Subgroups Generated by a Pair of 2-Tori in GL(4,K), II

The present paper is the next in a large series of works dedicated to the geometry of microweight tori and long root tori in the Chevalley groups that was announced in [1]. Namely we describe the subgroups generated by a pair of 2-tori in GL(4, K ) corresponding to degenerate cases in the sense below.

Recall that 2-tori in GL(n, K ) are the subgroups conjugate to the diagonal subgroup of the following form

{ diag(e, e, 1,..., 1), e G K * } .

From the general theory viewpoint 2-tori are microweight tori corresponding to the fundamental weight ^ 2 in the extended Chevalley group of type A _n- i .

In [2] we proved the reduction theorem for the pairs of m -tori. It follows from it that any pair of 2 -tori (X, Y ) can be embedded in GL(6, K ) by simultaneous conjugation. We call an orbit of a pair of 2-tori (X, Y) the orbit in GL(n, K) , if the pair (X, Y) is embedded in GL(n, K ) by simultaneous conjugation and it can not be embedded in GL(n — 1, K ) . It follows from the reduction theorem that n can take values 3, 4, 5 or 6.

• (0 2025 Nesterov, V. V. and Zhang, M.

2.    Notation

The orbits of 2-tori in GL(3, K ) coincides with the orbits of 1-tori and are described in [3] (see also Lemma 1 [2]). The orbits and spans of 2-tori in GL(6,K) were classified in [2]. In paper [4] we described the orbits and spans of 2-tori in GL(5, K ) .

The case of GL(4,K) is the most difficult and requires cumbersome calculations. In [5] we classified the orbits of 2-tori in GL(4, K ) . It turned out that there are more than 80 types of such orbits. Of course in many cases pairs of 2-tori belongning to the different orbits have the same span. However calculations of all spans take a lot of pages. So we divided their into two papers. In this paper we treat degenerate cases, and in the next paper we consider undegenerate cases.

The context of this problem and many references reader can find in the surveys [1, 6, 7] and in the detailed introduction of [4].

The idea of this research belongs to wonderful mathematician and person Nikolai Aleksandrovich Vavilov. The starting point of it was his work [3]. The next three papers were written by N. A. Vavilov jointly with the first author. Unfortunately N. A. Vavilov passed away. The authors are very grateful to him for setting the problem and numerous inspiring discussions.

The authors are very pleased to publish this paper in the journal devoted to anniversary of professor V. A. Koibaev. The first author met him in the early 90s and remembered him as a very benevolence and responsiveness person.

This paper is a direct continuation of [5] and we use the same notation as before. But for reader’s convenience we recall them here.

Let K be a field and K * = K \{ 0 } be the multiplicative group of it. Further, G = GL(n, K ) is the general linear group of degree n over K . By D = D (n,K) we denote the subgroup of diagonal matrices in G , and N = N (n, K ) denotes the subgroup of monomial matrices in G .

The quotient group N/D is isomorphic to S _n , the symmetric group on n letters. Denote by W = W n the group of permutation matrices in G . We identify S _n and W _n via the isomorphism n H w _n , where w _n is the matrix whose entry in the position (i,j) is 6 i, _n j .

Let V = K n be the right vector space of columns of height n over K . Usually we identify a matrix g G G with the corresponding linear map of the space K ⁿ . Here g acts on the left . To stress that we are using this geometric viewpoint, in such cases we call elements of G transformations .

By e 1 , . . . , e _n we denote the standard base of K ⁿ . Here e i is the column, whose i -th component equals 1, whereas all other components are equal to 0. The dual space V * = ⁿ K is left vector space of rows of length n . By f 1 , . . . , f _n we denote the standard base of ⁿ K . It is dual to e i ,..., e _n with respect to the standard pairing, V * x V —> K .

Denote by e ij a standard matrix unit, i. e. the matrix whose entry in the position ( i, j ) is 1 and all the remaining entries are zeroes. Next, X ij (£) = e + £e j for £ G K and 1 C i = j C n denotes elementary transvection . For given i = j we consider the corresponding unipotent root subgroup X ij = { x^ (£),£ G K } . The subgroup E(n,K ) of G , generated by all X ij , 1 C i = j C n , is called the elementary subgroup of G . In case of the field, it coincides with the special linear group SL(n,K) .

Similarly, by d i (e) = e + (e — 1)e ii we denote an elementary pseudo-reflection . For a given i we consider the corresponding 1 -torus

Q i = {d i (e), e G K * } .

Clearly, GL( n, K ) is generated by E ( n, K ) and Q 1 .

Let g G G . The largest subspace W C V , such that g\ w = id is called the axis of g . Similarly, the subspace U = {gv — v : v G K ⁿ } is called the centre of g . Clearly, dim U = m and dim W = n — m .

The elementary 2-torus Q = Qu0,w0 = {diag(E,E, 1,... , 1), e G K*} is defined by the subspaces Uo = (61,62) and Wo = (f1,f2). It means, that elements of it are d0(E) = 6 + 61(e — 1)f1 + 62(e — 1)f2,   e G K*•

It is clear that gQUWg 1 = QgU,Wg-1, g G GL(n,K)•

Therefore any 2-torus (see [2]) is conjugated to the elementary 2-torus Q . The elements of an arbitrary 2 -torus are the elements of the following form

d(e) = e + u1(e — 1)v1 + u2(e — 1)v2, e G K*, where Ui = gei, Vi = fig-1, 1 C i C 2, for some matrix g G GL(n,K). Thus each 2-torus is completely determined by the subspaces U = (u1,U2) and W = (v1,V2).

The subspace U is precisely the centre of Q UW , in the sense of being the centre of every d(e) G Q uw , e = 1 . Similarly, the subspace W ¹ orthogonal to W C n K with respect to the canonical pairing ⁿ K x K ⁿ —> K , is precisely the axis of Q uw , in the above sense. Oftentimes we loosely refer to W itself as the axis of Q UW .

Consider a pair of 2 -tori X and Y with centers U1 and U2 and with axes W1 and W ₂ , respictively. In paper [2] we introduce the following invariants for a pair of m -tori. r = r(X, Y) = dim(U 1 + U 2 ) , s = s(X,Y ) = dim(W 1 + W 2 ) , p = p(X,Y ) = dim(U 1 П W 2 ¹ ) , q = q(X,Y ) = dim(U 2 A W 1 ¹ ) . Clearly that in the case of orbits in GL(4,K) we have 2 C r,s C 4 , 0 C p,q C 2 .

If a pair of 2-tori (X, Y ) in GL(4, K ) has invariants r = s = 4 we refer to it as undegenerate case and if at least one of invariants r or s less than 4 we refer to this pair as degenerate case. In this paper we calculate spans of degenerate cases. In the last work we treat with undegenerate cases.

3.    Degenerate Cases

Let X , Y be 2 -tori in GL(4 , K ) with centers U 1 , U 2 and axes W 1 , W 2 , respectively. Choose some bases in these subspaces, U1 = (u1,U2 ) , U2 = ( U 3/ U 4 ) • W 1 = (w1,W2 ) , W 2 = ( w 3 ,W 4 ) •

In the previous work [5] we described all orbits (X, Y ) under action by simultaneous conjugation: (gXg ^-1 ,gYg^-1 ) , g G G . For this purpose we listed all bases (u 1 ,U 2 ,U 3 ,U 4 ) and (w 1 ,W2,W3 , W 4 ) corresponding to different orbits. Summarize these results in the next lemma.

Lemma 1. Let X and Y be 2-tori in GL(4, K ). Assume that r = 2, 3 and r C s, then the orbit (X, Y ) is determined by one of the following bases.

For r = 2, s = 2, u1 = e1 , u2 = e2, w1 = f1 , w2 = f2,

(r2s2a) U 3 = 6 1 , U 4 = 62, W3 = f 1 , W 4 = f2.

For r = 2 , s = 3 ,

U1 = 6 1 , U2 = 6 2 , W1 = f1 + f3, W2 = f 2 + f

(r2s3a)

U3 = 61, U4 = 62, W3 = f1, W4 = f2, where X = 0 or 1.

For r = 2, s = 4,

U1 = e i , U2 = e 2 , W 1 = f l + f 3 , W 2 = f 2 + f 4,                  , _л .

(r2s4a)

и з = e i , U 4 = e 2 , W 3 = f l , W 4 = f 2 .

For r = 3, s = 3, ui = ei, U2 = e2, W1 = fl + f4, W2 = f2 + ef4,                / о о ч

(r3s3a) из = el + e3, U4 = e2 + Абз, W3 = fl, W4 = f2, where в € K, А = 0 or 1.

u i = e l , U2 = e 2 , W 1 = f l + af3, W2 = f2 + ef 3 ,               , „   ,

(r3s3b) из = ei + ез, U4 = 62 + Аез, W3 = fi, W4 = f2, where А = 0 or 1, a € K*, в € K •

For r = 3, s = 4, p = 0, ui = ei, U2 = e2, wi = fi + afз + f4, W2 = f2 + вfз,             , „ , ,

(r3s4a) из = ei + 6з, U4 = e2 + А6з, wз = fl, W4 = f2, where А = 0 or 1, a € K, в € K* •

For r = 3, s = 4, p = q = 1, ui = ei, U2 = e2, Wi = fi + А1fз + f4, W2 = f2 + А1А2fз,           . 1 1 .

(plqla) uз = ei + А262, u4 = 6з, Wз = fl, W4 = fз, where А1,2 = 0 or 1.

In fact there are other orbits. They are obtained in two ways from listed bases. The first one is to consider a pair (Y, X) instead of (X, Y ) . These bases (orbits) we denoted in [5] adding prime ‘ . It is clear that the span {Y, X ) is not changed. The second way is to interchange rows and columns. In this case each set of bases with invariants (r, s) , r < s , corresponds to a new set of bases with r > s . Then the new span {X, Y ) is obtained from old one by matrix transposition.

4.    Calculation Spans

In this section we prove our main results. They are listed in tables 1–3. Note that if the field K = F 2 a torus coincides with the identity matrix. Thus this case is excluded.

For the description of spans we use new notations throughout the rest of the paper. X Ykm = ^{x ij (Y i^ x km^ ), ^{E €} K } , X Y^ ,pq = {x ij ⁽ Yl ^E)x km ⁽ Y2 ^E ) ^x pq ^{( e ) E € K }} , Q ij = {d i (e)d j (e), e € K } , where Y t € K * , t = 1, 2 . Note that we miss Y i = 1 .

In all calculations we have deal with the generators of 2-tori X = { x(e), e € K ) and Y = { y(n), П € K ) . The generators x(e) and y(n) are constructed with the help of bases listed in Lemma 1. Sometimes we consider the generators conjugated to them. By H we denote the subgroup generated by x(e) and y(n) H = {x(E),y(n), E,n € K * ) .

As usually [g i ,g 2 ] = g i g 2 g ^-i g ^-i denotes a commutator of two elements g i and g 2 . Also put z^{( E,} n) = [x( ^E ),y(n)L zx&n^") = ^[z(E, n ^{),x (}O, z y (^n,^ = ^[z(E, n ^), y ⁽ ^ ^)] -

Now we are ready to formulate and prove our results. Note we do not write down the spans corresponding to cases with r > s . The reader can do it himself using remark at the end of the previous section.

Theorem 1. Let X, Y be a pair of 2-tori in GL(4,K), K = F 2 . Suppose that r = 2 and r C s, then up to simultaneous conjugation X and Y generate one of the following subgroup H, listed in Table 1.

Table 1: For r = 2 .

	base	H
1	(r2s2a)	Q 12
2	(r2s3a) , A = 1	Q 12 X 13 , 23
3	(r2s3a) , A = 0	Q 12 X 13
4	(r2s4a)	Q 12 X 13 , 24

<1 (1) For the base (r2s2a), we obtain directly that H = Q12 .

• (2)    For the base (r2s3a), the generators have the following form:

x(e) = d i (e)d 2 (e)x i3 (“~) X 23 (“““"^) ,   у ( п ) = di(n)d2(n).

Straightforward calculation shows that z(e,n) = Х 1з ( — О о )х 2з ( — “O o ), where O o = (e - 1)(n — 1) . Then from the decomposition of x(e) and y(n) we get that Q 12 ,X ^ 3 13 C H • Finally, we conclude that if “ = 1 , H = Q 12 X 13 , 23 .

• (3)    For the base (r2s3a), if “ = 0 , due to the previous argument, we obtain that H = Q 12 X 13 ^.

• (4)    For the base (r2s4a), by calculating z(e,n) , we directly deduce that the subgroups Q 12 and X 13 , 24 are contained in H . Thus we conclude that H = Q 12 X 13 , 24 . >

Theorem 2. Let X, Y be a pair of 2-tori in GL(4, K ), K = F 2 . Suppose that r = s = 3, then up to simultaneous conjugation X and Y generate one of the following subgroup H , listed in Table 2.

Table 2: For r = 3 , s = 3 .

	base	H
1	(r3s3a) , A = 0	Q 23 X 14 X 3 e 4 , 24 X 12
2	(r3s3a) , A = 1	Q 23 X 14 X 34 , 24 X 12 , 13
3	(r3s3b) а = - 1 , A = 1 , в = - 1 , Char K = 2 (r3s3b) , а = - 1 , A = 1 , в = 1 (r3s3b) , а = 1 , A = 1 , в = — 1 (r3s3b) , а = 0, - 1 , A = 1 , в = 0, - 1 , 1 + а + в = 0 , а + в = 0	Q 13 X 12 X 13 X 23
4	(r3s3b) , а = - 1 , A = 1 , в = - 1 , Char K = 2 (r3s3b) , а = - 1 , A = 1 , в = - 1,0,1 (r3s3b) , а = 0, - 1 ; A = 0 , в = 0 or A = 1 , в = 0 (r3s3b) , а = 0, - 1 , A = 0 , в = 0 (r3s3b) , а = - 1,0,1 , A = 1 , в = - 1 (r3s3b) , а = 0, - 1 , A = 1 , в = 0, - 1 , 1 + а + в = 0 , а + в = 0	GL(2 , K )
5	(r3s3b) , а = - 1 , A = в = 0 (r3s3b) , а = - 1 ; A = 0 , в = 0 or A = 1 , в = 0 (r3s3b) , а = 0, - 1 , A = 1 , в = 0, - 1 , 1 + а + в = 0	Q 13 Q 23 X 12

< (1) For the base (r3s3a), a simultaneous conjugation by the element W 12 W 23 leads to the following generators:

x(e) = d 2 (e)d 3 (e)x 24 (“~) X34 (“

-

E

—) ,   y(n) = d 2 (n)d 3 (n)x i2 (n - 1)x i3 (“(n - 1))-

Straightforward calculation shows that

z(e,n) = x i2

( - ^ 0 A    (-^0 A    ( -⁽¹ + ^e ^)^ 0 A (ay (вау

I ~£'П ) X13 V ~П^ X X14 \£------J X24(-^0 )X34(—в^о), where 0o = (e — 1)(n -1). Calculate a commutator subgroup generated by z(e, n), we get that X14 < H.

Suppose that Char K = 2 , then

.          n \           /—2(e — 1)(n — 1П    f—2в(е — 1)(n — 1П

^{(e n)z} (л 2^—7;= 4    <—2n——) x³ 4 к—2n——; -

/ у ( у        Г ^{— 2(e — 1)(n — 1) 2} A     Г ^{— 2A(e — 1)(n — 1) 2} A , .

z(e, ^n)z(£ , ^{2 — n)} = x ¹² ^     e^ — 2)    ) x¹³<---- 2----Г^{14 W}'

Suppose that Char K = 2, then zue)zU 1 + £ + n) = xuWxu (з-ц+т+П)) x13 (eM1+e + n)) - z(e,n)z(£,£)z (£,  ^    A = x14(*)x24 ( / ,    , A x34 ГТТ^ГГA ’

A en + £ + nJ             \e(n + 1) + -J \£{+ + 1) + -J where O1 = (e +1) (n(n + 1) + £2(n + 1) + e(n2 + 1)) • In all cases the subgroups Хв4 24, X^ 12 and X14 are contained in H. Thus we can conclude that if A = 0, H = Q23X14X34 24X12.

• (2)    For the base (r3s3a), if A = 1 , due to the previous argument, we obtain that H = ^Q 23 ^X 14 ^X 3 4 _{i 24} ^X 12 , 13 .

• (3)    For the base (r3s3b), a = — 1 , A = 1 , в = — 1 . Simultaneous conjugation by W 23 W 12 X 12O yields the following generators:

x(e) = d 1 (£)d 3 (e)x 12 (“7“} X 32 (“(“T ² )) - у(-) = d 1 (n)d 3 (n)x 23 (n — 1)-

If Char K = 2 , the group ( X, Y ) is embedded in a group of upper triangular matrices, and the generators have the following form:

x(e) = d 1 (e)d 3 (e)x 12

('

y(n) = d 1 (n)d 3 (n)x 23 (n — 1)-

(A)

A direct calculation reveals that z(e, n)

= x x2 (0 2 )x 13 ( '          X 23 ^j , where 0 2 =

(e + 1)(n + 1) . By calculating a commutator subgroup generated by z(e,n) , we get that

[z(e 1 , - 1 ),z(£ 2 ,n 2 )] coincides with the subgroup X 13 .

Calculate the following products:

z(£, n)z(£, £

)z £,

^{e(e +} “h A =     ( ^{(e + 1)(} n ^{+ 1)(e +} n ^)(e 2 ^{+ 1)}

e + e ² + n + £ 2 n /    ^12 \       £ + e ² n + e ² + n

A X 13 ⁽*^)-

₍        \ Г ^{1 + e} 2 ⁺ n ^{+ £} n A_ Г ^{(e + 1)(} n ^{+ 1)(e +} n ^)(e2 ^{+ 1)} A _n

^{z(e,n)z(e,e)z}           £ + 1       ) x ²³^ e ² n(1 + e ² + n + en)     J ^{X 1} 3 ⁽*^)-

From the above, we derive that X 12 , X 13 , X 23 are the subgroups of H . Thus H = Q 13 X 12 X 13 X 23 , if Char K = 2 . We will consider the case of Char K = 2 below.

• •    For the base (r3s3b), а = — 1 , A = 1 , в = 0, — 1 . Let в = 1 , the action of simultaneous conjugation by w i2 w i3 x i2 (1) results in the generators given below:

x(e) = d i (E)d 3 (e)x i2 ^““) , y(n) = d i (n)d 3 (n)x 23 (n — 1)•

Applying the similar arguments as to that for generators (A) yields that H = Q i3 X i2 X i3 X 23 .

• •    For the base (r3s3b), а = 1 , A =1 , в = — 1 . Conjugating this base by the element w i2 w i3 x i2 (1) , we have new generators

x(e) = d i (e)d 3 (e)x i2 (““) ’ у(п) = d i (n)d 3 (n)x 23 (n — 1)•

Similar to the generators (A), we obtain that H = Q 13 X 12 X 13 X 23 .

• •    For the base (r3s3b), а = 0, — 1 , A = 1 , в = 0, — 1 , 1 + а + в = 0 . Let а + в = 0 . Conjugating the generators by the element W 23 W 12 X 21 ( — 1)x i2 (1) produces the following new generators:

  x(e) = di (e)d3 (e)x i2 ^^^

  
  
  
  

  ε

  
  

  —) ,    y ⁽ n) = d i ⁽ n ^)d 3 ⁽ n ^)x 23 ⁽ n ^— “)•

  
  

An argument analogous to that for generators (A) shows that H = Q i3 X i2 X i3 X 23 .

• (4)    For the base (r3s3b), а = — 1 , A = 1 , в = — 1 . Suppose that Char K = 2, 3 , conjugating the original base by the element x i2 ( — 2 )w i2 x i2 (1) , we get the new generators

x(e) = d i (e)d 2 (e)x 23 (“(“у—)) , y(n) = d i (n)d 2 (n)x 32 (n — 1).

For E = 1 , n =1 , z x (e, 1/2,n) = X 23 ( ^_ 3( ^£ — 1)e(n — 1)) • We get that X 23 is contained in H . It follows from the decomposition of x(e) and y(n) that X 32 , Q 12 C H .

Define a map ф : Q 12 ^ { diag(a, 1) : a G K * } C GL(2, K) by ф diag(a, a, 1,1) = diag(a, 1) . It is easy to see that ф is a surjective and injective homomorphism. Moreover ( X 32 ,X 23 > = SL(2,K) . In fact, ({ diag(a, 1) : a G K * } , SL(2,K) ) = GL(2,K) , therefore ( X ₃₂ ,Q ₁₂ ,X ₂₃ ) = GL(2,K) . Finally we obtain that H = GL(2,K) , if Char K = 2, 3 .

If Char K = 3 , the original generators become the following form:

x‘(e) = di (e)d2(e)xi3 J 2      ') X23 Г ') , y'(n) = di (n)d2 (n)x3i(n — 1)x32(n — 1).

Conjugating this base by the element x 12 (1) w 12 x 12 (1) we get new generators

x(e) = d i (e)d 2 (e)x 23 ^““) ,  y(n) = di(n№(n)x32(n — 1)•

Put P (e, n) = x(e)y(n)x ( 22 П +2+2 П +21) • For 2 + e + П + en = 0 , 1 + 2e + n + 2en = 0 , 0 = 1 , we have

[P ^P (2,0)]= X32 (2E^

(20 + 1) + 2 (n ² + 2) (0 + 2) 2n + 2)(e(2n + 2) + 2n + 1)

.

It follows that X 32 is contained in H . Then we extract the subgroups X 23 and Q 12 from the decomposition of generators. Finally, we conclude that H = GL(2,K) , when Char K = 3 .

• •    For the base (r3s3b), a = - 1 , A = 1 , в = 0, — 1 . Let в = 1 . Conjugate by Х 2з ( 1 — e ) x32 ( ^в ¹ ) d 1 (в)w 13 x i2 ( в ) . Thus the generators become the following:

x(£) = d 2 (e)d 3 (фзх ^—— 1^ —^) , y(n) = d 2 (n)d 3 (n)x i3 (в(п - 1))-      (B)

Now suppose that Char K = 2 . Let в = 2 • For £ =1 , П = 1 , it follows directly from calculation that

z x '       ’ ⁿ) = x ³¹ (

£ (2 в — 1)( £ —

β

1)(n — 1) ) .

Let в = 2 , put

f ^(£, П) = z y ^^£, n,

1 + £ — n + £П — 1 + £ + n + £n

)

= d / (£n + £ — n + 1)(£n + £ + n — 1) A d /             4£ 2 n             A

A             4£ ² n             ) 3 2(£n + £ — n + 1)(£n + £ + n — 1) /

/ ( e ² — 1 ) (n — 1) 2 (£n + ^e — П + 1) (^£2П - £ 2 + 2£n + 2£ + n - 1 )\

^X x ³¹                         1&V                      ) ^-

Then we have [f (£ i ,n i ),f (£ 2 ,n 2 )] = X 31 .

In the case of characteristic 2, z _x (£,в(в + 1),£) = X 31 (£(£ + 1) ² /в) -

Next, from the decomposition of the generators we get that the subgroups X 13 , X 31 and Q 23 are contained in H . Similarly, we conclude that H = GL(2,K) .

• •    For the base (r3s3b), a = 0, — 1 , A = 0 , в = 0 . A simultaneous conjugation by the element w i3 W i2 X 2i ( a ) xi2 ( — д ) leads to the following generators:

  x(£) = d2 (£)d 3 (£)X 31 (—

  
  

  _

  
  

  ε

  
  

  —) ,   y ⁽ n) = d 2 ⁽ n ^)d 3 ⁽ n ^)x i3 ( ^p 1) ) -

  
  

An argument similar to the generators (B) gives that H = GL(2,K) .

• •    For the base (r3s3b), a = 0, — 1 , A = 1 , в = 0 , conjugating by the element x i3 (1)w 23 , we can extract the subgroups Q 13 , X 12 and X 21 and then H = GL(2,K) .

• •    For the base (r3s3b), a = 0, — 1 , A = 0 , в = 0 , the generators have the following form:

  x(£) = d1(£)d2 (£)x i3 ^^a( -

  
  
  
  

  ε

  
  

  —) ,    y ⁽ n) = d i ⁽ n ^)d 2 ⁽ n ^)x 3i ⁽ n ^— !)•

  
  

The similar calculations to the generators (B) yield H = GL(2, K) .

• •    For the base (r3s3b), a = 0, — 1,1 , A = 1 , в = — 1 . Performing conjugation by x i3 (a — 1)X 31 (1 — a)w 23 W i2 X i2 (a) gives rise to the generators listed below:

  (a

  x(£) = d i (£)d 3 (£)X 32 I ---

  
  
  
  

  ^^Оё —¹ )

  
  

  y(n) = d i (n)d 3 (n)x 23 (n — 1)-

  
  

Now suppose that Char K = 2. Let a = 2, for the element 6 G K, we have zx

ε,

α

a -

,θ

= ^x 32 ^{((2a — 1)£} ( ^{£ — 1)(6 —} 1)) -

Let а = 2 , after similar calculations to the case в = 1 of the base (r3s3b), а = - 1 , A = 1 , в = 0, - 1 , we have the subgroup X 23 .

Assume that Char K = 2 , z x (e, О ру , e) = X 32 (e(e + 1)²) • Thus in all cases, it follows that X 32 , Q 13 and X 23 are contained in H .

• •    For the base (r3s3b), а = 0, - 1 , A = 1 , в = 0, - 1 , 1 + а + в = 0 . Under condition а + в = 0 , conjugate this base by the element Х 2з ( - a + e ) x 32 ( ^в в а ) d 1 (в)w 13 X 12 ( - в ) • We have new generators

x(e) = d2 (e)d 3 (e)x 3i ( ^{(а +} ^^ ——) ,  y(e) = d 2 (e)d 3 (e)x 13 (в(е - 1)).

Similar to the generators (B), we can conclude that X 13 , Q 23 and X 31 are subgroups of H .

• (5)    For the base (r3s3b), а = - 1 , A = в = 0 . With the help of simultaneous conjugation by W 12 X 12 ( - 1)w 23 , the generators of the group X and Y have the following form:

  x(e) = d 2 (e)d 3 (e),

  
  

  y(n) = d i (n)d 3 (n)x i2

  
  

  (’Л

  
  

We calculate directly that z(e,n) = X 12 ( - (e - 1)(n - 1)/e), it follows from the decomposition of the generators that the subgroups Q 13 , X 12 and Q 23 are contained in H . Finally we can obtain that H = Q 13 Q 23 X 12 .

• •    For the base (r3s3b), а = - 1 , A = 0 , в = 0 . Conjugate this base by the element Х 1з ( - в)w 2з X 2з ( — 1)x 12 (в)w 13 W 23 . Then we get a new base

  ( а

  x(e) = d 1 (e)d 3 (e)x 32 I —

  
  

  _-

  
  

  ^---— ) ,   У ⁽ П) = d 1 ⁽ n ^)d 3 ⁽ П ^)х 23 ⁽ П ^- !)•

  
  

Then z(e,n) = X 23 ((e - 1)(n - 1)/e) • We consecutively get that the subgroups Q 12 , Q 13 and X 23 are contained in H . Then under conjugation of the element w 13 w 23 , we have that ^H = Q 13 Q 23 X 12 .

• •    For the base (r3s3b), а = - 1 , A = 1 , в = 0 . Conjugate this base by the element w 12 x 12 ( - 1)x 13 (1)w 23 . Then we get the same generators as in the base (r3s3b), а = - 1 , A = в = 0 .

• •    For the base (r3s3b), а = 0, - 1 , A = 1 , в = 0, - 1 , 1 + а + в = 0 . Simultaneous conjugation by Х 1з ( - 1 - а)х 21 (1)x 13 (1 + а)d 1 ( - 1 - а))w 1з X 12(1+ a) yields the following generators:

  ( ^!T^!)

  
  

x(e) = d2 (e)d 3 (e), y(n) = d 1 (n)d 3 (e)x 12

Direct calculation shows that z(e,n) = X 12 ((e - 1)(n - 1)/e) • We consecutively get that the subgroups Q 13 , Q 23 and X 12 are contained in H . Finally we can conclude that H = Q 13 Q 23 X 12 ^{. ⊲}

Theorem 3. Let X, Y be a pair of 2-tori in GL(4, K ), K = F 2 . Suppose that r = 3, s = 4, then up to simultaneous conjugation X and Y generate one of the following subgroup H , listed in Table 3. In cases (4) and (5) below, we also suppose that K = F 3 .

Table 3: For r = 3 , s = 4 .

	base	H
1	(r3s4a) , a = - 1 , A = 0 , в = 0	XYX 14 Х^ 3,4
2	(r3s4a) , a = 0 , A = 1 , в = — 1	Q 12 Q 23 X 13 X 24
3	(r3s4a) , a = 0, — 1 , A = 1 , в = 0, — 1 , 1 + a + в = 0	XYX 14 X 1S , 24
4	(r3s4a) , a = 0 , A = 0 , в = 0	Q^X^ X 23 Х Г з в 2 4 X 14
	(r3s4a) , a = — 1 , A = 1 , в = - 1 , Char K = 2
	(r3s4a) , a = — 1 , A = 1 , в = 1
	(r3s4a) , a = 1 , A = 1 , в = — 1
	(r3s4a) , a = 0, — 1 , A = 1 , в = 0, — 1 , 1 + a + в = 0 , a + в = 0
5	(r3s4a) , a = 0, — 1 , A = 0 , в = — 1	GL(2 , K ) X 24
	(r3s4a) , a = 0 , A = 1 , в = 0, — 1
	(r3s4a) , a = 0, — 1,1 , A = 1 , в = — 1
	(r3s4a) , a = 0, — 1 , A = 1 , в = 0, — 1 , 1 + a + в = 0 , a + в = 0
6	(r3s4a) , a = — 1 , A = 1 , в = — 1 , Char K = 2	GL(2 , K ) X 14
7	(r3s4a) , a = — 1 , A = 1 , в = 0, — 1,1	GL(2,K)X 2 1 4- - e 4
8	(p1q1a)	Q 12 Q 23 X 24

<1 (1) For the base (r3s4a), a = — 1 , A = 0 , в = 0 . Conjugate this base by the element W 12 X 31 ( - 1) . Then the generators of the group X and Y have the following form:

x(-) = d i (-)d 3 (-)x i2 ( ^{в ( -} _£ 1) ) X 13 (—

—

ε

~^ ^x23 ⁽¹ - ^-)x 24 ⁽ c - ^1)x 34 ^ — - —^ ,

y ⁽ n) = d i (n)d 2 ⁽ n)-

For -, n = 1 , we have

z(-,n) = xi3 (—^) xM    ‘) X23 ^ X24 (^o) , where Qo = (- — 1)(n — 1). We find that zx (е,П, -1/-) = х1з(в^з)х2з(—^з)х24(^з), where #3 = —(-2 — 1)(n - 1)/-. It follows that X^3 2З 24 C H. After it we get X14. Since Y commutes with X^3 2З 24, we conclude that H = XYX^X^ 23 24.

• (2)    For (r3s4a), A = 1 , a = 0 , в = — 1 . The generators obtaining after conjugation by X 24 ( — 1)x 34 (1)x 23 ( — 1)x i4 ( — 1)x 32 (1)x 2i ( — 1)w i3 are as follows:

  x(-) = d 2 (-)d 3 (-),

  
  

y(n) = d i (n)d 2 (n)x i3 ^П—~ ) X24 ^П—~^- ) •

For -,n = 1 , we have z(-,n) = x i3 ( — (- — 1)(n — 1)/-) X 24 ((- — 1)(n — 1)) • Put Ы-,П) = z(-,n)y ((-n — П + 1)/-) • We have that [f i (- i , n i ), f i (- 2 , П 2 )] = X 24 . Next we get X i3 . Finally, we conclude that H = Q i2 Q 23 X i3 X 24 .

• (3)    For the base (r3s4a), A = 1 , a = 0, — 1 , в = 0, — 1 , 1 + a + в = 0 . Conjugate this base by the element w 23 x 23 ( — 1)w i2 x i2 (1) . Then the generators of the group X and Y have the following form:

  x(-) = d i (e)d 3 (e)x i2 ^ ⁽1 ^{+ a} - ^(-—

  
  

—) X34 (-Г”) ’ У(п) = d i (n)d 2 (n)x 23 ^П—~ ) •

Simple calculations show that

/ x (^{— (1} + ^a) # 0 z(- n) = x i3 I -----------

A /(1 + a)(- — 1)0o A    f—#oA    f—#oA x14 ε x23 ε    x24    ε , where Oo = (e — 1)(n - 1). Moreover, we put f1(e,n) = z(e,n)y ((en - П + 1)/e) • And

[f i ( ^£ i , П 1 ), f i ( ^£ 2 , П 2 )] equals Хц. Next we get Х 1+ а з24 . Finally, we conclude that H =

XY X 14 X

1+ α 13 , 23 , 24 ^.

• (4)    For the base (r3s4a), а = 0 , A = 0 , в = 0 . Simultaneous conjugation by W 23 W 12 leads

to the following generators:

x(e) = d 1 (e)d 3 (e)x i2 в^^—^ ) X34 (^) ’

У(п) = d i (n)d 3 (n)x 23 (П ~ 1).

Performing the straightforward calculation, we obtain that z(e, n) equals

X 12 ( - S« « )X 13 ( °   '      ' ) X 14 ( — ^.T ^{1) O} 0 ) X 23 (^°) X 24 (^) W- O _O ).

where Oo = (e — 1)(n — 1) .

Next, suppose that Char K = 2 ,

[z(e, n), z(e, 3 — n)] = Х1з(в°4)х24(—O4), [z(e, n), z(n, e)] = Х14(в°5), where . _ (e — 1)2 (2n3 — 9n2 + 13n — 6)      _ 2(e — 1)2(n — 1)2(e — n)

° 4                                                    , °5                                          •

e(n — 3)n                             en

Take e = 1 , n = 1, 2 , 2, 3 , e = n . Hence for some ° G K * , Х 1з ( — в°)х 24 (°) and хц(0) lie in H .

On the other hand, for n =1, 2, | ,

/ \ ( nA        f^—2 ^{e (} n^{— 1) O}oA     f^{— 2(} n^{— 1) O}oA

^z(e ’ ^n)z V’ 2    . J = x ¹² к     2 n — 1 ) x ³⁴ к 2 n — 1 ;

X Х 1з ( — eq 1 (e,n))x 24 (q 1 (e,n))x 14 (q 2 (e,n)),

z(e,n)z (e,2 — n) = X13 (^ x24    °' xu °    2            ° ) , en        en        en        en(n - 2)

where q 1 (e, n) and q 2 (e, n) are rational functions which have pole at point n = 2 • But we know that X ₁ ^- ₃ ^β _{, 24} and X 14 are contained in H . Hence we can extract the elements of the groups X ₁ ^β _{2 , 34} and X 23 , after it we get the whole groups.

Let Char K = 2 . Then

^[z(e, n ^{),z(e, 1} + n ^)] = x 13

/ e(e + 1)°oA     /(e + 1)° o

V e(n + 1) r24^ e(n + 1)

[z f . , en Al         /^в ° 0 ( ^e 2 ⁽ n ^{+ 1)+ e} n ^{2 +} n ^{+ 1})\

L ^z(e,n),z(j ^{+ 1,} e +        x¹^          (e + 1) 2 n          J.

One readily calculates that

^z(e ' ^n)z(^e '■..

where ° 6 =

= Х 12 (в° б )х з4 (° б )х 1з ( — eg 1 (e, n))x 24 (g 1 (e, n))x 14 (g 2 (e, n)),

( η +1)( ε +1)( ε + η +1) ε ² + η +1

, g1(e,n) and g2(e,n) are rational functions which have poles at points e = —1, n = 1 + e2. Take arbitrary e and n except for e + n + 1 = 0, e = 1 and n = 1, we get all elements of the group X1β2,34 .

For e = 1 , 'n + n + 1 = 0 , multiplying z(e, n)z (e + 1, ^2 + 2 +ij by suitable elements from X^^ and X i4 , we can get an element X 23 (ⁱ ; +iaj+ ⁺ )4 ⁺ y ⁾) - Take arbitrary e and n except for e + n +1 = 0 and n = 1 , we get the subgroup X 23 . In all cases we get the subgroups X ^ 34 , X 23 , X]^^ , X i4 . Finally, we conclude that H = Q i3 X ^ 234 X 23 X !3 ^e 4 X i4 .

• •    For the base (r3s4a), a = - 1 , A =1 , в = - 1 . If Char K = 2 , conjugating this base by the element W 23 W i2 X i2 (1) , the group ( X, Y ) is embedded in a group of upper triangular matrices, and the generators have the following form:

x(e) = di(e)d_?J(e)xi2 (“'“') X 34 (■“) ,   У(п) = d i (nM(п)х 2з (п — 1)-

The generators are special case of the base (r3s4a), a = 0 , A = 0 , в = 0 . Let в = - 1 , we obtain that H = Q i3 X — 2¹34 X 23 X i3 , 24 X i4 , if Char K = 2 . We will consider the case of Char K = 2 below.

• •    For the base (r3s4a), a = - 1 , A = 1 , в = 0, - 1 . W 23 w i2 X i2 (1) leads to the following generators:

  Let в = 1 . Conjugation by the element

  
  

  y(n) = di (n)d 3 (п)х 2з (п - 1)-

  
  

x(e) = di (e)d 3 (e)xi2 (^} X34 (^} ,

The generators are special case of the base (r3s4a), a = 0, A = 0, в = 1, we have that

H = ^Q i 3 ^X i2 , 34 ^X 23 ^X _i3 , ₂₄ ^X i 4 .

• •    For the base (r3s4a), a = 0, - 1 , A = 1 , в = - 1 . If a = 1 , conjugating this base by the element W 23 W i 2 X 12 (1) , the group ( X, Y ) is embedded in a group of upper triangular matrices, and the generators are special case of the base (r3s4a), a = 0 , A = 0 , в = 0 . Let в = - 1 , we obtain that H = Q 3 X 3 , 24 X 4 X 23 X ^- _{2 , 34} .

• •    For the base (r3s4a), a = 0, - 1 , A = 1 , в = 0, - 1 , 1 + a + в = 0 . If a + в = 0 , conjugating by W 23 W i2 X 2i ( - 1)x 12 (1) , the generators become the following:

  x(e) = d i (e)d 3 (e)x i2 (-(-

  
  

  _-

  
  

  ε

  
  

  ") X" C -')'24 C -¹)'

  
  

  y(n) = di (n)d 3 (n)x 23 (П - 1)-

  
  

Direct calculation yields that z(e, n) equals

X i2 (aO o )x i3 ( ^a0 0 ⁽¹

+ ' ⁽ П ^- W εη

A     (Oo(a(e x14

ε

1) + £)\ x23

(^n°) x24 (-г) Х34(-ад’

where 0 0 = (e - 1)(n - 1) .

Suppose that Char K = 2 . Calculate the following product:

zc ■ ¹ - ’)’ ^z(^, ’ ²- ⁱ).

= X i3 ^a)

' ’I I'

we get that the subgroup X“3 24 is contained in H. Then for e + n = 3, e = n, we have z α ,ε ,z α ,η a+1      a+1

= X 13

/ ⁰ 0 ⁽ n ^{- e)} \ (a + 1)en

x 24

Oo(n - ') a(a + 1)'n

A     (0. '2

x 14

3e - n + 3n 2

(a + 1) ² en

.

Multiplying the product [z( 0 +1 , e) , z( 0 +1 ,П ] by suitable elememt of X ^ 24 we get the whole group X 14 . For e = 2, - 2 , e = 3a/(3a + 2) , we have

( e — 1^ ( e — 1^      ( a(E — 1)(e + 2A (

^{( e — 1)( e} + ²⁾ (E — 2) e ²

z

V ^E—)z[^E—2 )= x ¹³ — ( e — 2) e 2  ) x ²⁴ 4

/ (e — 1)(3a(E — 1) + 2e) \      / 2a(E — 1) A      / 2(e — 1) A

^X x ¹⁴ I-------2 2-------) x ¹² I 2 ) x ³⁴ <^—(7—2)7 ) ’

Multiplying the product z(e, ^^—1) z(e, f — 2) by suitable elements of X 1324 and X 14 , we can extract an element of the subgroup X ^— “ 34 . Next, multiplying z(E,n) by suitable elements of X 12 ° 34 , X 14 and X 13 24 , we extract an element X 23 ((e — 1)(n — 1)/еп) • From the decomposition of y(n) , we get the subgroups Q 13 and X 23 .

Note that the characteristic of the field plays role only in extracting the subgroup X^, for this reason for Char K = 2 it is sufficient to show that X ^— “ 34 is contained in H . In fact, multiplying the product z ( e, ^^ ) z ( e + 1, ^ 2 : ^—1 ) by suitable elements of X " ₃ 24 and X 14 , we can extract an element of the subgroup X ^— “ 34 . Therefore we conclude that H = Q 13 X 12 “ 4 X 14 X 23 X f3 , 24 , if a + в = 0 .

• (5)    For the base (r3s4a), a = 0, — 1 , A = 0 , в = — 1 . The generators are as follows:

X(E) = d1(E)d2(E)X13 (^4-4) X14 (^) X23 (^^ ,

у(П = d 1 (n)d 2 (n)x 31 (n — 1).

Now suppose that Char K = 2 , a = — 2 . Then z _x (e, ^a + ¹ ,n) equals

( ^{(1 + 2a)E} #o A     ( ^{(2a + 1) e}# o A    / # o ^{(1 — 2aE — e )} A

X13 ((2a + 1)e^g) X14 ( ------a------ ) X23 (--a------ ) X24 ( -------a2-------J , z. (  1  ,0+1 X = X13 ( _ A x14 (fclA x23 (4h l)A.

2a + 1   a               2a + 1         2a + 1        2a + 1

It follows that X 13 ° 14 23 is contained in H . Multiplying z_x ( e, 1 0" , n ) by suitable element from X 13 ° 14 23 , we can get an element X 24 ( — # o (2aE + E — 1)/a ² ) • Take e = — 1/(2a +1) , we obtain all elements of X 24 .

Suppose that a = — 2 , direct calculation shows that

Z x

= X 13 (1 — n)x 14 (2 — 2n)x 23 ( — 2 + 2n),

, _r 1 . - 1 therefore X 13 2 14 . 23

C H . Calculate the product:

zx (E, 3, n) x13(—2e#0)x24(—4e#0)x23(4E^0) = x24(—4 (2E2 — 3e + 1) (n — 1)), where 0o = (e — 1)(n — 1). Take e = 2, we get all elements of X24.

In the case of characteristic 2,

( a + 1 A ( a + 1 E2n + П + 1Л 4 A f^2 A f^2 A z. c—rz. a+1-—■ —E2—г x ' “r14 Ur23 Ur where #2 = (e + 1)(п +1). It follows that X"3 14 23 is subgroup of H. Multiplying zx (e, ^+1, n) by suitable element from X"3 14 23, we get an element of the subgroup X24. From the decomposition of the generators, we consecutively get Q13 and X31 .

Now describe the subgroup (Х “ з 14 2з , X 3₁ У Put t(e) = x ₁3 (e)x ₁₄ (e/a) х 2з (e/a) , s(n) = X 31 (n) . Note that the elements t(e) and s(n) lie in H . Consider the map ф from the group generated by t(e) and s(n) to SL(2,K) defined by ф(t(e)) = x ₁₂ (e) , ф(s(n)) = x ₂₁ (n)• It is clear that ф is a surjective homomorphism. Thus {t(e), s(n),e,n € K ) is isomorphic to Kerф * SL(2,K) .

We set w(^ ) = t(£)s( - £ ^-1 )t(£) , h(£) = w ( ^)w(1) ^-1 . It is clear that

Kerф = (w(£)t(e)w( — £)s(£ ² e), h(£)h(Z)h(£ 1 Z ¹ ), £, Z € K *} •

Direct calculation yields that Ker ф = X 24 . Then we get ( Х^ О^ з ,Х з1 ) is isomorphic to SL(2, K)X 24 . Therefore we conclude that H = GL(2,K)X 24 .

• •    For the base (r3s4a), a = 0 , A = 1 , в = 0, — 1 . Conjugating by x i2 (1) results in the generators given below:

  x(e) = d 1 (e)d 2 (e)x 13 ( ^в(£^ 1) ) X14 (“4) X 23 (—

  
  

  _—

  
  

  ε

  
  

  —) , y ⁽ n) = di ⁽ n)d2 ⁽ n ^)x 3i ⁽ n ^— 1)-

  
  

A minor variation of the argument of the base (r3s4a), a = 0, — 1 , A = 0 , в = - 1 establishes that H is generated by the subgroups Q 12 , X 24 , X 31 and X ^ 3 ^ >3 14 . Further we consider the subgroup generated by a pair of subgroups X 31 and X ^ 3 ^ >3 14 , a similar argument for < Х ₃₁ ,х в 3 ^в ₂3 , ₁₄ ) yields that H = GL(2,K)X ₂₄ .

• •    For the base (r3s4a), a = 0, — 1 , A = 1 , в = - 1 . Let a = 1 . Suppose that Char K = 2 , a simultaneous conjugation by the element X 12 (1) leads to the following generators:

  ( a x(e) = d1 (e)d 2 (e)x13 ( —

  
  

  _—

  
  

  ) ^X14 ^) X23 (M ’

  
  

y(n) = d 1 (n)d > (n)x 31 (n — 1)-

For a = 2, n € K*, straightforward calculations show that zx (e, ary , n) equals zz0 n \    ((2a — 1)e^oA     / (2a — 1)e^o\     / ^0((2a — 1)e — 1)\

X13 ((2a — 1)e0o) X14 --------:---- X23--:---- x>4--~(----- ,

A a — 1   / A a — 1   / A (a — 1)2/

(a — 1 A (     0o   A /(2a — 1)0oA zy [ —e,n) = x24 v (O—na) x31 к (a — 1)n),

/ 1 a A2       / 4(a — 1)(n — 1) A / 4(n — 1) A / 4(n — 1) A zxU •,— ' = x13к——r14<—ю—т2, where 0o = (e — 1)(n — 1). It follows that X^ 04 >3 ^ H, then from the decomposition of the generators and Zy ((a — 1)/a,e,n) we obtain Q12, X31 and X24. Straightforward calculations show that H = (Q12,x1374 j23,X31,X24) = GL(2,K)X24.

Suppose that a = 2 , simultaneous conjugation by X 12 (1) yields the following generators:

x(e) = d 1 (e)d 2 (e)x 13

(M X 14      ) X 23       •

y(n) = d 1 (n)d 2 (n)x 31 (n — 1)-

The calculation straightforwardly gives that Z y ( — 1,e,n) = X 24 (4(e — 1)(n — 1)) . Put

\ i \ ^{en — e} + ^{n + 1}\ z \     / \     / \ f^{— en} + ^{e — n + 1}A / \

f ( ^e , ^{n )} = ^x ( ^e)y j _{( g} + 1)( n — 1) ) '      g ^(e ’ ^П) = ^y(e) x к ( e — 1)( n + 1) ) ■'

Multiplying [f (e,n),f (n,£)] and [g(£, n), g(n, £)] by suitable element from X 24 , we get the ¹ , - 1

elements from X 31 and X-^ 14 23 . Thus we can conclude that H = GL(2, K )X 24 .

In the case of characteristic 2, we have

^r(e - ⁿ⁾ = z * (*■ a+r■ 0=x ¹³ " ^{x i} 4 (o+t) x ² 3 (ah) ^x 2 ⁴ (и?) ■

/   £ 2 + n ² « + П ² + « V     (  « 7   4    2      « 7      \    (     « 7      4

r(n’«)r\£’     £2 + T ) x13 U+tJ x14 к (a + T)(£ + tJ x23 <(a + T)(£ + T) J ’ where «2 = (£ + T)(n + T), «7 = £(n(« + T) + « + T) + n(n + T)(« + T). We get the subgroup X13+14 23, after it we get that X31 and Q12 are contained in H. In all case we can conclude that H = GL(2,K)X24.

• •    For the base (r3s4a), a = 0, — T , A = T , в = 0, — T , T + a + в = 0 . If a + в = 0 , conjugating this base by the element X 23 ( а + в ) x 32 ( ^а ++ ^в ) d 1 (в)w 13 X 12 ( - д ) , then generators have the following form:

x(£> = _{i!Hdl ( E )} X24 (-И) x₃1    '      ^T ) x₃₁ (£—2) ,

y ⁽ n) = ^d 2 ⁽ n)d3 ⁽ n)x13 ^(e( n - ^T ))-

Suppose that Char K = 2 , when a + в = — 2 , we have

Г ( a + в + TA ( a + в + T\1       (8Д / «8 A

[z v’    ■   ? z r ■   ) J = x31 Ы x34 lo+в), a + в          a + в             в«8

[z    ■ ■ ’£) 4a+T+T ’n) J =x13 (a+7) ’ where «8 = (e — T)(n — T)(2a + 2в + T)(e — n). When a + в = — 2, we put

^{f (£,} n) = ^z *

ε, η

— e ² n ² + £ 2 — 2en — 2e + n ² — 2n + T

’ e (e ² n ² — 2e ² n + £ 2 — 2en — 2£ — n ² + T)

,

g ^(£, n) = z y ^^£, n

en + £

,

£n + £ + n

^{n +}T).

Calculations [f(£ 1 ,n 1 ),f(£ 2 ’n 2 )] and [g(£ 1 ,n 1 ),g(£ 2 ’n2y] result in the subgroups X 13 and 1_

V ² в

^X 31 , 34 ^.

In the case of characteristic 2, for £ = n, « = (£ + T)(n + T), rz C a + ■л              41 = x31 /«(^4 x34        4,

L \ ^{a + в} /   \ ^{a + в}          у ^в / у ^{a + в} /

Ц ° AX ° Al = X 13 («C

[ у a + в + T / ya + в + T /J у a + в /

Therefore, in all cases it follows from the decomposition of the generators that the subgroups a + в

^X 13 , Q 23 , ^X ₃₁ ^e ₃₄ and X 24 are contained in H . As previous arguments, the subgroup

I          ^а+в \

(X 13 ,X 31 ^e 34^ is isomorphic to SL(2,K) , and then we obtain that H = GL(2,K)X 24 , if a + в = 0 .

• (6)    For the base (r3s4a), a = — 1 , A = 1 , в = — 1 - If Char K = 2, 3 , new generators become the following form:

x(e) = d i (e)d 3 (e)x i2 (1^) X 34 (^) X 32 (^^^) ,

y(n) = d 1 (n)d 3 (n)x 23 (n - 1).

Due to straightforward calculations, we find that

[z • (^-32’^e)       = x ¹² (v) x ³⁴ (^-23   x ³²   '3   ■

Zy (2,e,n) = xi4 (-4) . (300), where 0q = (e —

1 _- 1

1)(n — 1) . It follows that X Л’, 123432

is contained in H . We consecutively get

Q 13 and X 23 from the decomposition of the generators. Further, we can extract an element of X 14 through multiplying Z y (2,e,n) by suitable element from X 23 . Hence for some 0 G K * , we may get the whole group X 14 . _{1 1}

Now describe the subgroup ^X i^34 ² 32 ,X 23^ . Consider a map ф from the subgroup generated by t(e) and s(n) to SL(2,K) defined by ф(t(e)) = Х 2з (е) , ф(s(n)) = X 32 (n) . Note that t(e) = X 32 (e)x i2 (f) X 34 (—f) , s(n) = Х 2з (п) . A straightforward calculation shows that Ker ф = X 14 . Therefore we conclude that H = GL(2,K)X i4 , if Char K = 2, 3 .

If Char K = 3 , the original generators become the following form:

x‘(e) = di(e)d2(e)xi3 ^^-^) xi4 (^) X23 (2(^7^) , y'(n) = di(n)d2(n)x3i(n — 1)x32(n - 1)-

Conjugating this base by the element W 23 W i2 X i2 (1) we get new generators

x(e) = di(e)d3(e)xi2 (^^"^) x34 (^) X32 (^) ,

y(n) = d i (n)d 3 (n)x 23 (n - 1)-

Due to straightforward calculations, we find that z _x (e, 2, n) = xi4(2(2e +1)(2 + n)). It follows that X 14 is contained in H . Put

^Q(e, n) = x ^

en + 2e + 2n + 2

2en + 2e + n + 2

\            ( 2(e + 2) 2 (n + 2) \

) MxMx¹⁴ ( e(en + 2e + 2„ + 2)) ^-

Calculate the commutator

[^ ( ^^)] = X 23 ( + ) ^-

Thus we obtain the subgroup X 23 . It follows that from the decomposition of the generators we extract the subgroups X _i ² _{2 , 34 , 32} and Q i3 .

Now describe the subgroup X _i ² _{2 , 34 , 32} X 23 . As above argument, we may get X X 22 3 4 3 2 , X 23) is isomorphic to SL(2 , K )X i4 , and then conclude that H = GL(2 , K )X i4 , when Char K = 3 .

• (7)    For the base (r3s4a), а = - 1 , A = 1 , в = 0, — 1 . Let в = 1 , suppose that Char K = 2 . New generators are obtained with the help of conjugation by d i (e)w i3 X 12 (|) • Then

  x(e) = d 2 (e)d 3 (e)x 21

  
  

  () X34        ),

  
  

y(n) = d 2 (n)d 3 (n)xi2 ^—— -^—— ) Х 1з (в(п - 1))-

When в = 2 , for the element n € K * , by straightforward calculation, we see that

/ 1 в V / 4(n - 1) \    / 4(n - 1) \ zx U   ' ~1,n) = x21 V 2   ■ Г34 г 2    ?

(в -1   a /(2в - m a   /в(2в - m a / 0o a   (-0o \ zy In) = x121 —n— J x131 п(в -1) J x24'     ) x341~J ’ where ^o = (e - 1)(n - 1). It follows that X2i,34 C H• Note that we also get X2i,34, when

Char K = 3 . In fact, when Char K = 3 , we have

/ 1 в Л2_    (0 -1 A < 0 -1 a zx ^в + 2 ’в - 1’J x21 кв + J x34 V + J '

β- 1

Then from the decomposition of the generators we obtain Q23 and Х12в13. After it we have the β                                                                            β-1                         β subgroup X214-34. Straightforward calculations show that (X21,34, X12A3) - SL(2,K)X243β4, β and finally we conclude that H = GL(2, K)Х214з4.

Suppose that в = 2 , a simultaneous conjugation by the element W 12 d1(2)W 13 X 12 (2) leads to the following generators:

x(e) = d 1 (e)d 3 (e)x 12 ("Ц—) X 34 (“7^) ’ у(п) = d1(n№ (n)x 21 (“y^) X 23 (^2“) '

We have z _y ( - 1, n, 0) = x ₁₄ ( - 2(n - 1)(0 - 1))x 34 ( - 2(n - 1)(0 - 1))- Put

\      \ (^{(1 - e)(1 + n)}\ n               f \ Л^{1 + e)(1 - n)}\ f \

f ^(e ’ ⁿ⁾ = x ^(£)y I (e +1)(n - 1)} x ⁽ⁿ⁾ ’   g ^(e ’ ⁿ⁾ = ^y(e) x I(e - 1)(n + 1)) y ⁽ⁿ⁾ -

Multiplying [f (e,n),f (n,e)] and [g(e, n), g(n, e)] by suitable element from X 14 , 34 , we get the elements, x ₂₁ (0 9 )x ₂3 ( - 0 9 ) , x ₁₂ ( - 0 1o )x ₁₄ (0 1o ), where

^{( e} n ^{- 1) 2} (e ^{2 -} n ²)           ₀ =       ^{2( e} n ^{- 1) 2} (^e 2 ^- n ²)

(e - 1)e(e + 1)(n - 1)n(n + 1) ’     10    (e - 1)e(e + 1)(n - 1)n(n + 1) ^-

Take e,n = - 1 , en = 1 and e ² = n ² , we may get all elements of X 21123 and X 12114 . From the decomposition of the generators, we get that Q 13 and X ₁₂ ¹ _{, 34} are subgroups of H . And ( Q ₁₃ ’X ₁₂1₃₄ ’X 21123 ) = GL(2,K)X ₁₄ , ₃₄ . Thus we can conclude that H = GL(2, K)X ₁₄ , ₃₄ .

In the case of characteristic 2, we use the following generators:

x(e) = d 2 (e)d 3 (e)x 21

x«      )’

y(n) = d2(n)d3(n)x12 (——-^—— ) X1з(в(n - 1))-

Then we have r^ i) = . (^,e-n) = x12 (^) x13 ((в-ln) x24 (err) x34 (в) ■

,     / e2 + e + n20 + П2 + П0 + n\/

0 2 (0 + 1) εθ

\    ( в^ 2 (0 + 1)\

) x¹³< (в - iw/

r(n 0)r V’-----(^

β where 02 = (e + 1)(n + 1). We get the subgroup Х^^, after it we obtain X21,34 and Q23 '

are contained in H , and ( Q 23 , X 2134 , X 13 12 / = GL(2, K)X 24 34 . Thus we can conclude that

H = GL(2,K )X2^ .

• (8)    For the base (p1q1a). Put

g 1 = W 12 , g2 = X 12 ( - 1)X 24 (1)W 12 , g 3 = W 12 X 13 (1), g 4 = X 14 ( - 1)X 23 (1)X 12 ( - 1)W 12 •

We have four cases depending on the value of A i , i = 1, 2 . (I) A 1 = 0 , A 2 = 0 , (II) A 1 = 0 , A 2 = 1 , (III) A 1 = 1 , A 2 =0 and (VI) A 1 = 1 , A 2 = 1 . We conjugate the corresponding generators by g 1 , g 2 , g 3 and g 4 respectively, and get the new generators, which are upper triangular matrices. Then we calculate commutator subgroup z(e, n) generated by new generators x(e) and y(n) . Moreover, it follows from the decomposition of x(e) and y(n) that in all cases the subgroups Q 12 , Q 23 and X 24 are contained in H . Thus we conclude that H = Q 12 Q 23 X 24 . >