Thompson subgroups and large abelian unipotent subgroups of Lie-type groups Текст научной статьи по специальности «Математика»

УДК 512.5

Thompson Subgroups and Large Abelian Unipotent Subgroups of Lie-type Groups

Vladimir M. Levchuk*

Institute of Mathematics and Computer Science, Siberian Federal University, Svobodny, 79, Krasnoyarsk, 660041

Russia

Galina S. Suleimanova^

Khakas Technical Institute, Abakan, 665017

Russia

Received 10.12.2012, received in revised form 11.12.2012, accepted 25.12.2012 Let U be a unipotent radical of a Borel subgroup of a Lie-type group over a finite field. For the classical types the Thompson subgroups and large abelian subgroups of the group U were found to the middle 1980's. We complete a solution of well-known problem of their description for the exceptional Lie-types.

Keywords: Lie-type group, unipotent subgroup, large abelian subgroup, Thompson subgroup.

Introduction

It is well-known that similarly to A.I.Mal'cev's schema from [1] the problem of enumeration of the large abelian subgroups of a Lie-type group G over a finite field is reduced to the same problem for the unipotent radical U of the Borel subgroup of G. The problem has been under active investigation since 1970's. For classical types the sets A(U) of large abelian subgroups of U were found by the middle 1980's, as well as the subsets AN(U) of normal subgroups and Ae(U) of elementary abelian subgroups and, also, the Thompsons subgroups

J(U) = (A | A e A(U)), Je(U) = (A | A e Ae(U)).

In 1986 A.S.Kondratiev singled out in his survey [2, (1.6)] the following problem:

Problem (A): Describe the sets A(U),AN(U),Ae(U) and the Thompson subgroups J(U) and Je(U) for the remaining cases of G.

The present paper summarizes the investigations of this problem, carried out by E.P.Vdovin [3,4], the authors [5-8] and G.S.Suleimanova [9-12].

1. Preliminaries

A Chevalley group $(K) over a field K, associated with a root system is generated by the root subgroups Xr = xr (K), r e the root subgroups taken for the positive roots r e generate the unipitent subgroup U = U$(K). A twisted group m$(K) is defined as the centralizer in $(K) of a twisting automorphism 0 of order m = 2 or 3. For a twisted group we

* vlevchuk@sfu-kras.ru tsuleymanova@list.ru © Siberian Federal University. All rights reserved

have U = Um$(K) := m$(K) n U$(K). Besides, d is a superposition of a graph automorphism t G Aut $(K) and a field automorphism a : t ^ i (t G K), and for the only continuation " on $ of a symmetry of a Coxeter graph of order m we have 0(Xr) = t(Xr) = Xf (r G $) [13,14].

Given a group-theoretic property P, every P-subgroup of the highest order is called a large P-subgroup. Developing the A.I.Maltsev's approach [1], E.P.Vdovin has mainly calculated [4, Concluding table] the orders a(U) of large abelian subgroups of finite groups U = UG(K) (G = $ or G = m$) and those of Thompson subgroups.

Having described the maximal normal abelian subgroups of U, the authors ( [5,6,8]) also described the large normal abelian subgroups of finite groups U by showing that they form the set An(U). (In general, a large normal P-subgroup of a finite group is not a large P-subgroup.) It allows us [8] to complete (for types G2, 3D4 and 2E6) the calculation of the orders a(U). In [6] the problem (A) is reduced to the question:

Describe the groups U, in which every large abelian subgroup is G(K)-conjugate to a normal subgroup of U and enumerate all the exceptional large abelian subgroups of the remaining groups U ( [15,16]).

See the exceptions in [6,9] and [12]. In [8] the authors proved

Theorem 1. Either every large abelian subgroup of U is G(K)-conjugate to a normal subgroup of U or G(K) is of type G2, 3D4,F4 or 2E6.

G.S.Suleimanova described the exceptional large abelian subgroups for the type F4 in [9,11] and those for the type 2E6 in Section 2 and [12]. In Section 3 we complete this description for the types G2 and 3D4.

Twisted groups m$(K) required further development of the methods [1]. For the types 3D4 and 2E6 there exists a homomorphism Z from the lattice of the root system $ to the lattices of the systems of types G2 and F4 respectively; the preimages of the elements of Z($) being the " -orbits in $.

Let m$ = Z($). For any a G Z($) is defined the root subgroup Xa of m$(K). Similarly to [17] and [8], we have Xa = xa(Ka), Ka := Ker (1 — a) when Z-1(a) is an "-orbit of length 1; in the remaining cases Xa = xa(K). Then U = UG(K) = (Xr | r G G+), where G = $ or m$. The standard central series is U = U1 D U2 D ■ ■ ■ [13]. Let {r}+ be the set of all s G G+ for which the coefficients in the decomposition of s — r in n(G) are all nonnegative. Set

T(r) := (Xs | s G{r}+), Q(r) := (Xs | s G {r}+, s = r) (r G G).

If H C T(r1)T(r2).. .T(rm) and the inclusion fails under every substitution of T(r4) by Q(rj) then L(H) = {r1,r2, ■ ■ ■ , rm} is said to be the set of corners of H. Also, L1(H) denotes the set of first corners for all elements of H.

We fix a regular order of the roots, compatible with the root height function [13, Lemma 5.3.1]. Each element 7 of U permits a unique compatible (canonical) decomposition into a product of root elements xr(jr) (r G G+), [14, Lemma 18]. The coefficient Yr is called an r-projection of the element 7. Obviously, the first corner of 7 corresponds to the first multiplier in its canonical decomposition.

We use usual notation from [13]: h(x) for diagonal automorphisms, nr for monomial elements and the subgroups Ur = (Xr | r G G+ \ {r}), r G n(G). For the root systems of types En and F4 simple roots are denoted by a1,a2, ■ ■ ■ , similarly to [18, Tables V-VIII].

Refining [4, Table 4], the following theorem completes the description of Thompson subgroups.

Theorem 2. Let K be a finite field and U = UG(K). Then:

a) J(U) = Je(U) = U in UG2(K), IK| > 2, and in U3D4(K), |Kff | > 2;

b) Je(U) = 1 and J(U) = T(a) in U3D4(8);

c) Je(U) = 1, J(U) =< a > x < anb >,a = xa(1)x2a+b(1) in UG2(2);

d) J(U) = Je(U) = Uai in U2Ee(K);

e) J(U) = Je(U) = Uar n Uas in UEg(K).

2. Large Abelian Subgroups of Groups U of Type 2E6

In this section we suppose U = U2E6(K). For the root system $ of type E6 the corresponding system m$ = Z($) = G is of type F4. According to Section 1, the group U is generated by the root subgroups Xa = xa(KCT) for all long roots a e G+ (the first type) and Xa = xa(K) for all short roots a e G+. For the system of type F4 see also the diagram of the positive roots from [17] or [8]. Use the notation abcd for the root a«i + ba2 + ca3 + da4 of the system of type F4 (ai,a2,a3,a4 are simple roots) similarly to [18, Table VIII]. For certain maps ~ : K ^ K and * : K ^ K we choose the following subgroups in the group U:

{xim(t)xi23i(i) 11 e K}{xii2i(u)xi22i(U) | u e K} • t(0122), (1)

Xiiii(Kff){xii2i(t)xi22i(i) | t e K}Xi23i(K)T(0122), (2)

Xnn(Kff )Xii2i(Kff )Xi22i(K)Xi23i(K)T (0122), (3)

{xii2i(t)xi22i(t) | t e K}Xi23iT(0122), (4)

{xii2i(t)xi22i(ct) | t e K}Xi22i(K)Xi23iT(0122), c e K, (5)

T(0122)U6, (6)

(T(0121) n E6(Kff))Ur. (7)

Let G(K) = 2E6(K). The aim of this section is to prove that, up to G(K)-conjugacy, the subgroups (1) - (6) for 2K = K and the subgroup (7) for 2K = 0 are all large abelian subgroups of U.

We need the following lemma [8, Lemma 4.4].

Lemma 1. If [xr(F), xs(V)] C Q(r + s) in the group U for F, V C K, FV = 0, then r + s is of the first type, r and s are not of the first type, and, up to a diagonal automorphism conjugacy, F C K, V C K.

According to A.I. Mal'cev [1], a subset ^ of $+ is said to be abelian if r+s e $ for all r, s e A subset ^ of $+ is said to be p-abelian (E.P. Vdovin [4]), if for all r, s e ^ either r + s e $ and the structure constant Ciirs in the Chevalley formula is zero in characteristic p, or r + s e $. The maximal 2-abelian subsets for type F4 are listed in [4, Table 3]. The table contains 7 rows and 13 columns, so the subsets are denoted by ^¿j (i is the row number and j is the column number). In particular, ^2,i2 = {0121}+ := and also = {1111}+ U {0122}+, = {0011,0111,1111,1231} U {0122}+ are ^2,i3, ^6,io, respectively. Using [4] we easily obtain the following lemma.

Lemma 2. Each maximal 2-abelian subset of the root system of type F4 either coincides with one of the sets , (1 ^ j ^ 13) of order 10, or is W-conjugate to one of the sets 'f j of order 11 for i =1, 2 or 3.

Let m(x) := Li(x) (x e U). The following three lemmas is proved in [12].

Lemma 3. Let A be a large abelian subgroup of U of type 2E6. Then for all x, y e A, x, y = 1, the subset {m(x),m(y)} of $ of type F4 is 2-abelian and, if m(x) + m(y) e $, then the m(x)-projections of all elements of A with the first corner m(x) are contained in a 1-dimentional Ka -module.

Lemma 4. Let A be a large abelian subgroup of U of type 2E6, ^ be a maximal 2-abelian subset of the root system of type F4 and let L1(A) C Then:

a) if {r1 ,r2,r3} C r'i + rj G $ and C11,ri,rj = ±2 for all i = j then the subset {r1,r2,r3} is contained in L1(A),

b) if r + s G $ for the roots r, s G C11r,s = ±2, and the pair r, s is not contained in any triple from a), then r G L1(A) or s G L1(A);

c) if ^ contains a root r such that (r + n $ = 0 then r G L1(A).

Lemma 5. If ^ is a maximal 2-abelian subset of type F4, for which there exists a large abelian subgroup A such that m(A) C then ^ is W-conjugate to when 2K = 0 or to when 2K = K. Furthermore, all such sets ^ are exhausted, respectively, by the sets

^2,9, ^2,12, ^3,1, ^3,7, ^3,12, ^5,1, ^5,3, ^5,6, ^5,8, ^5,9, ^5,13, ^6,4, ^6,11, ^7,1i ^2,10, ^2,11, ^2,13, ^3,13, ^5,2, ^5,4, ^5,5, ^5,7, ^6,7, ^7,2, ^7,3.

Now we consider a large abelian subgroup A of U = U2E6(K).

Lemma 6. If A has a simple corner p then A C T(p) and p = a1.

Proof. In the canonical decomposition of elements of U we use the regular order of the system $, defined by the order a1 < a2 < a3 < a4 of the simple roots. Note that the inverse order is also regular.

Suppose that A has at least two simple corners p < q. None of the sets from Lemma 5 contains a1, so we get p > a1. Suppose that q = a4. If we replace the regular order of $ by the inverse order then L1 (A) will contain a4 and, by Lemma 5, L1(A) is contained in one of the sets ^5,4, ^5,5, ^5,6 for 2K = K or ^5,3, ^5,6, ^5,8, ^5,9 for 2K = 0. Each of these sets contains the root 1242. This root is the only root in $+ of height 9. By Lemma 4, c), the root 1242 is contained in L1(A). It is clear, that if there exists an element in A n T(r) with the corner r, then r G L1(A) for any regular order of $. Therefore, the roots a2 and a4 can not be corners in A simultaneously. If q = a3 then similarly 1242 G L1(A). Therefore, the case p = a2,q = a3 is also impossible.

In the remaining case p = a3, q = a4, by Lemmas 5, 4 and [4, table 3], we get 1122 G L1(A) for the inverse order of $. Therefore, for the initial order, the set L1(A) contains a root of height 6. However, this root and the root a3 can not be contained in L1(A) simultaneously, by Lemma 5 and [4, table 3]. Therefore, in all cases the subgroup A can have only one simple corner.

Let A have a simple corner a4 and A C Xai U2. Suppose that A ^ T(a»). Then A has the corner 0011 for i = 2, so An4 has the simple corners a2 and a3. If i = 3 then A has the corner 1100, therefore, Ani has the corners a2 and a3. If i = 4 then A has a corner r G {1100,0110,1110,0120,1120,1220}. In the first and second cases An2 has the corner a1 or a3, respectively, and the corner a4. The third case is reduced to the second case by n1-conjugacy. In the fourth case An3 has a corner a2 and An3 C T(a2); this gives a contradiction with the proved above. The fifth case is reduced to the fourth case by a n1 -conjugacy. The sixth case is reduced to the fifth case by a n2-conjugacy. □

Similarly we obtain the following two lemmas.

Lemma 7. If A has a corner r of height 2, then A C T(r) and r = 1100.

Lemma 8. If A has a corner r of height 3, then:

a) A C T(r) for r = 0111;

b) A C T(1110)T(0120) for r = 0120;

c) A C T(1110)T(0122) for r = 1110 and A C X1110U4.

Lemma 9. Each large abelian subgroup of U is G(K)-conjugate to a subgroup of U2.

Proof. Let A be a large abelian subgroup in U and A ^ U2. By Lemma 6, there exists a simple corner p = ai in A and A C T(p). If p = a2 then Li(A) is contained in one of the sets '2,9, '3ji for 2K = 0 or '2,io, '2jii for 2K = K, and 1100,1222 e Li(A), by lemma 5. Therefore for arbitrary non-zero elements t, u e Ka and suitable tj,uj e K there exist elements x, y e A such that

x = xiioo(t)xoiio(ti)xiiio(t2)xoi2o(t3)xoiii(t4) mod U4, (8)

y = xi222(u)xi232(ui)xi242(u2) mod Uio. ()

If ti =0 for all elements of the form (8) then in the inverse order we have 0110 e Li(A). However, the sets '2jii and '3ji do not contain this root. Since

[x, y] = xi342(ut3 ± (uiti + uiii)) mod Un,

we have t3 = 0 for ti =0. So, there exists an element x' e A such that

x' = xiioo(t)xiiio(t2)xoiii(t4) mod U4

and hence

x' = xiioo(t)xiiio(t2)xoiii(t4)xii2o(t5)xiiii(t6)xoi2i(t7) mod U5 for some t5, t6, t7 e K. We may cancel t2 by Xa3-conjugacy. Moreover,

(x')x-3 (y) = xiioo(t)xiiio(t2 + ty)xoiii(t4)xii2o(t5 ± tyy ± fey)^)

xiiii(t6)xoi2i (t7) =

= xiioo(t)xiiio(ty)xoiii(t4)xii2o(t5 ±tyy)xiiii(t6)xoi2i(t7) mod U5 (y e K).

If y e K and K" = (y) then K| = (yy). Therefore we can choose y such that tyy = ^t5. So, we can transform x' by an U-conjugacy to the form

x' = xiioo(t)xiiio(ty)xoiii(t4)xiiii(t6)xoi2i(t7) mod U5 (y e K).

Then

(x')"3 = xiiio(ty)xoiii(t7)xii2o(t)xiiii(t6)xoi2i(t4) mod U5 (y e K).

Suppose that t7 = 0. Then in the inverse order we get 0111 e Li(A"3) and £i(A"3) = '3ji for 2K = 0 or Li(An3) C '2 io for 2K = K. In the first case 1111 e A(A"3). Therefore, in the initial order, the set £i(A"3) (= '2,9) must contain a root of height 4; so we get a contradiction with [4, Table 3]. In the second case, due to inclusion 0111 e Li(A"3) we get that the 1231-projection of the set of all elements z e A"3 with m(z) = 1231 can not coincide with K, by lemma 1. Hence, in the initial order, Li(A"3), which is contained in '2jii, must contain the root 1111. Since in the inverse order Li(A"3) C '2,io then the set '2,io must contain a root of height 4, and we also get a contradiction with [4, Table 3]. Consequently,

(x')"3 = xiiio(ty)xii2o(t)xiiii(t6)xoi2i(t4) mod U5 (y e K).

By U-conjugacy, we get t6 = 0 and for some u e K we obtain the equality

(x')"3"4 = xoi2o(ui)xii2o(u2)xiiii(u3)xoi2i(u4) mod U5 (U3 = 0).

We may assume that ui =0 because otherwise 0120 e m(A"3"4) and a2 e A (A"3"4), see [4, Table 3]. Moreover, u2 = 0, since otherwise 1120 e Li(A"3"4) and a2 £ Li(A"3"4). Also, u4 = 0, since otherwise in the inverse order of G we have 0121 e Li(A"3"4) and a2 e Li(A"3"4). Note that 1220 e '2,9 n '2,io n '2jii n '3ji. By Lemma 4, 1220 e Li(A"3"4) and the 1220-projection of the elements y with m(y) = 1220 coincides with Ka. Thus, if A"3"4 has a corner

a2, then we may assume that the 1220-projection of (x')n'3n4 is zero, up to a multiplication by a suitable element y. Applying the U- and n3-conjugation to (x')n3n4, we get an element of the form

X1121(v1)xo122(v2) mod U6 (v1 = 0). Hence 1121 G L1(An3n4n3). It follows a2 G L1(An3n4n3) and An3n4n3 C U2. If p = a3 then we get 1231 G L1(A), by Lemma 5. The relation

1 = [A n U7, A] = [A n U7, Xa3 n (AU2)] mod U9

shows that A n U7 = A n T(1231). Up to Xa4-conjugacy, A n (X1231U9) has an element 7 with the corner 1231. Since 1 = [7, A] mod U9, we have A C Xa3T(0110) and An4 C U2. Similarly we consider the case p = a4. □

Analogously we proved

Lemma 10. Any large abelian subgroup of U is G-conjugate to a subgroup of U3 and even to a subgroup of U4.

Finally we get that either 2K = K and the subgroup A is G-conjugate to ones from (1) - (6) or 2K = 0 and A is G-conjugate to the normal subgroup (7).

Remark 1. Taking into account that (1) - (7) are abelian subgroups, we obtain the equalities A(U) = Ae(U) and J(U) = Je(U) = Uai for the group U = U2E6(K). All large abelian subgroups of the group UF4(K) are described in [9] and [11].

3. Large Abelian Subgroups of Groups U of Type G2 and

3D4

According to § 1, the root elements xr (t) of the groups U of type G2 and 3D4 match the roots of the system G2. Choosing its simple roots a and b such that |a| < |b|, we use a hyper-central automorphism qd (d G K) of a group U (see [17]), for which ?d(xb(t)) = xb(t)x3a+b(2dt) mod U5 (t G K). We set

a := xa(1)x2a+b(1), &(t) := x0+6(t)x2a+b(tc). (9)

We now prove the following theorem.

Theorem 3. Each large abelian subgroup of the group U = UG2(K) is G2(K)-conjugate to one of the following subgroups:

a) a normal large abelian subgroup of U;

b) an image under some automorphism <;d (d G K) of a subgroup, which is (Xana)-conjugate to U3 or Xa+bU4 for 6K = K;

c) {xb(t)x3a+b(t) | t G K}pd(K)U5 (d G K) for even IK| > 2; (10)

d) <a,ft(1)>U4 for K| =4.

The proof of the theorem is based on a number of lemmas.

In [5,7,8], the normal large abelian subgroups of U are described as large normal abelian ones. The following lemma follows from [8].

Lemma 11. If the group U is of type G2 then the set AN(U) consists of

U3 and f3c(K)U4 (c G K) for even |K| > 2, U3 for 6K = K, U2 for 3K = 0, (a> x (£1(1)) for K| = 2.

Up to diagonal automorphisms, normal large abelian subgroups of the group U3D4(K), are exhausted by the groups:

U3 and pc(Kff) • X20+5(K) • U4 (c G K) for even |K| > 2, U3 for 2K = K, (a) x <ft(1)> x X20+5(K^) for |Kff | = 2.

Corollary 1. The order a(U) of large abelian subgroups of the group U = UG(K) of type G2 or 3D4 equals |U"31, except the cases |K| = 2 or 3K = 0 for the group UG2(K) where a(U) = |K|4 and the group U3D4(8) where a(U) = 26.

Due to [6, Theorem 2], the group U of type G2 satisfies the following isomorphisms: U/U3 ~ UA2(K) and U/U4 ^ UB2(K). The following lemma is well known for the group UA2(K) ^ UT (3,K).

Lemma 12. Let A be a maximal abelian subgroup and Z be the center of the group U$(K). Then A = {xa (t)xb(ct)|t G K}Z (c G K) or T(b) for the type A2. For the type B2 we have A = T(b) or A is B-conjugate either to XaZ or for the cases 2K = K and 2K = 0 to the subgroup, respectively,

{x0(t)x6(t)x0+6((t2 - t)/2 | t G K)}Z, (xa(1)x6(1))Z. (11)

Proof. The center Z of the group U of type B2 equals U3 for 2K = K or U2 for 2K = 0. If there exists an element 7 G A having two corner, then up to B-conjugation we may suppose that Y = xa(1)xb(1). Choosing an arbitrary element p = xa(t)xb(t')xa+b(t'') mod U3 of A, we find

1 = [p, y] = xa+b(t' - t) mod U3, t' = t (t G K);

[P,Y] = [xa(t),x6(1)][x6(t),Xa(1)][Xa+b(t''),Xa(1)] = X2a+b(2t'' + t - t2).

(The signs of the structural constants are chosen according to [6, Theorem 2].) If 2K = 0 then t2 -1 = 0 and p G (y)Z When 2K = K we have t'' = (t2 -1)/2 and hence A is the first subgroup in (11). □

Setting n := 1 + a + a2 for the type 3D4 we require the following lemma.

Lemma 13. If 2K = K, then Ker (1 + a) = 0. In the general case we have:

K = K+ K, K n K= 2Kff, Kn = K, Ker (n) = K.

Proof. If v = -v, then -y = -v = v, v = -y G KCT and 2v = 0. If 2K = K then Ker (1 + a) = 0. Since for any KCT-linear transformation of the field K the sum of the rank and defect equals 3, the remaining statements of the lemma easily follow from relations

K D K1+ff + Kn D Kff2 = K, 0 = 1 - a3 = (1 - a)n = n(1 - a). □

The order of a subgroup A of a group U = UG(K) of type G2 or 3D4 may be estimated using the orders of intersections of the projections A4:

A n Uj = xr (A,) mod Ui+1, 1 < ht(r) = i < 5; (12)

|A| = |A : A n U21 • |A21 • |A31 • |A41 • |A51. (13)

Lemma 14. Let A be an abelian subgroup of U. Then there exist elements da,db G K and an additive subgroup F C K such that dbFA4 = 0, and

A = y(F) • (A n U2), y(t) = xa(dat)x6(d6t) mod U2 (t G F). (14)

For the type 3D4 and G2 we have (A2A3)n = 0 and 3A2A3 = 0, respectively. When daF 9 1 we have A1+CT = = 0 and 2A2 = 3A3 = 0, respectively.

Proof. Recall that (AU2)/U3 is an abelian normal subgroup of the factor group U/U3, which is isomorphic to a subgroup of the unitriangular group UT(3, K). By Lemma 12 we obtain (14), where y(F) is the system of representatives of cosets of the subgroup An U2 in A. The equalities [A n Ui,A n Uj] = 1 mod Ui+j+1 and (12) imply dbFA4 = 0 and

(A2A3)n =0, (daFA3)n =0, (daFA2)1+T = 0 for the type 3D4, 3A2A3 = 0, 3daFA3 = 0, 2daFA2 =0 for the type G2.

When daF 9 1, we have = A\ =0 and 2A2 = 3A3 = 0 respectively. □

Lemma 15. If an abelian subgroup A of U has two corners, then |A| < a(U).

Proof. Using the notation of lemma 14 and the representation (14) of the subgroup A, we have F 9 1 and da = db = 1 up to a diagonal automorphism. Furthermore, |A : A n U2| = |F| and A4 = 0.

By Lemma 14, for the type G2 we have 2A2 = 3A3 = 0. Hence, A2 = 0 when 3K = 0 and if 6K = K then A3 =0 as well. In both cases, |A| < a(U) due to (13) and Corollary 1. Since (AU4)/U4 is an abelian subgroup of a factor group U/U4 ~ UB2(K), using Lemma 12 in the case 2K = 0 we have:

|F| = 2, |A| = |F| • |A2| • |U5| < 2 • |K|2 < a(U).

For the type 3D4 we have F C Ka, and, by Lemma 14, = A\ = (A2A3)n = 0, and

hence A2 C Ker (1 + a). When 2K = K, using Lemma 13 we find:

A2 =0, |A31 < |Ker(n)| = |K|2, |A| = |F| • |As| • |U51 < |K|4 < a(U).

If 2K = 0 then by Lemma 13 we have A3 C K1+<T and A2 C KT. If |A| > |U3| then

|A| = |F| • |A21 • |A31 • |kt | = |U31, F = A2 = Kt, A3 = K1+T.

Thus, we may assume that a 2a+b-projection of 7(F) is contained in KT. Since [y(F), AnU3] = 1, KT also contains the a + b-projection of y(F). Hence,

(y(F)> C U3D4(K) n UD4(Kt) ~ UG2(Kt)

and, by Lemma 12 we have |F| = 2 = |KT|. Then |A| = |U3| = 25 < 26 = a(U). The lemma is proved. □

The following lemma easily follows from the commutator relations for U.

Lemma 16. If A1 := Xa+bX2a+bU and A2 := XbU4 then T(b) = A1A2. If U is of type G2 and 3K = 0 then the center Z of U is X2a+bU5, and the centralizer C(A1) is T(b); otherwise, Z = U5, C(A1) = A2 and C(A2) = A1. Furthermore, if U is of type G2 and 3K = K then A1 ~ A2 ~ UT(3, K), else if U is of type 3D4 then A2 ~ UT(3, Kt).

Lemma 17. A large abelian subgroup A of UG2(K) is one of the following:

a) U2 or its (Xana U Xbnb) -conjugates when 3K = 0;

b) a subgroup B-conjugate to ((a) x (£1(1)>) • U4 for |K| = 2 or 4;

c) a subgroup B-conjugate to M1 • M2 for 3K = K, |K| > 2, Mi being an arbitrary maximal abelian subgroup of Ai, i = 1, 2.

When 6K = K, the subgroup M1 • M2 coincides with U3 or Xa+bU4 up to an automorphism of the form <;d and to (Xana)-conjugacy, and when 2K = 0, it is G(K)-conjugate to U3, £d(K)U4 or to

{xb(t)x3a+b(t) 11 G K}/3d(K)U5 (d G K). (15)

Proof. Clearly, A contains the center Z. If A ^ U2, then there exists a corner r = a or b of A and a representation (14) with dr = 1 and d,f = 0; furthermore, r + wf.(r) G G+ and wr induces a substitution Wr on G+ \ {r}:

Wa = (b 3a + b)(a + b 2a + b)(3a + 2b), Wb = (a a + b)(3a + b 3a + 2b)(2a + b).

For the type G2, when i = ht(wf(r)) and 3K = 0 we have Ai =0 by lemma 14. Hence, Corollary 1, Lemma 12 and (13) give

T(r) D A D C(T(r)) = XWr(f)Z = Xwr(r-)X2a+bU5; A = 7(K)XWr(f)Z, y(K) = {xr(t)xWf{r)(ct) | t G K} mod C(T(r)).

Having cancelled the scalar c G K with Xr-conjugation, we map A into n-1U2nF.

Let 3K = K. Then (XaU3)/U5 ~ UT(3,K), and if 2K = K, then T(a)/U4 ~ UT(3,K). By Lemma 14, either r = a, A D U4 and A3 = 0 = 2A2, or r = b and A4 = A2A3 = 0. When two out of three projections A2, A3 and A4 are zero, the remaining projection and F are both equal to K, since |A| > |U3|. Hence

A = y(K)U4 when r = a, A = 7(K)3(K)U5 when r = b,

3(t) being the coset representatives of U5 in A n U2 where 3(t) = xq(t) mod Q(q) for the angle q of A n U2. When r = b we define {q, s} := {a + b, 2a + b}. Due to Lemmas 12 and 16, there exist maps ', '' and c,d G K, such that

7 (t) = xb(t)xs(t')x3a+b(ct), 3 (v) = xq (v)xs(dv)x3a+b(v") G A (t,V G K), 1 = [7(t),3(v)] = [xb(t),x3a+b(v")][xs(t'),xq(v)] = x3a+2b(±3vt' ± v''t),

and hence t' = 1' • t and v" = (±3 • 1')v for a suitable choise of the signs. If q = 2a + b then d = 0 and Xr-conjugacy cancels the scalar 1'; when q = a + b, the scalar 1' is similarly defined up to addition of squares from K. Up to B-conjugacy of A we have 1' =0 and A = (A n A1)(A n A2), A n Ai being arbitrary maximal abelian subgroups of Ai, i = 1,2.

When 6K = K, the exceptional automorphism from [17, Theorem 1] of the group U cancels the scalar c in A n A2, and the U-conjugacy implies either n-1Ana = U3 or Xa+bU4. With a glance of Lemma 12, when r = a we are able to cancel the a + b- and 2a + b-projections in y(F) by means of U-conjugacy; thus we transform A to the form

Xa U4 = n-1(Xa+bU4)nb = (^nb)-1 (XbX2a+bU .

If 2K = 0 then by means of diagonal h(x)-conjugacy we achieve c =1 (when x(a) = u G K*, x(a) = u G K*, x(b) = u-1 and x(3a + b) = u2), obtaining A in the form (15). Similarly, when r = a, we obtain a subgroup

{xa(t)x2a+b(t) | t G K}U4 = n-1£1(K)U4nb = (nanb)-1Xb31(K)U5nanb.

Finally, we find the subgroups A = y(F)3d(A2)U4, where

7(t)= xa(t)x2a+b(ct) (t G F), A2 =0, 2K = 0, c, d G K.

The relations

1 = [7(t), 3d(v)] = x3a+b((t2 + td)v)x3a+2b((v2 + cv)t) show that for all t G F and v G A2 we have

(t + d)tA2 =0, (v + c)vF = 0, F = {0, d}, A2 = {0, c}, |A| = 4|K|2.

By corollary 1, we obtain |K| =2 or 4. Clearly, if |K| = 2 then A < U, and up to diagonal conjugacy A has the form

((xa(1)x2a+b(1)> X (^(1))) • U4. (16)

□

For the type 3D4 the description is similar. If A C T(a) and hence T(a) D A D C(T(a)) = U4, then A has the form

P(A2)x2a+b(A3)U4, p(v) := Xa+6(v)x2a+b(v ) (v G A2) (17)

for some map ~ : A2 ^ K. Due to Lemmas 13 and 14 the commutativity of A is equivalent to the inclusion A2A3 C Ker(n) = K1-T. Due to the maximality of A, the projections of A2 and A3 are both KT-modules, as well as Ker (n). If one of the projections are zero or equals K then we have either A = U3 or A = P(K)U4 for~ from End(K +); besides,

[p(t),p(v)] = X3a+2b(±(tv - iv)w), (tv - iv)n = 0 (t,v G K).

Thus, xa(d)-conjugation transforms the subgroup Xa+bU4 into P(K)U4, where

t = dt + (tw - tv)n = [d(— - vt= - tv)]n = (d • 0)n =0 (t, v G K).

When both KT-modules A2 and A3 are nonzero, their dimension is 1 or 2. Up to na- and diagonal conjugacy, the dimension of A2 is less or equals the dimension of A3, and 1 G A2. Therefore we may choose s G A2) such that

A3 C (K + Ks)A3 = A3 + SA3 C K1-T.

If the dimension of A3 is 2 then the inclusions turn into equalities, and multiplication by s induces a KT-linear transformation of a 2-dimensional module K1-T with a characteristic root s. Since the field K does not contain a quadratic extension of the subfield KT, A2 is a 1-dimensional KCT-module. Hence A2 = K and A3 = K1-T. It follows that |A| = |U3| or |K| =8 and A is B-conjugated to a normal subgroup of U. Moreover we now find the Thompson subgroups.

Lemma 18. For the group UG2(K), |K| > 2, and U3A(K), |K| > 2, we have J(U) = Je(U) = U. Besides, Je(U) = 1 and J(U) = T(a) in U3D4(8) and

Je(U) = 1, J(U) =< a > x <anb >, a = xa(1)x2a+b(1) in UG2(2). Remark 1 from § 2, [10] and Lemma 18 give Theorem 2. The investigations are supported by the RFBR (project 12-01-00968)

References

[1] A.I.Mal'tsev, Commutative subalgebras in semisimple Lie algebras, Izv. Akad. Nauk SSSR. Ser. Mat., 9(1945), no. 4, 291-300 (in Russian).

[2] A.S.Kondratiev, Subgroups of finite Chevalley groups, Uspekhi Mat. Nauk, 41(1986), 65-118 (in Russian).

[3] E.P.Vdovin, Maximal orders of abelian subgroups in finite Chevalley groups, Math. Notes, 69(2001), no. 4, 475-498.

[4] E.P.Vdovin, Large abelian unipotent subgroups of finite Chevalley groups, Algebra and Logic, 40(2001), no. 5, 292-305.

[5] V.M.Levchuk, G.S.Suleimanova, The normal structure of unipotent subgroups in Lie-type groups and related questions, Doklady Mathematics, 77(2008), no. 2, 284-287.

[6] V.M.Levchuk, G.S.Suleimanova, Automorphisms and normal structure of unipotent subgroups of finitary Chevalley groups, Proceed. of the Steklov Inst. Math., Suppl. 3, 2009, 118-127.

[7] V.M.Levchuk, G.S.Suleimanova, The normal structure of the unipotent subgroup in Lie-type groups and its extremal subgroups, J. Math. Sci., 185(2012), no. 3, 448-457.

[8] V.M.Levchuk, G.S.Suleimanova, Extremal and Maximal Normal Abelian Subgroups of a Maximal Unipotent Subgroup in Lie Type Groups, J. Algebra, 349(2012), no. 1, 98-116.

[9] G.S.Suleimanova, On conjugacy in a Chevalley group of large abelian subgroups of the unipotent subgroup, J. Math. Sci., 169(2010), no. 5, 696-704.

[10] G.S.Suleimanova, Conjugacy in a finite Chevalley group of type Eg of large abelian unipotent subgoups, J. Siberian Federal Univ., Math. & Physics, 4(2011), no. 4, 536-540.

[11] G.S.Suleimanova, Classes of conjugated in a Chevalley group of type F4 large abelian subgroups in the unipotent subgroup, Vladikavkaz Math. J., 13(2011), no. 2, 45-55 (in Russian).

[12] G.S.Suleimanova, Exceptional large unipotent abelian subgroups in groups of Lie type, Vestnik SibSAU, 44(2012), no. 4, 82-86 (in Russian).

[13] R.Carter, Simple groups of Lie type, New York, Wiley and Sons, 1972.

[14] R.Steinberg, Lections on Chevalley groups, Yale University, 1967.

[15] V.M.Levchuk, G.S.Suleymanova, T.Yu.Voitenko, Some questions for the unipotent subgroup of the Chevalley group, Abstr. of int. conf. "Algebra and its applications", Krasnoyarsk, SibFU, 2007, 168-169.

[16] C.K.Gupta, V.M.Levchuk,Yu.Yu.Ushakov, Hypercentral and monic automorphisms of classical algebras, rings and groups, J. Siberian Federal Univ., Math. &Physics, 1(2008), no. 4, 280-290.

[17] V.M.Levchuk, Automorphisms of unipotent subgroups of Chevalley groups, Algebra and Logic, 29(1990), 97-112 (I), 211-224 (II).

[18] N.Bourbaki, Groupes et algebres de Lie (Chapt. IV-VI), Hermann, Paris, 1968.

Подгруппы Томпсона и большие абелевы унипотентные подгруппы групп лиева типа

Владимир М. Левчук Галина С. Сулейманова

Пусть U — унипотентный радикал подгруппы Бореля группы лиева типа над конечным полем. Для классических типов подгруппы Томпсона и большие абелевы подгруппы групп U были описаны к середине 1980-х годов. Мы завершаем решение известной проблемы их описания для исключительных лиевых типов.

Ключевые слова: группа лиева типа, унипотнтная подгруппа, большая абелева подгруппа, подгруппа Томпсона.