It is these 2-cohomology groups that our method was developed for.

385

As far as we know, there are no instances of calculating 2-cohomologies for Ω− (8, 3) that are not computerbased. On the other hand, it is known that this cohomology group is nontrivial because the monster group F1 includes a subgroup which is a nonsplit extension of the form 38 · Ω− (8, 3). (3) In the present paper, we consider the case p > 2 only. The case with even q is less interesting since L is then reducible, and 2-cohomologies with coefficients in all composition factors of L are known. More specifically, L includes a submodule M such that M ∼ = V  , where V  is a = Λ2 (V ) and L/M ∼ 2 module obtained from V by twisting by a field automorphism. In particular, H (G, V  ) ∼ = H 2 (G, V ). All groups H 2 (SL(n, q), Λi (V )) were found by Bell in [6], and also in the previous papers by McLaughlin, Sah, Dempwolff, and Greis cited therein. 1. THE CASE n  4 We recall the G¨aschutz and Schur–Zassenhaus theorems. THEOREM 1.1 (G¨ aschutz, see [1, Thm. 15.8.6; 7, Assert. 10.4]). Let G be a finite group, P be its Sylow p-subgroup, M be a module over G, and M be a finite Abelian p-group. Further, let E:

 −→ G −→ 1 1 −→ M −→ G i

ρ

be some extension of M by G. The extension E is split if and only if its restriction to P is split, i.e., if  such that ρ maps P isomorphically onto P . there is a subgroup P ⊆ G THEOREM 1.2 (Schur–Zassenhaus, see [8, Thm. 20.2.6; 1, Cor. 15.2.1; 2, Chap. 2, Sec. 8; 7, Assert. 18.1]). Let G be a finite group, N  G, and (|N |, |G : N |) = 1. Then there is a subgroup K such that G = N  K, and every two such subgroups are conjugate via an element of N . LEMMA 1.3. Let G be a finite group, M be a G-module which is a finite Abelian p-group, N  G be a normal p -subgroup, and ρ i  −→ G −→ 1 E : 1 −→ M −→ G  such that ρ be an extension. Suppose CM (N ) = 0. Then E is split, and for two subgroups G1 , G2 ⊆ G i(m) maps Gi isomorphically onto G, there is exactly one element m ∈ M with G2 = G1 .  via i. Let N  = ρ−1 (N ). It is obvious that N  G  and Proof. We identify M with its image in G ∼  N/M = N . Since (|M |, |N |) = 1, it follows by the Schur–Zassenhaus theorem that M has a complement in  N , and any two complements are conjugate via an element of M . Let N1 be one of these complements. We  = M  G1 (this obviously implies that ρ maps G1 isomorphically define G1 = NG (N1 ) and claim that G onto G).  Then N g = g −1 N g is also a complement  = M G1 . Let g be an element of G. First, we prove that G 1 −1  . Thus, N g = N m for some m ∈ M . Therefore, N gm = (N g )m−1 = (N m )m−1 = N1 , which of M in N 1 1 1 1 1  we have yields l = gm−1 ∈ NG (N1 ) = G1 . Hence, g = lm ∈ G1 M . Since g is an arbitrary element of G,  = G1 M , whence G  = M G1 . G Next, we show that M ∩ G1 = 1. Let m be an element of the intersection. Then m normalizes N1 . If n is an arbitrary element of N1 , then [m, n] ∈ M ∩ N1 = 1. This yields m ∈ CM (N1 ) = 1, i.e., M ∩ G1 = 1.  = M  G1 . The first assertion of the lemma is proved. Thus, G  Put N2 = N  ∩ G2 . We show that N2 We argue for the second. Let G2 be any complement of M in G.  . Clearly, M ∩ N2 = 1. Let g ∈ N  . We have g = ml for some m ∈ M , l ∈ G2 . is a complement of M in N     ∩ G2 = N2 . Therefore, N  = M N2 , and Since g ∈ N and m ∈ M ⊆ N , it follows that l ∈ N , whence l ∈ N . hence N2 is a complement of M in N 386

Obviously, N2 is normal in G2 , and so G2 ⊆ NG (N2 ). If this inclusion is strict, then NG (N2 ) must contain an element of M . On the other hand, NM (N2 ) = CM (N2 ) = 1. Thus, G2 = NG (N2 ).  , we conclude that N2 = N m for some m ∈ M . ConseSince N1 and N2 are complements of M in N 1  quently, G2 = NG (N1m ) = NG (N1 )m = Gm 1 . Lastly, assume that m ∈ M is another element for which

m
m−1 m−1 = (Gm = G1 , which implies that l = m m−1 normalizes G1 . Moreover, l G2 = Gm 1 . Then G1 1 )  ∩ G1 and is therefore in NM (N1 ) = CM (N1 ) = 1. This means that m = m, i.e., m is normalizes N1 = N defined uniquely. Remark. In cohomological terms, the conclusion of the lemma implies that H i (G, M ) = 0 for i = 0, 1, 2. It is well known that this is also true for other i  0. The proof is given to make our argument as grouptheoretic as possible. In what follows, we need the following: LEMMA 1.4. Let a p -group Q act on a p-group P , N  P be a Q-invariant normal subgroup, and x ∈ P/N be a Q-invariant element. Then there exists a Q-invariant element y ∈ P such that x = yN . Moreover, if CN (Q) = 1, then y is defined uniquely. Proof. We embark on the first assertion. Assume the opposite, letting (P, N, Q, x) be a counterexample for which the product |P ||N | is minimal possible. Obviously, N = 1. Let P1 ⊆ P be the preimage of a subgroup x ⊆ P/N in P . Then (P1 , N, Q, x) is a counterexample. Since |P ||N | is minimal, we conclude that P1 = P , and so P/N = x. Now let A be a minimal nontrivial Q-invariant normal subgroup of P contained in N . Suppose A = N . Then (P/A, N/A, Q, x) meets the conditions ((P/A)/(N/A) is identified with P/N ). In view of |P/A||N/A| < |P ||N |, there is z ∈ P/A which is Q-invariant and is such that x = z(N/A). Consider a quadruple (P, A, Q, z), which is also not a counterexample. Hence, there is y ∈ CP (Q) for which z = yA. It is clear that x = yN . Consequently, (P, N, Q, x) is not a counterexample, a contradiction. We have A = N . Thus, N is a minimal Q-invariant normal subgroup of P . Since N ∩Z(P ) = 1, it follows that N ⊆ Z(P ). gy1 . As Q acts trivially on P/N , Hence, P is Abelian. Take a preimage y1 ∈ P of x and put y = (1/|Q|) g∈Q

the image of y in P/N coincides with an image of y1 , i.e., with x. It is also obvious that y is Q-invariant. We pass to the second assertion. Let y and y  be two Q-invariant preimages of x. Then the element y −1 y  is Q-invariant and lies in N . Therefore, CN (Q) = 1 implies y −1 y  = 1, which yields y  = y. 2 We now turn to constructing preimages of generators. Hereinafter, it is assumed that q = pm , p  3, and m  2; in particular, q  9. Fix a basis e1 , . . . , en for V . By the matrix of an element g ∈ SL(V ) we mean its matrix w.r.t. this basis. Let E:

 −→ G −→ 1 1 −→ L −→ G i

ρ

be an extension such as in the Introduction. Suppose T ⊂ G is a subgroup of diagonal matrices. Since T is a p -group, it follows by the Schur–  which ρ maps isomorphically onto T , and two such Zassenhaus theorem that there is a subgroup of G subgroups are conjugate via an element of L. Fix one of these subgroups and denote it by T. Let N ⊂ G be a subgroup of monomial matrices. It is obvious that T  N . In order to apply Lemma 1.3 to N , it is necessary to establish that CL (T ) = 0. Simultaneously, we prove the same fact for certain proper subgroups of T , for we will need this in our further reasoning. Denote by εi (λ) the diagonal matrix with λ ∈ F ∗ for the ith entry and with 1 for others. Thus, T consists of elements ε1 (λ1 ) . . . εn (λn ), where λi ∈ F ∗ and λ1 . . . λn = 1.

387

LEMMA 1.5. Let T12 , T123 , and T12,34 be subgroups of T , consisting of all elements that act scalarly on subspaces e1 , e2 , e1 , e2 , e3 , and on both of the subspaces {e1 , e2 , e3 , e4 }, respectively. Then: (1) CL (T ) = CL (T12 ) = 0 for all n  3; (2) CL (T123 ) = CL (T12,34 ) = 0 for all n  4. Proof. (1) It suffices to show that CL (T12 ) = 0. Elements of the form ei ej , where 1
i
j
n, constitute a basis for L. Every subspace ei ej  is T -invariant, and so we need only state that T12 acts nontrivially on each of these. There are three cases to consider: j
2; i
2, j  3; i  3. (a) Let 1
i
j
2. Consider an element h = ε1 (λ)ε2 (λ)ε3 (λ−2 ), where λ ∈ F ∗ . The element is contained in T12 and multiplies ei ej by λ2 . Since q  9, there is λ ∈ F ∗ such that λ2 = 1. Consequently, h acts nontrivially on ei ej . (b) Let 1
i
2, j  3. We may assume i = 1 and j = 3. Then the element ε1 (λ)ε2 (λ)ε3 (λ−2 ) multiplies e1 e3 by λ−1 and, therefore, acts nontrivially on e1 e3  for λ = 1. (c) Let i, j  3. Consider an element h = ε1 (λ−1 )ε2 (λ−1 )εi (λ2 ). The element multiplies ei ej by λ2 or by λ4 for i = j and i = j, respectively. It is now sufficient to take λ with λ4 = 1. (2) First, consider T123 . We argue similarly to how we did for the case with T12 . Again there are three cases: j
3; i
3, j  4; i  4. (a) Let 1
i
j
3. The element h = ε1 (λ)ε2 (λ)ε3 (λ)ε4 (λ−3 ) multiplies ei ej by λ2 and, therefore, acts nontrivially on ei ej  if λ = ±1. (b) Let i
3, j  4. We may assume that j = 4. The same element as in (a) multiplies ei e4 by λ−2 , and again it suffices to take λ = ±1. (c) Let i, j  4. We may assume that i = 4. The same h multiplies ei ej by λ−3 or by λ−6 for j > 4 and j = 4, respectively. We need only take λ with λ6 = 1. Next, consider T12,34 . We now handle four cases: (a) 1
i
j
2 or 3
i
j
4; (b) 1
i
2, 3
j
4; (c) i
4, j  5; (d) i, j  5. In case (a) we define h = ε1 (λ)ε2 (λ)ε3 (λ−1 )ε4 (λ−1 ), where λ = ±1. In cases (b) and (c) we take h = ε1 (−1)ε2 (−1) and h = ε1 (−1)ε2 (−1)ε3 (−1)ε4 (−1), respectively. Lastly, in case (d) we put h = ε1 (λ)ε2 (λ)εi (λ−2 ). This element is contained in T12,34 and multiplies ei ej by λ−2 or by λ−4 , and it is sufficient to take λ with λ4 = 1. 2  ⊆G  such that ρ maps N  isomorphically onto N . Let Lemma 1.5 implies that there is a subgroup N  . Then ρ maps T isomorphically onto T . Hence, T = Tv for some (unique) T be the preimage of T in N −1  =N  v−1 . Then N  ⊇ T and ρ maps N  isomorphically onto N . v ∈ L, which yields T = Tv . Put N For 1
i < j
n, denote by Tij the subgroup of elements of T acting scalarly on the subspace ei , ej . Then Tij is conjugate to T12 via an element of N , and so CL (Tij ) = 0. Denote by Tij the preimage of Tij in T. Given 1
i < j
n and t ∈ F , put xij (t) = 1 + teij (where eij is an ordinary matrix unit). Also ij = ρ−1 (Xij ). Since T normalizes Xij , T normalizes X ij . Furthermore, let Xij = {xij (t) | t ∈ F } and X ij , N = L, Q = Tij , and x = xij (t), we infer that Tij centralizes Xij . Applying Lemma 1.4 with P = X   there is an element y ∈ Xij which is Tij -invariant and is such that ρ(y) = xij (t). Moreover, in view of ij (t). We write n  for the preimage CL (Tij ) = CL (Tij ) = 0, the element y is defined uniquely. Denote it by x . of n ∈ N in N LEMMA 1.6. The following assertions hold: xij (u) = x ij (t + u) for all t, u ∈ F ; (1) x ij (t) ij (t)n = x kl (u). (2) if 1
i < j
n, 1
k < l
n, t, u ∈ F , n ∈ N , and xij (t)n = xkl (u), then x

388

Proof. (1) Both elements x ij (t) xij (u) and x ij (t+u) commute with Tij and are ρ-preimages of xij (t+u); so they coincide. ij (t) = x kl (u) = 1, and we are done. (2) For t = 0 or u = 0, we have xij (t) = xkl (u) = 1, hence x Therefore, we assume that t, u = 0. In this case it is obvious that CT (xij (t)) = Tij and CT (xkl (u)) = Tkl . Since n normalizes T , we have Tijn = CT (xij (t))n = CT (xij (t)n ) = CT (xkl (u)) = Tkl , whence Tijn = Tkl . Hence, a Tij -invariant preimage of xij (t), under conjugation by n , is mapped into a Tkl -invariant preimage ij (t) is sent to x kl (u), as required. 2 of xkl (u), i.e., x Let U ⊂ G be a subgroup of upper unitriangular matrices. Then U is a Sylow p-subgroup of G. The subgroup is generated by xij (t). As defining relations we may take the following: (1) xij (t)xij (u) = xij (t + u), t, u ∈ F ; (2) [xij (t), xik (u)] = 1; (3) [xik (t), xjk (u)] = 1; (4) [xij (t), xjk (u)] = xik (tu) (in (2), (3), and (4), it is assumed that 1
i < j < k
n and t, u ∈ F ); (5) if {i, j}∩{k, l} = ∅, then [xij (t), xkl (u)] = 1 (the commutator, note, is defined to be [a, b] = aba−1 b−1 , not a−1 b−1 ab). Below we prove that the same relations hold also for x ij (t). This will readily imply that E is a split  extension. Indeed, since x ij (t) satisfy the same relations as xij (t), there is a homomorphism σ : U −→ G such that σ(xij (t)) = x ij (t). Hence, the composition ρσ : U −→ G fixes xij (t), whence ρσ = idU . Hence,   which ρ maps isomorphically onto U . By the G¨ U = σ(U ) is a subgroup of G, aschutz theorem, this implies that E is split. Therefore, it remains to prove that x ij (t) satisfy relations (2)-(5) (for (1), see above). For n  4, this will be done right here, and for n = 3, in the next section. First, we confine ourselves to treating the following: (2a) [x12 (t), x13 (u)] = 1; (3a) [x13 (t), x23 (u)] = 1; (4a) [x12 (t), x23 (u)] = x13 (tu); (5a) [x12 (t), x34 (u)] = 1. Indeed, under the action of a suitable element of N , any relation of the form (2)-(5) can be translated ij (t) by into one of the form (2a)-(5a). If (2a)-(5a) hold for elements x ij (t), then we derive (2)-(5) for x  and using Lemma 1.6. acting by a suitable element of N For example, consider the relations in (4a). Since [x12 (t), x23 (u)] = x13 (tu), it follows that 23 (u)] = v x13 (tu) for some v ∈ L. Take an element h ∈ T123 . As h ∈ T12 , the element  h [ x12 (t), x commutes with x 12 (t). Likewise,  h commutes with x 23 (u) and x 13 (tu). Hence,  h ought to also commute with v. This is true for all h ∈ T123 ; so v ∈ CL (T123 ) = CL (T123 ) = 0. Therefore, [ x12 (t), x 23 (u)] = x 13 (tu), as required. The relations in (2a), (3a), and (5a) can be treated in a similar way. Thus, Proposition 5 is proved for n  4. 2. THE CASE n = 3 Let n = 3. As with n  4, the following hold: [ x12 (t), x 13 (u)] = v(t, u),

(1)

[ x13 (t), x 23 (u)] = w(t, u),

(2) 389

[ x12 (t), x 23 (u)] = r(t, u) x13 (tu),

(3)

where v(t, u), w(t, u), r(t, u) ∈ L. To prove Proposition 5, it is sufficient to state that v, w, and r are identically equal to zero. First, we prove this for v and w. We derive some relations which should be satisfied by v(t, u) and w(t, u). For 1
i
j
3, it might be convenient to denote an ordered pair like (i, j) merely by ij. A subgroup T consists of elements h(α, β) = ε1 (α)ε2 (β)ε3 ((αβ)−1 ) = diag(α, β, γ), where γ = α−1 β −1 . The subgroup acts by conjugation on each of the subgroups Xν , where ν = 12, 13, 23. Given h = h(α, β) and x = x12 (t), it is easy to verify that h(x) = hxh−1 = hx12 (t)h−1 = x12 (αβ −1 t). Similarly, hx13 (t)h−1 = x13 (αγ −1 t) = x13 (α2 βt) and hx23 (t)h−1 = x23 (βγ −1 t) = x23 (αβ 2 t). This, together with Lemma 1.6(2), implies  h x12 (t) h−1 = x 12 (αβ −1 t),  h x13 (t) h−1 = x 13 (α2 βt),  h x23 (t) h−1 = x 23 (αβ 2 t). Now we conjugate relation (1) by  h. We have h−1 ,  h x13 (u) h−1 ] =  hv(t, u) h−1 . [ h x12 (t)

(4)

The left-hand side in the last relation is [ x12 (αβ −1 t), x 13 (α2 βu)] = v(αβ −1 t, α2 βu); the right-hand one is h(v(t, u)) = h(α, β)v(t, u). Hence, (4) yields the identity v(αβ −1 t, α2 βu) = h(α, β)v(t, u).

(5)

w(α2 βt, αβ 2 u) = h(α, β)w(t, u),

(6)

r(αβ −1 t, αβ 2 u) = h(α, β)r(t, u).

(7)

In a similar way, we can prove that

The derived identities may be interpreted in another way: namely, the maps from X12 × X13 , X13 × X23 , and X12 × X23 to L defined via (x12 (t), x13 (u)) → v(t, u), (x13 (t), x23 (u)) → w(t, u), and (x12 (t), x23 (u)) → r(t, u), respectively, are T -equivariant (i.e., are consistent with the action of T ). x13 (u) x12 (t)−1 = Maps v and w satisfy certain additive conditions. Indeed, it follows from (1) that x 12 (t) v(t, u) x13 (u). Since the left-hand side is multiplicative w.r.t. u, the right-hand side ought to also be multiplicative, i.e., x13 (u1 + u2 ) = v(t, u1 ) x13 (u1 ) · v(t, u2 ) x13 (u2 ). v(t, u1 + u2 ) (Generally, we say that an expression f (t) is additive or multiplicative w.r.t. t if f (t1 + t2 ) = f (t1 ) + f (t2 ) and f (t1 + t2 ) = f (t1 )f (t2 ), resp.) The right-hand side in the last relation is v(t, u1 ) x13 (u1 )v(t, u2 ) x13 (u1 )−1 · x 13 (u1 ) x13 (u2 ) = (v(t, u1 ) + x13 (u1 )v(t, u2 )) · x 13 (u1 + u2 ). Substituting it into the preceding relation and canceling by x 13 (u1 + u2 ) on the left, we see that v(t, u1 + u2 ) = v(t, u1 ) + x13 (u1 )v(t, u2 ).

(8)

Furthermore, [ x13 (u), x 12 (t)] = [ x12 (t), x 13 (u)]−1 = −v(t, u). Likewise, we can derive v(t1 + t2 , u) = v(t1 , u) + x12 (t1 )v(t2 , u). 390

(9)

In much the same manner, we prove w(t, u1 + u2 ) = w(t, u1 ) + x23 (u1 )w(t, u2 ),

(10)

w(t1 + t2 , u) = w(t1 , u) + x13 (t1 )w(t2 , u).

(11)

We have thus established the required equivariance and additivity relations. Let R be a set of all pairs ij. Order R ∪ {0} as follows: 0 < 11 < 12 < 13 < 22 < 23 < 33. For



Lν . Now assume Hν = Lν
and H0 = 0. Then ν = ij ∈ R, put eν = ei ej and Lν = eν . Then L = ν∈R

ν
≤ν

0 = H0 ⊂ H11 ⊂ H12 ⊂ . . . ⊂ H33 = L. Clearly, both T and U preserve this flag, and U acts on its factors identically. Let δ ∈ R ∪ {0} be minimal with v(t, u) ∈ Hδ for all t and u. If δ = 0, then v(t, u) ≡ 0. If we assume that δ = 0 we will arrive at a contradiction. For ν ∈ R, denote by vν (t, u) the Lν -component of v(t, u). As T preserves all Lν , the map from X12 ×X13 to Lν defined via (x12 (t), x13 (u)) → vν (t, u) is T -equivariant. Since v(t, u) ∈ Hδ , the Lδ -components of v(t, u) and x13 (s)v(t, u) coincide, for any s ∈ F . Hence, (8) yields vδ (t, u1 + u2 ) = vδ (t, u1 ) + vδ (t, u2 ). Similarly, vδ (t, u) is also additive w.r.t. t. Thus, ϕ : (x12 (t), x13 (u)) → vδ (t, u) is a T -equivariant bihomomorphism from X12 × X13 to Lδ . For finding all such bihomomorphisms, we can use the following: LEMMA 2.1. Let (λ, µ) be one of (12, 13), (12, 23), (13, 23); ν ∈ R. Then a nonzero T -equivariant bihomomorphism Xλ × Xµ −→ Lν may exist only if (λ, µ, ν) = (12, 13, 23), (12, 23, 13), or (13, 23, 12). Proof. For the sake of simplicity, we only handle the case (λ, µ) = (12, 13), letting ν be such that the above-stated bihomomorphism exists. An element h(−1, 1) acts as inversion on X12 , and identically on X13 ; so it must act like inversion on Lν . Thus, ν = 12 or 23. Furthermore, h(1, −1) acts as inversion on X12 and X13 ; hence it should act identically on Lν . Consequently, ν = 23. 2 Let (λ, µ, ν) be one of the triples stated in the last lemma. We want to clarify for which q the required bihomomorphism does in fact exist. Any map f : X12 × X13 −→ L23 has the form f (x12 (t), x13 (u)) = ϕ(t, u)e2 e3 , for some map ϕ : F × F −→ F . Obviously, f is a bihomomorphism iff ϕ is biadditive. Furthermore, h(α, β) acts on X12 , X13 , and L23 by multiplication by αβ −1 , α2 β, and α−1 , respectively. Therefore, the T -equivariance of f is equivalent to the condition that ϕ(αβ −1 t, α2 βu) = α−1 ϕ(t, u)

(12)

for any α, β ∈ F ∗ . For (λ, µ, ν) = (12, 23, 13) and (13, 23, 12), the last-mentioned condition should be replaced by (13) ϕ(αβ −1 t, αβ 2 u) = β −1 ϕ(t, u) and ϕ(α2 βt, αβ 2 u) = αβϕ(t, u),

(14)

respectively. To find all biadditive maps satisfying one of conditions (12), (13), or (14), we make use of the following: PROPOSITION 2.2 [9, Lemmas 3.1, 3.2]. Let F be a finite field with q = pm elements.

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(1) Every biadditive map ϕ : F × F −→ F can be uniquely represented as  i j ϕ(x, y) = aij xp y p , aij ∈ F. 0
i,j
m−1

Conversely, every map of this form is biadditive. (2) Let m1 , m2 , m ∈ Z. A map ϕ satisfies the condition ϕ(λm1 x, λm2 y) = λm ϕ(x, y) for all λ ∈ F ∗ if and only if aij = 0 for all pairs (i, j) that do not satisfy m1 pi + m2 pj ≡ m (mod q − 1). In particular, if the congruence has no solutions with 0
i, j
m − 1, then every biadditive map satisfying ϕ(λm1 x, λm2 y) = λm ϕ(x, y) is trivial. LEMMA 2.3. Consider a congruence 3pl ≡ ε (mod q − 1) on a set of quadruples (p, q, l, ε), where p  3 is a prime, q = pm , m  2, 0
l
m − 1, and ε = ±1. Then solutions to this congruence are exactly the quadruples (3, 3m , m − 1, 1). Proof. Obviously, the above quadruples are indeed the solutions. We claim that there are no other solutions. Let (p, q, l, ε) be a solution. Then 3pl −ε is a multiple of q −1. Furthermore, 3pl −ε  3p0 −1 = 2, and so 3pl − ε  q − 1. Hence, 3pm−1 + 1  3pl − ε  q − 1 = pm − 1. Thus, 3pm−1 + 1  pm − 1, which yields pm−1 (p − 3)
2. For p  5, the left-hand side of the last inequality is greater than or equal to 52−1 (5 − 3) = 10, a contradiction. Hence, p = 3. Furthermore, if l
m − 2, then 3pl − ε  q − 1 implies 3 · 3m−2 + 1  3m − 1, so 2 · 3m−1
2, and hence m
1, a contradiction. We have l = m − 1. Now the congruence turns into 3 · 3m−1 ≡ ε (mod 3m − 1); therefore ε = 1. 2 LEMMA 2.4. Any biadditive map ϕ satisfying one of conditions (12), (13), or (14) is trivial for q = 3m . If q = 3m and ϕ satisfies (14), then ϕ(t, u) = at1/3 u1/3 for some a ∈ F . Proof. Condition (12) is equivalent to the fact that ϕ(λt, λ2 u) = λ−1 ϕ(t, u) and ϕ(λ−1 t, λu) = ϕ(t, u) i j aij tp up . If aij = 0, then pi + 2pj ≡ −1 (mod q − 1) and for all λ ∈ F ∗ . Expand ϕ(t, u) = 0
i,j
m−1

−pi + pj ≡ 0 (mod q − 1). Summing up these congruences, we derive 3pj ≡ −1 (mod q − 1). The last congruence has no solutions (under the stated restrictions on p and q); hence ϕ is trivial. Similarly, (13) and (14) result in the following respective systems of congruences: pi + pj ≡ 0 (mod q − 1), 2pi + pj ≡ 1 (mod q − 1), −pi + 2pj ≡ −1 (mod q − 1), pi + 2pj ≡ 1 (mod q − 1). The first of these systems entails 3pj ≡ −1 (mod q − 1), which is impossible. The second system implies 3pi ≡ 3pj ≡ 1 (mod q − 1), which is possible only if q = 3m , i = j = m − 1 (which does in fact solve the system). Thus, a map ϕ satisfying one of (12), (13), (14) is trivial for q = 3m . If q = 3m and ϕ satisfies (14), m−1 m−1 then ϕ(t, u) = am−1,m−1 t3 u3 = at1/3 u1/3 , where a = am−1,m−1 . 2 COROLLARY 2.5. Let (λ, µ, ν) be one of (12, 13, 23), (12, 23, 13), (13, 23, 12). A nontrivial T equivariant bihomomorphism f : Xλ × Xµ −→ Lν may exist only if (λ, µ, ν) = (13, 23, 12), q = 3m , in which case it is of the form f (x13 (t), x23 (u)) = at1/3 u1/3 e1 e2 , for some a ∈ F . Now we complete proving that v(t, u) ≡ 0. Assume the opposite. Let δ ∈ R be minimal with v(t, u) ∈ Hδ for all t and u. Then f : (x12 (t), x13 (u)) → vδ (t, u) is a nontrivial T -equivariant bihomomorphism from X12 × X13 to Lδ . There are no such bihomomorphisms, and so we arrive at a contradiction. In fact, we have thus proved the following: LEMMA 2.6. The identity v(t, u) ≡ 0 holds. 392

LEMMA 2.7. The identity w(t, u) ≡ 0 holds. Proof. Assume the opposite. Let δ ∈ R be minimal with w(t, u) ∈ Hδ for all t and u. As above, then, the map X13 × X23 −→ Lδ defined via (x13 (t), x23 (u)) → wδ (t, u) is a nontrivial T -equivariant bihomomorphism. This implies that δ = 12, q = 3m , and w12 (t, u) = at1/3 u1/3 e1 e2 . Therefore, w(t, u) = at1/3 u1/3 e1 e2 + g(t, u)e21 for some map g : F × F −→ F . Consider    0 1 n = diag , −1 = e12 + e21 − e33 . 1 0 Obviously, n ∈ N and n2 = 1. It is easy to calculate that nx13 (t)n = x23 (−t) and nx23 (t)n = x13 (−t). Conjugating [ x13 (t), x 23 (u)] = w(t, u) by n , we derive [ nx 13 (t) n, n x 23 (u) n] = n w(t, u) n. The left-hand side is [ x23 (−t), x 13 (−u)] = [ x13 (−u), x 23 (−t)]−1 = −w(−u, −t). The right-hand side is nw(t, u). Hence, −w(−u, −t) = nw(t, u). We express the last relation in coordinates. Since w(−u, −t) = a(−u)1/3 (−t)1/3 e1 e2 + g(−u, −t)e21 , we have −w(−u, −t) = −au1/3 t1/3 e1 e2 − g(−u, −t)e21 . Furthermore, nw(t, u) = (e12 + e21 − e33 )(at1/3 u1/3 e1 e2 + g(t, u)e21 ) = at1/3 u1/3 e1 e2 + g(t, u)e22 . Hence, −w(−u, −t) = nw(t, u) is equivalent to the fact that g(t, u) ≡ 0 and −at1/3 u1/3 ≡ at1/3 u1/3 , which yields a = 0. Therefore, w(t, u) ≡ 0, a contradiction. 2 Thus, the maps v and w are identically equal to zero. It remains to prove that r(t, u) ≡ 0. LEMMA 2.8. The identity r(t, u) ≡ 0 holds. Proof. Assume the contrary. Let δ ∈ R be minimal with r(t, u) ∈ Hδ for all t and u. Also suppose 12 , X 13 , X 23 , V , which is that γ ∈ R ∪ {0} is an immediate predecessor of δ. Consider A = ρ−1 (U ) = X the complete preimage of U . Obviously, Hα  A for all α ∈ R ∪ {0}. x23 (u) x12 (t)−1 = r(t, u) x13 (tu) · x 23 (u). The left-hand side is multiplicative Relation (3) yields x 12 (t) w.r.t. u; so the right-hand side is also multiplicative. Note that x 23 (u) and x 13 (tu) are multiplicative 13 , and X 23 commute modulo Hγ . This implies that r(t, u)Hγ ∈ Hδ /Hγ w.r.t. u, and all elements of Hδ , X is multiplicative w.r.t. u. Since r(t, u)Hγ = rδ (t, u)Hγ and Lδ ∩ Hγ = 1, it follows that rδ (t, u) is additive w.r.t. u. Additivity w.r.t. t can be proved in a similar way. Clearly, the map (x12 (t), x23 (u)) → rδ (t, u) is T -equivariant. Hence, it is a biadditive, T -equivariant, and nontrivial map from X12 × X23 to Lδ , which is impossible. 2 Proposition 5 is completed, thus proving Theorem 1. Acknowledgments. I am grateful to I. D. Suprunenko for support. REFERENCES 1. M. Hall, The Theory of Groups, Macmillan, New York (1959). 2. M. Suzuki, Group Theory I, Springer, Berlin (1982). 3. S. MacLane, Homology, Springer, Berlin, 1963. 393

4. C.-H. Sah, “Cohomology of split group extensions. II,” J. Alg., 45, 17-68 (1977). 5. A. S. Kleshchev, “Cohomology of finite Chevalley groups with coefficients in modules with 1-dimensional weight spaces,” Comm. Alg., 22, No. 4, 1197-1218 (1994). 6. G. W. Bell, “On the cohomology of the finite special linear groups. I, II,” J. Alg., 54, 216-238, 239-259 (1978). 7. M. Aschbacher, Finite Group Theory, Cambridge Univ. Press (1986). 8. M. I. Kargapolov and Yu. I. Merzlyakov, Fundamentals of Group Theory [in Russian], 3d edn., Nauka, Moscow (1982). 9. V. P. Burichenko, “2-cohomologies of the group Ω− (4, q) with coefficients in a natural module,” Mat. Sb., 189, No. 9, 29-42 (2007).

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