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Automorphisms and isomorphisms of Chevalley groups of typ e G2 over lo cal rings with 1/2 and 1/3
E. I. Bunina
M.V. Lomonosov Moscow State University Russia, 119992, Moscow, Leninskie Gory, Main Building of MSU, Faculty of Mechanics and Mathematics, Department of Higher Algebra email address: helenbunina@yandex.ru Abstract. We prove that every isomorphism of Chevalley groups of type G2 over commutative local rings with 1/2 and 1/3 is standard, i. e., it is a composition of a ring isomorphism and an inner automorphism. Key words: Chevalley groups, local rings, isomorphisms and automorphisms Introduction An associative commutative ring R with a unit is called local, if it contains exactly one maximal ideal (that coincides with the radical of R). Equivalently, the set of all non-invertible elements of R is an ideal. We describe automorphisms of Chevalley groups of type G2 over local rings with 1/2 and 1/3. Note that for the root system G2 there exists only one weight lattice, that is simultaneously universal and adjoint, therefore for every ring R there exists a unique Chevalley group of type G2 , that is G(R) = G ad (G2 , R). Over local rings universal Chevalley groups coincide with their elementary subgroups, consequently the Chevalley group G(R) is also an elementary Chevalley group. Theorem 1 for the root systems Al , Dl , and El was obtained by the author in [12], in [14] all automorphisms of Chevalley groups of given types over local rings with 1/2 were described. Theorem 1 for the root systems B2 and G2 is obtained in the paper [13], but we repeat it here for the root system G2 with an easier proof. Similar results for Chevalley groups over fields were proved by R. Steinberg [50] for the finite case and by J. Humphreys [36] for the infinite case. Many papers were devoted to description of automorphisms of Chevalley groups over different commutative rings, we can mention here the papers of Borel­Tits [10], Carter­Chen Yu [16], Chen Yu [17]­[21], A. Klyachko [40]. 1The work is supported by the Russian President grant MK-2530.2008.1 and by the grant of Russian Fond
of Basic Research 08-01-00693.
1

1


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E. Abe [1] proved that all automorphisms of Chevalley groups under Noetherian rings with 1/2 are standard. The case Al was completely studied by the papers of W.C. Waterhouse [63], V.M. Petechuk [43], Fuan Li and Zunxian Li [39], and also for rings without 1/2. The paper of I.Z. Golubchik and A.V. Mikhalev [30] covers the case Cl , that is not considered in the present paper. Automorphisms and isomorphisms of general linear groups over arbitrary associative rings were described by E.I. Zelmanov in [68] and by I.Z. Golubchik, A.V. Mikhalev in [31]. We generalize some methods of V.M. Petechuk [44] to prove Theorem 1. The author is thankful to N.A. Vavilov, A.A. Klyachko, A.V. Mikhalev for valuable advices, remarks and discussions. 1. Definitions and main theorems. We fix the root system of the type G2 (detailed texts about root systems and their properties can be found in the books [37], [11]). Let e1 , e2 , e3 be an orthonorm basis of the space R3 . Then we numerate the roots of G2 as follows: 1 = e1 - e2 , 2 = -2e1 + e2 + e3 are simple roots; 3 = 1 + 4 = 21 + 5 = 31 + 2 = 6 = 31 + 22 = 2 = e3 - e1 , 2 = e3 - e2 , e1 + e3 - 2e2 , 2e3 - e1 - e2

are other positive roots. Suppose now that we have a semisimple complex Lie algebra L of type G2 with Cartan subalgebra H (detailed information about semisimple Lie algebras can be found in the book [37]). Then in the algebra L we can choose a Cheval ley basis {hi | i = 1, 2; x | } so that for every two elements of this basis their commutator is an integral linear combination of the elements of the same basis. Namely, 1) [hi , hj ] = 0; 2) [hi , x ] = i , x ; 3) if = n1 1 + · · · + n4 4 , then [x , x- ] = n1 h1 + · · · + n4 h4 ; 4) if + , then [x , x ] = 0; / 5) if + , and , are roots of the same length, then [x , x ] = cx+ ; 6) if + , is a long root, is a short root, then [x , x ] = ax+ + bx+2 + . . . . Take now an arbitrary local ring with 1/2 and 1/3 and construct an elementary adjoint Chevalley group of type G2 over this ring (see, for example [49]). For our convenience we briefly put here the construction. In the Chevalley basis of L all operators (x )k /k ! for k N are written as integral (nilpotent) matrices. An integral matrix also can be considered as a matrix over an arbitrary commutative


3

ring with 1.Let R be such a ring. Consider matrices n â n over R, matrices (x )k /k ! for , k N are included in Mn (R). Now consider automorphisms of the free module Rn of the form exp(tx ) = x (t) = 1 + tx + t2 (x )2 /2 + · · · + tk (x )k /k ! + . . . Since all matrices x are nilpotent, we have that this series is finite. Automorphisms x (t) are called elementary root elements. The subgroup in Aut(Rn ), generated by all x (t), , t R, is called an elementary adjoint Cheval ley group (notation: E ad (, R) = E ad (R)). In an elementary Chevalley group there are the following important elements: -- w (t) = x (t)x- (-t-1 )x (t), , t R ; -- h (t) = w (t)w (1)-1 . The action of x (t) on the Chevalley basis is described in [15], [58], we write it below. Over local rings for the root system G2 all Chevalley groups coincide with elementary adjoint Chevalley groups E ad (R), therefore we do not introduce Chevalley groups themselves in this paper. We will work with two types of standard automorphisms of Chevalley groups G(R) and G(S ) and with one unusual, "temporary" type of automorphisms. Ring isomorphisms. Let : R S be an isomorphism of rings. The mapping x (x) from G(R) onto G(S ) is an isomorphism of the groups G(R) and G(S ), it is denoted by the same letter and called a ring isomorphism of the groups G(R) and G(S ). Note that for all and t R the element x (t) is mapped to x ((t)). Inner automorphisms. Let g G(R) be an element of a Chevalley group under consideration. Conjugation of the group G(R) with the element g is an automorphism of G(R), that is denoted by ig and is called an inner automorphism of G(R). These two types of automorphisms are called standard. There are central and graph automorphisms, which are also standard, but in our case (root system G2 ) they can not appear. Therefore we say that an isomorphism of groups G(R) and G(S ) is standard, if it is a composition of two introduced types of isomorphisms. Besides that, we need also to introduce temporarily one more type of automorphisms: Automorphisms­conjugations. Let V be a representation space of the Chevalley group G(R), C GL (V ) be a matrix from the normalizer of G(R): C G(R)C
-1

= G(R).

Then the mapping x C xC -1 from G(R) onto itself is an automorphism of the Chevalley group, which is denoted by i and is called an automorphism­conjugation of G(R), induced by the element C of the group GL (V ). In Section 5 we will prove that in our case all automorphisms­conjugations are inner, but the first step is the proof of the following theorem: Theorem 1. Suppose that G(R) = G(, R) and G(S ) = G(, S ) are Cheval ley groups of type G2 , R, S are commutative local rings with 1/2 and 1/3. Then every isomorphism of the groups G(R) and G(S ) is a composition of a ring isomorphism and an automorphism­conjugation. Sections 2­4 are devoted to the proof of Theorem 1.


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2. Changing the initial isomorphism to a special isomorphism. In this section we use some arguments from the paper [44]. Definition 1. By GL n (R, J ) we denote the subgroup of such matrices A from GL n (R), that satisfy A - E Mn (J ), where J is the radical of R. Prop osition 1. By an arbitrary isomorphism can construct an isomorphism = ig-1 , g some subgroup GL n (S ), with the property that subring of R, generated by 1, is mapped under between Cheval ley groups G(R) and G(S ) we GLn (S ), of the group G(R) GL n (R) onto any matrix A G(R) with elements of the to the matrix from A · GL n (S, JS ).

Proof. Let JR be the maximal ideal (radical) of R, k the residue field R/JR . Then the group G(R, JR ) generated by all x (t), , t JR , is the greatest normal proper subgroup in G(R) (see [2]). Therefore under the action of the group G(R, JR ) is mapped to G(S, JS ). By this reason the isomorphism : G(R) G(S ) induces an isomorphism : G(R)/G(R, JR ) = G(R/JR ) G(S/JS ). The groups G(R/JR ) and G(S/JS ) are Chevalley groups over fields, so that the isomorphism is standard, i. e., it is = ig , g G(S/JS ) (see [49], § 10). Clear that there exists a matrix g GL n (S ) such that its image under factorization S by JS is g . Note that it is not necessary g N (G(S )). Consider the mapping = ig-1 . It is an isomorphism of the group G(R) GL n (R) onto some subgroup in GL n (S ), with the property that its image under factorization the rings by their radicals is the isomorphism . Since the isomorphism acts identically on matrices with all elements generated by the unit of k , we have that ane matrix A G(R) with elements from the subring of R, generated by 1, is mapped under the action of to some matrix from the set A · GL n (S, JS ).
1 Let a G(R), a2 = 1. Then the element e = 2 (1 + a) is an idempotent of the ring Mn (R). This idempotent e defines a decomposition of the free R-module V = Rn :

V = eV (1 - e)V = V0 V1 (the modules V0 , V1 are free, because every pro jective module over a local ring is free). Let 1 V = V 0 V 1 be a decomposition of the k -module V with respect to a, and e = 2 (1 + a). Then we have Prop osition 2. Modules (subspaces ) V 0 , V 1 are the images of V0 , V1 under factorization by J . Proof. Denote the images of V0 , V1 under factorization R by J by V0 , V1 , respectively. Since V0 = {x V | ex = x}, V1 = {x V | ex = 0}, we have e(x) = 1 (1 + a)(x) = 1 (1 + a(x)) = 2 2 1 (1 + a(x)) = e(x). Then V0 V 0 , V1 V 1 . 2


5

Let x = x0 + x1 , x0 V0 , x1 V1 . Then e(x) = e(x0 ) + e(x1 ) = x0 . If x V0 , then x = x0 . Let b = (a). Then b2 = 1 and b is equivalent to a modulo J . Prop osition 3. Suppose that a G(R), b G(S ) a2 from the subring of R, generated by the unit, b and elements of a are generated by unit, we can consider it is a decomposition of V with respect to a, V = V0 V1 Then dim V0 = dim V0 , dim V1 = dim V1 . = b2 = E , a is a matrix with elements a are equivalent modulo JS (since al l as an element of Mn (S )), V = V0 V1 is decomposition of V with respect to b.

Proof. We have an S -basis of V {e1 , . . . , en } such that {e1 , . . . , ek } V0 , {ek+1 , . . . , en } V1 . Clear that
n n

aei = aei = (
j =1

aij ej ) =
j =1

aij ej .

Let V = V 0 V 1 , V = V 0 V 1 be decompositions of k = S/JS -module (space) V with respect to a and b. Clear that V 0 = V 0 , V 1 = V 1 . Therefore, by Proposition 2, the images of V0 and V0 , V1 and V1 under factorization by JS coincide. Take such {f1 , . . . , fk } V0 , {fk+1 , . . . , fn } V1 , that f i = ei , i = 1, . . . , n. Since a matrix that maps the basis {e1 , . . . , en } to {f1 , . . . , fn } is invertible (it is equivalent to the unit matrix modulo JS ), we have that {f1 , . . . , fn } is an S -basis of V . Cleat that {f1 , . . . , fk } is an S -basis of V0 , {vk+1 , . . . , vn } is an S -basis of V1 . 3. Images of w
i

Consider Chevalley groups G(R) and G(S ) of type G2 , their adjoint representations in the groups GL 14 (R) and GL 14 (S ), in bases of weight vectors v1 = x1 , v-1 = x-1 , . . . , v6 = x6 , v-6 = x-6 , V1 = h1 , V2 = h2 , corresponding to the Chevalley basis of G2 . We suppose that by the isomorphism we constructed the isomorphism = ig-1 , described in the previous section. Recall that it is an isomorphism of the group G(R) GL n (R) on some subgroup of GL n (S ), with the property that its image under factorization ring by their radical is a ring isomorphism . Consider the matrices h1 (-1), h2 (-1) in G(R). They are h1 (-1) = diag [1, 1, -1, -1, -1, -1, -1, -1, -1, -1, 1, 1, 1, 1], h2 (-1) = diag [-1, -1, 1, 1, -1, -1, 1, 1, -1, -1, -1, -1, 1, 1]. By Proposition 3 we know that every matrix hi = (hi (-1)) in some basis is diagonal with ±1 on its diagonal, and the number of 1 and -1 coincides with their number for the matrix hi (-1). Since h1 and h2 commute, there exists a basis, where h1 and h2 have the same form as h1 (-1) and h2 (-1). Suppose that we came to this basis with the help of the matrix g1 . Clear that g1 GL n (S, JS ). Consider the mapping 1 = i-1 . It is also an isomorphism of g1 the group G(R) onto some subgroup of GL n (S ) such that its image under factorization rings by their radicals is , and 1 (hi (-1)) = hi (-1) for i = 1, 2. Instead of we now consider the isomorphism 1 .


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Remark 1. Every element wi = wi (1) moves by conjugation hi to each other, therefore its image has a block-monomial form. In particular, this image can be rewritten as a block-diagonal matrix, where the first block is 12 â 12, and the second is 2 â 2. Consider the first basis vector after the last basis change in GL 14 (S ). Denote it by e. The Weil group W acts transitively on the set of roots of the same length, therefore for every root i of the same length as the first one, there exists such w(i ) W , that w(i ) 1 = i . Similarly, all roots of the second length are also conjugate under the action of W . Let k be the first root of the length that is not equal to the length of 1 , and let f be the k -th basis vector after the last basis change. If j is a root conjugate to k , then let us denote by w(j ) an element of W such that w(j ) k = j . Consider now the basis e1 , . . . , e14 , where e1 = e, ek = f , for 1 < i 12 either ei = 1 (w(i ) )e, or ei = 1 (w(i ) )f (it depends of the length of k ); we do not move e13 , e14 . Clear that the matrix of this basis change is equivalent to the unit modulo radical. Therefore the obtained set of vectors also is a basis. Clear that the matrices 1 (wi ) (i = 1, 2) on the basis part {e1 , . . . , e12 } coincide with the matrices for wi in the initial basis of weight vectors. Since hi (-1) are squares of wi , then there images are not changed in the new basis. Besides, we know (Remark 1) that every matrix 1 (wi ) is block-diagonal up to decomposition of basis in the first 12 and last 2 elements. Therefore the last part of basis consisting of 2 elements, can be changed independently. Denote the matrices wi and 1 (wi ) on this basis part by wi and 1 (wi ) respectively, and the 2-generated module generated by e13 and e14 , by V . Lemma 1. For the root system G2 there exists such a basis that 1 (w1 ) and 1 (w2 ) in this basis are w1 and w2 . Namely, they are equal to -1 3 01 and 10 . 1 -1

Proof. Since w1 is an involution and V11 has dimension 1, there exists a basis {e1 , e2 }, where 1 (w1 ) has the form diag [-1, 1]. In the basis {e1 , e2 - 3/2e1 } the matrix has the obtained form for G2 . Let the matrix 1 (w2 ) in this new basis be ab . cd Make the basis change with the help of 1 (1 - a)/c - 0 1 + 2(1c a) (it is possible, since c is equivalent to the unit modulo radical). Under such basis change the matrix 1 (w1 ) is not change and the matrix 1 (w2 ) becomes 1b cd .


7

Since this last matrix is an involution, we have c (1 + d ) = 0, 1 + b c = 1. Therefore, d = -1, b = 0. Besides, -1 1 01 10 c -1
3

=

10 c -1

-1 1 01

3

,

consequently 3c (3c - 1)(c - 1) = 0. Since c 1 mod J , 3c - 1 2 mod J , 3 R , we have c - 1 = 0. Therefore we can now come from the isomorphism 1 under consideration to an isomorphism 2 with all properties of 1 and such that 2 (w1 ) = w1 , 2 (w2 ) = w2 . We suppose now that an isomorphism 2 with all these properties is given. 4. Images of xi (1) and diagonal matrices. Recall that w1 = -e1,2 -e2,1 +e3,9 +e4,10 -e9,3 -e10,4 +e5,7 +e6,8 -e7,5 -e8,6 +e11,,11 +e12,12 -e13,13 +e14,14 +3e13,14 ; w2 = -e3,4 -e4,3 +e1,5 +e2,6 -e5,1 -e6,2 +e7,7 +e8,8 +e9,11 +e10,12 -e11,9 -e12,10 + e13,13 -e14,14 +e14,13 ; x1 (1) = E - e1,2 - 2e1,13 + 3e1,14 - e4,6 - e4,8 - e4,10 + 3e5,3 + 2e6,8 + 3e6,10 - 3e7,3 - 2e7,5 + 3e8,10 + e9,3 + e9,5 - e9,7 + e13,2 ; x2 (1) = E + e2,6 - e3,4 + e3,13 - 2e3,14 - e5,1 + e10,12 - e11,9 + e14,4 . - - Since x1 = 2 (x1 (1)) commutes with h1 (-1) and with w31 +22 = w2 w1 w2 w1 1 w2 1 , we have that x1 can be decomposed to the blocks {v1 , v-1 , v6 , v-6 , V1 , V2 } and {v2 , v-2 , v3 , v-3 , v4 , v-4 , v5 , v-5 }; on the first block the matrix is y y1 y2 y3 -y3 y4 - 324 y y5 y6 y7 -y7 y8 - 328 y9 y10 y11 y12 y13 y14 ; -y9 -y10 y12 y11 -y13 3y13 + y14 3(y20 -y18 ) y15 y16 y17 y17 + 3y19 y18 2 0 0 y19 y19 0 y20 on the second block it is y21 y29 y37 y45 y52 y44 y36 y28 y22 y30 y38 y46 y51 y43 y35 y27 y23 y31 y39 y47 y50 y42 y34 y26 y24 y32 y40 y48 y49 y41 y33 y25 -y25 -y33 -y41 -y49 y48 y40 y32 y24 -y26 -y34 -y42 -y50 y47 y39 y31 y23 -y27 -y35 -y43 -y51 y46 y38 y30 y22 -y28 -y36 -y44 -y52 y45 y37 y29 y21 .

- - Similalrly, since x2 = 2 (x2 (1)) commutes with h2 (-1) and w21 +2 = w1 w2 w1 w2 1 w1 1 , we have decomposition of x2 on the blocks {v1 , v-1 , v3 , v-3 , v5 , v-5 , v6 , v-6 } and {v2 , v-2 , v4 , v-4 , V1 , V2 };


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on the first block the matrix is z1 z9 -z12 -z4 z17 z 25 -z 28 -z20

z2 z10 -z11 -z3 z18 z26 -z27 -z19

z3 z11 z10 z2 z19 z27 z26 z18

z4 z5 z6 z12 z13 z14 z9 -z16 -z15 z1 -z8 -z7 z20 z21 z22 z28 z29 z30 z25 -z32 -z31 z17 -z24 -z23

z7 z15 z13 z6 z23 z31 z30 z22

z8 z16 z14 z5 z24 z32 z29 z21

;

on the second block it is z33 z34 -z35 z35 z38 -z39 z39 z37 z42 z43 z44 z41 z44 z43 -z41 -z42 0 0 z47 + z48 z47 + z z51 z52 z47 z48

48

z36 2z36 z40 -2z40 z45 z46 . z45 + z46 -z46 2z49 + z50 0 z49 z50

Therefore we have 104 variables y1 , . . . , y52 , z1 , . . . , z52 , where y1 , y6 , y11 , y16 , y18 , y20 , y21 , y30 , y34 , y36 , y39 , y48 , z1 , z10 , z12 , z21 , z30 , z32 , z33 , z36 , z38 , z43 , z50 , z52 are equivalent to 1 modulo radical, y2 , y32 , z34 are equivalent to -1, y4 , y50 , are equivalent to -2, y37 , -y52 are equivalent to 3, all other elements are from the radical. We apply step by step four basis changes, commuting with each other and with all matrices wi . These changes are represented by matrices C1 , C2 , C3 , C4 . Matrices C1 and C2 are blockdiagonal, with 2 â 2 blocks. On all 2 â 2 blocks, corresponding to short roots, the matrix C1 is unit, on all 2 â 2 blocks, corresponding to long roots, it is 1 -z51 /z -z51 /z52 1
52

.

On the last block it is unit. Similarly, C2 is unit on the blocks corresponding to long roots, and on the last block. On the blocks corresponding to the short roots, it is 1 -y15 /y -y15 /y16 1
16

.

Matrices C3 and C4 are diagonal, identical on the last block, the matrix C3 is identical on all places, corresponding to short root, and scalar with multiplier a on all places corresponding to long roots. In the contrary, the matrix C4 , is identical on all places, corresponding to long roots, and is scalar with multiplier b on all places, corresponding to short roots. Since all these four matrices commutes with all wi , i = 1, 2, then after basis change with any of these matrices all conditions for elements x1 and x2 still hold. At the beginning we apply basis changes with the matrices C1 and C2 . After that new y15 in the matrix x1 and z51 in the matrix x2 are equal to zero (for the convenience of notations we do not change names of variables). Then we choose a = -1/z52 (it is new z52 ) and apply the third basis change. After it z52 in the matrix x1 becomes to be 1. Clear that z51 is still zero.


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Finally, apply the last basis change with b = 1/y16 (where y16 is the last one, obtained after all previous changes). We have that y15 , z51 , z52 are not changed, and y16 is now 1. Now we can suppose that y15 = 0, y16 = 1, z51 = 0, z52 = 1, and we have 100 variables. - - Introduce x1+2 = 2 (x1 +2 (1)) = w2 x1 w2 1 , x1+1+2 = 2 (x21 +2 (1)) = w1 x1+2 w1 1 , x1+1+1+2 = - - 2 (x31 +2 (1)) = w1 x2 w1 1 , x1+1+1+2+2 = 2 (x31 +22 (1)) = w2 x1+1+1+2 w2 1 . Now we will use the following conditions, that are true for elements wi and xi : C on1 = (x2 x
1+2

=x

1+2 x2

),

C on2 = (h1 x2 h1 x2 = E ); C on3 = (h2 x1 h2 x1 = E ); C on4 = (x2 x C on5 = (x2 x C on6 = (x C on8 = ( C on7 = (x1 x
3 w1 1+1+2

=x

1+1+2 x2

); ); ; );

1+1+1+2

=x

1+1+1+2+2 x1+1+1+2 x2 1+1+1+2

1+1+1+2 x1 1+1+2

= x1 x

=x

= x1 w

3 1+1+1+2 x1+1+2 x1 3 1 x1 w1 x1 ).

Note that every matrix condition is 196 polynomial identities, where all polynomials have integer coefficients and depends of yi , zj . Temporarily numerate all variables as v1 , . . . , v100 . Suppose that one of our polynomials can be rewritten in the form (vk0 - v k0 )A + v1 B1 + · · · + v
k0 - 1

B

k0 -1

+v

k0 +1

B

k0 +1

+ ··· + v

100

B

100

= 0,

where v k0 is such an integer number that is equivalent to vk0 modulo radical, the polynomial A in invertible modulo radical, Bi are some polynomials (the variable vk0 can enter in all polynomial and also in A). Then + vk0 +1 Bk0 +1 + · · · + v100 B100 , A we can substitute the expression for vk0 in all other polynomial conditions. If we can choose 100 such conditions that on every step we except one new variable, then on the last step we obtain the expression (vk100 - v k100 )C = 0, where C is some rational expression of variables v1 , . . . , v100 , invertible modulo radical. Therefore, we can say that vk100 = v k100 , and consequently all other variables are equal to the integer numbers equivalent them modulo radical. The existence of the obtained 100 conditions is equivalent to the existence of such 100 conditions that the square matrix consisting of all coefficients of these conditions modulo radical has an invertible determinant. Since it is very complicated to write a matrix 100 â 100, we will sequentially take the obtained equations, but for simplicity write coefficients A and Bi modulo radical (in the result these coefficients are just numbers 0, ±1, ±2, ±3). We write below how the variables are expressed from the conditions (in brackets we write the number of the condition and the position there): (C on1, 14, 3): y22 = 0; (C on1, 6, 10): y3 = 0; (C on1, 1, 1): y47 = -z7 ; (C on1, 1, 3): y46 = 0; (C on1, 1, 6): y40 = -z3 ; (C on1, 1, 7) y49 = -z7 + 3z39 ; (C on1, 1, 9): y43 = 0; (C on1, 1, 11): y51 = 0; (C on1, 2, 1): z15 = -3z20 ; v
k
0

=-

v 1 B1 + · · · + v

k0 -1

B

k0 - 1


10

(C on1, 2, 3): y41 = 0; (C on1, 2, 5): y5 = -3z18 ; (C on1, 2, 7): z44 = -3/2z20 ; C on1, 2, 9): y7 = -3z24 ; (C on1, 3, 1): z20 = y23 + 2z35 - z9 ; (C on1, 3, 3): z41 = 0; (C on1, 3, 5): z18 = z11 ; (C on1, 4, 1): y23 = -2z39 ; (C on1, 4, 2): z37 = -y24 ; (C on1, 5, 3): y35 = 0; (C on1, 7, 5): z3 = 0; (C on1, 11, 5): y9 = 0; (C on1, 11, 3): y24 = 0; (C on1, 10, 3): y27 = 0; (C on1, 9, 3): y38 = 0; (C on1, 4, 5): z11 = 0; (C on1, 5, 5): z2 = 0; (C on2, 3, 3): z33 = 1; (C on2, 14, 4): z50 = z38 ; (C on2, 11, 5): z19 = 0; (C on2, 3, 13): z38 = 1 - z40 ; (C on2, 10, 5): z27 = 0; (C on2, 5, 5): z10 = 1; (C on2, 6, 6): z1 = 1; (C on2, 1, 9): z7 = 0; (C on1, 11, 6): z26 = y10 + z4 + y33 ; (C on2, 7, 7): z43 = 1; (C on2, 9, 9): z23 = 2(z21 - 1); (C on2, 11, 12): z31 := -2z29 - z24 ; (C on3, 1, 1): y1 = 1; (C on3, 1, 2): y4 = 1 + 2y2 - y6 ; (C on3, 2, 2): y8 = 2(y6 - 1); (C on3, 4, 4): y30 = 1; (C on3, 14, 14): y20 = 1; (C on3, 13, 13): y18 = y6 ; (C on3, 12, 12): y11 = 1; (C on1, 13, 11): z6 = 0; (C on4, 2, 2): y42 = 6z39 - 6z35 + 3z9 + 3z10 + 3z4 + 3y33 ; (C on2, 5, 6): z4 = -2z9 ; (C on2, 9, 11): z21 = 1; (C on2, 12, 11): z22 = 0; (C on2, 10, 10): z30 = 1; (C on2, 14, 7): z39 = 0; (C on2, 11, 1): y33 = 2y28 - y10 + 2z9 - z17 ; (C on3, 12, 2): y13 = y10 ; (C on3, 10, 6): y25 = 0; (C on3, 10, 5): y26 = 0; (C on3, 7, 7): y48 = 1; (C on3, 3, 3) y28 = 1; (C on3, 8, 3): y44 = 0; (C on2, 5, 9): z5 = 2z16 - z14 ; (C on4, 10, 5): z24 = -2z29 ; (C on3, 7, 5): y39 = 1; (C on4, 5, 10): y45 = 3z14 ; (C on4, 11, 12): z16 = -3z28 - 3z25 - z13 ; (C on4, 9, 6): y31 = -2z9 + 3z17 ; (C on4, 11, 2): z9 = 3z28 ; (C on4, 5, 5): z14 = 0; (C on2, 1, 12): z8 = 0; (C on4, 2, 1): z13 = -3z25 ; (C on4, 6, 10): y52 = -y37 + 3y50 /2 + 3; (C on4, 10, 10): y28 = -6z28 + 3z17 ; (C on1, 13, 1): z48 = -z47 ; (C on1, 5, 4): y37 = 3/2 + 3/2y6 - 3z12 - 3y2 ; (C on4, 13, 13): y17 = 3/2y19 - z45 - z46 ; (C on1, 14, 1): z47 = 0; (C on2, 7, 4): z46 = 2z42 ; (C on4, 3, 3): y19 = -y12 ; (C on5, 3, 2): z17 = z25 + z35 + 2z28 ; (C on5, 3, 3): z29 = 0; (C on5, 3, 4): z40 = 0; (C on5, 3, 5): z35 = -z25 ; (C on5, 3, 6): z25 = -3z28 ; (C on5, 3, 13): z49 = 0; (C on5, 14, 12): z32 = 1; (C on5, 11, 4): z36 = -z34 ; (C on5, 9, 4): z34 = -1; (C on5, 3, 10): z45 = -1/2y12 - 2z42 ; (C on1, 3, 9): y12 = -9z28 ; (C on3, 11, 2): y10 = 0; (C on3, 7, 4): z42 = 0; (C on4, 14, 14): y14 = 0; (C on4, 11, 1): z12 = 1 + 6z28 ; (C on4, 11, 10): y29 = -9z28 ; (C on6, 7, 13): y6 = 1 - 9/2z28 ; (C on6, 9, 2): y32 = -1 + 4z28 ; (C on3, 9, 5): y50 = -2y34 - 8z28 ; (C on7, 7, 2): y2 = -8y34 +7-119/4z28 ; (C on8, 13, 1): y34 = 1+137/32z28 ; (C on8, 3, 9): y36 = 1; (C on8, 13, 2): z28 = 0. Thus, x1 = x1 (1), x2 = x2 (1), consequently 2 (x (1)) = x (1) for any root . Now look at the images (under 2 ) of h (t), t R . Let ht = 2 (h1 (t)). Since ht commutes with h1 , h2 , w6 (1) and x6 (1), we directly have
ht = d1 d3 0 0 0 0 0 0 0 0 0 0 0 0 d2 d4 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d5 0 0 0 d12 0 0 0 0 0 0 0 0 0 0 d7 0 0 0 d10 0 0 0 0 0 0 0 0 0 0 d9 0 0 0 d8 0 0 0 0 0 0 0 -d6 0 0 d11 d11 0 0 d6 0 0 0 0 0 0 0 -d8 0 0 0 d9 0 0 0 0 0 0 0 0 0 0 -d10 0 0 0 d7 0 0 0 0 0 0 0 0 0 0 -d12 0 0 0 d5 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d13 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d13 0 0 0 0 0 0 0 0 0 0 0 0 0 0 d14 0 0 0 0 0 0 0 0 0 0 0 0 0 (d13 - d14 ) d13 .

3 2

- - Use now the conditions h1 +2 (t) = w2 h1 (t)w2 1 , h21 +2 (t) = w1 h1 +2 (t)w1 1 and h1 (t)h1 +2 (t) = -1 -1 -1 h21 +2 (t). They give us w1 w2 ht w2 w1 = w2 ht w2 ht , therefore d2 d11 = 0 d2 = 0;


11

d3 d9 = 0 d3 = 0; d5 (1 - d13 ) = 0 d13 = 1; d6 = d8 = d10 = d12 = 0; d7 = 1/d5 ; d1 = d2 ; d4 = d2 ; d11 = 1/d9 ; d14 = 1. 11 9 Finally, use the fact that h31 +22 (t) = h1 +2 (t)h21 +2 (t) commutes with x1 . It gives us d5 = d3 , therefore ht = h1 (1/d9 ). 9 5. Final steps of the proof of Theorem 1. Now we have stated that for the root system G2 2 (x (1)) = x (1), 2 (h (t)) = h (s), , t R , s S . For every long root j there exists a root k such that hk (t)xj (1)hk (t)-1 = xj (t). Therefore, 2 (xj (t)) = xj (s). From the conditions written above and commutator formulas it follows 2 (x (t)) = x (s) for all . Denote the mapping t s by : R S . If t R , then t J , i. e., t = 1 + t1 , where / t1 R . Then 2 (x (t)) = 2 (x (1)x (t1 )) = x (1)x ((t1 )) = x (1 + (t1 )), . Therefore we can continue the mapping on the whole ring R (by the formula (t) := 1 + (t - 1) for t J ), and obtain 2 (x (t)) = x ((t)) for all t R, . Clear that is injective, additive, multiplicative on invertible elements. Since every element of R is a sum of two invertible elements, we have that is multiplicative on the whole R, i. e., it is an isomorphism of R onto some subring in S . Note that in this situation C G(S )C -1 = G(S ) for some matrix C GL 14 (S ). Show that S = S . Lemma 2. The Cheval ley group G(S ) generates M14 (S ) as a ring.
1 Proof. The matrix 2 (x2 (1) - 1)2 is e3,4 (ei,j is a matrix unit). Multiplying it to the suitable diagonal matrix we can obtain an arbitrary matrix of the form · e3,4 (since the invertible elements of S generate S ). According to the fact that all long root are conjugate under the action of the Weil group, multiplying e3,4 from left and right sides to suitable wi , wj , we obtain all ek,l , k , l = 3, 4, 9, 10, 11, 12. Then e14,4 = (x2 (1) - E ) · e4,4 + e3,4 , again with the help of multiplying (from the right side) to different elements of the Weil group we obtain all e14,l , l = 3, 4, 9, 10, 11, 12. Now e14,14 = -(e14,3 (x2 (1) - E ) + e14,4 )(w1 - E ); e3,14 = -1/2(x2 (1) - E )e14,14 . As above, we obtain all el,14 , l = 3, 4, 9, 10, 11, 12. Also we have e14,13 = e14,3 (x2 (1) - E ) + e14,4 + 2e14,14 ; e3,13 = e3,3 (x2 (1) - E ) + 2e3,14 + e3,4 (and directly obtain all el,13 , l = 3, 4, 9, 10, 11, 12); e3,2 = e3,13 (x1 (1)- E ). Now use the fact that all short roots are also conjugate under the action of W ; we obtain all ek,l , k = 3, 4, 9, 10, 11, 12, l = 1, 2, 5, 6, 7, 8. Since e13,13 = 1/4(h1 + E )(h2 + E ) - e14,14 , e1,13 = -1/2(x1 - E )e13,13 , e13,2 = e13,13 (x1 - E ), we have all el,13 , e13,l , l = 1, 2, 5, 6, 7, 8, and after that by multiplying matrix units we obtain all el,k , l, k = 1, . . . , 14, and therefore the whole ring M14 (S ). Lemma is proved.

Note that in the previous lemma we did not suppose 1/3 S . Lemma 3. If for some C GL 14 (S ) we have C G(S )C then S = S . Proof. Suppose that S is a proper subring of S .
-1

= G(S ), where S is a subring of S ,


12

Then C M14 (S )C -1 = M14 (S ), since the group G(S ) generates M14 (S ), and the group G(S ) = C G(S )C -1 generates M14 (S ). It is impossible, since C GL 14 (S ). Therefore we have proved that is an isomorphism from the ring R onto S . Consequently the composition of the initial isomorphism and some basis change with a matrix C GL 14 (S ), (mapping G(S ) onto itself ), is a ring isomorphism . Therefore, = iC -1 . So Theorem 1 is proved. 6. Proof of Theorem 2. In this section we still consider a Chevalley group G(R) of type G2 (it is simultaneously elementary and adjoint) over a local ring with 1/2 and 1/3. Theorem 2. The normalizer of G(R) in GL 14 (R) is · G(R). Proof. Suppose that we have some matrix C = (ci,j ) GL 14 (R) such that C · G(R) · C
-1

= G(R).

If J is the radical of R, then the matrices from M14 (J ) is the radical in the matrix ring M14 (R), therefore C · M14 (J ) · C -1 = M14 (J ), consequently, C · (E + M14 (J )) · C -1 = E + M14 (J ), i. e., C · G(R, J ) · C -1 = G(R, J ), since G(R, J ) = G(R) (E + M14 (J )). Thus, the image C of the matrix C under factorization R by J gives us an automorphism­ conjugation of the Chevalley group G(k ), where k = R/J is a residue field of R. Lemma 4. If G(k ) is a Cheval ley group of type G2 over a field k of characteristics = 3, then every its automorphism­conjugation is inner. Proof. By Theorem 30 from [49] every automorphism of a Chevalley group of type G2 over a field k of characteristics = 3 is standard, i. e., for this type it ia a composition of inner and ring automorphisms. Suppose that a matrix C is from normalizer of G(k ) in GL 14 (k ). Then iC is an automorphism of G(k ), so we have iC = ig , g G(k ), is a ring automorphism. Consequently, ig-1 iC = iC = and some matrix C GL 14 (k ) defines a ring automorphism . For every root we have (x (1)) = x (1), therefore C x (1) = x (1)C for all . Thus we have that C is scalar and an automorphism iC is inner. By Lemma 4 iC = ig , g G(k ). Since over a field every element of a Chevalley group is a product of some set of unipotents x (t)) and the matrix g can be decomposed into a product xi1 (Y1 ) . . . xiN (YN ), Y1 , . . . , YN k . Since every element Y1 , . . . , YN is a residue class in R, we can choose (arbitrarily) elements y1 Y1 , . . . , yN YN , and the element g =x

i1

(y1 ) . . . xiN (yN )


13

satisfies the conditions g G(R) and g = g . Consider the matrix C = g -1 d-1 C . This matrix also normalizes the group G(R), and also C = E . Therefore, from the description of the normalizer of G(R) we come to the description of all matrices from this normalizer equivalent to the unit matrix modulo J . Therefore we can suppose that our initial matrix C is equivalent to the unit modulo J . Our aim is to show that C G(R). Firstly we prove one technical lemma that we will need later. Lemma 5. Let X = t1 (s1 )t2 (s2 )x1 (t1 ) . . . x6 (t6 )x-1 (u1 ) . . . x-6 (u6 ) G(R, J ). Then the matrix X contains 15 coefficients (precisely described in the proof of lemma ), uniquely defining al l , s1 , s2 , t1 , . . . , t6 , u1 , . . . , u6 . Proof. By direct calculation we obtain that in the matrix X x therefore we find u2 ; x x
12,14 12,8 12,12

=

is now known. Now from the system of equations x
14,12 4,12

(u6 +u2 u5 ) we know u4 , u5 , u6 . Besides, from x10,12 = sts22 and x 3 s3 s2 12 1 find t2 and s3 s2 , and consequently we know s2 . 1 1 From x10,8 = - (us3-t2 u3 we find u1 . From the first equation we can express 1 s2

u3 s3 s2 12

, so we get u3 ; similarly from x

, s3 s2 12 u4 12,6 = s3 s2 12

=

=-

u2 , s3 s2 12 u5 , x12,4 = - s3 s2 12 - = (1s3t2 u2 ) we 10,10 1 s2

x

12,10

=-

by s1 . Therefore,

= (t2 t5 + 3t3 t4 + 2t6 ); (t5 + 3t1 t4 + 3t2 t3 - t3 t2 ) 1 1 = ; s2 -t1 + 2t2 u1 - t3 u2 + u4 t5 + 3u4 t1 t4 + 3u4 t2 t3 - u4 t3 t2 ) 1 11 1 1 = ; s2 (1 - 3t1 u1 - 3u3 t3 + 3u3 t1 t2 ) = ; s2 s2 1 = (-t3 - 2t4 u1 - t5 u2 + u4 t2 t5 + 3u4 t3 t4 + 2u4 t6 ); 1 = (t4 + t5 u1 + u3 t2 t5 + 3u3 t3 t4 + 2u3 t6 ),

x

x x x x

4,6

8,8

14,6 14,8

where s1 , t1 , t3 , t4 , t5 , t6 are variables, in every equation there is exactly one variable with invertible coefficient, and for all equations these variables are different, we can find all six variables. Now we know all obtained elements of the ring. Lemma is proved. Now return to our main proof. Recall, that we work with the matrix C , equivalent to the unit matrix modulo radical, and mapping the Chevalley group into itself. For every root we have (1) (2) C x (1)C
-1

= x (1) · g ,

g G(R, J ). (c1 ) . . . x (c6 ),

Every g G(R, J ) can be decomposed into a product t1 (1 + a1 )t2 (1 + a2 )x1 (b1 ) . . . x6 (b6 )x
-
1

-

6

where a1 , a2 , b1 , . . . , b6 , c1 , . . . , c6 J (see, for example, [2]).


14

Let C = E + X = E + (xi,j ). Then for every root we can write a matrix equation 1 with variables xi,j , a1 , a2 , b1 , . . . , b6 , c1 , . . . , c6 , each of them is from the radical. Let us change these equations. We consider the matrix C and "imagine", that it is some matrix from Lemma 5 (i. e., it is from G(R)). Then by some its concrete 15 positions we can "define" all coefficients , s1 , s2 , t1 , . . . , t6 , u1 , . . . , u6 in the decomposition of this matrix from Lemma 5. In the result we obtain a matrix D G(R), every matrix coefficient in it is some (known) function of coefficients of C . Change now the equations (1) to the equations (3) D-1 C x (1)C
-1

D = x (1) · g ,

g



G(R, J ).

We again have matrix equations, but with variables yi,j , a1 , a2 , b1 , . . . , b6 , c1 , . . . , c6 , every of them still is from radical, and also every yp,q is some known function of (all) xi,j . The matrix D-1 C will be denoted by C . We want to show that a solution exists only for all variables with primes equal to zero. Some xi,j also will equal to zero, and other are reduced in the equations. Since the equations are very complicated we will consider the linearized system. It is sufficient to show that all variables from the linearized system (let it be the system of q variables) are members of some system from q linear equations with invertible in R determinant. In other words, from the matrix equalities we will show that all variables from them are equal to zeros. Clear that linearizing the product Y -1 (E + X ) we obtain some matrix E + (zi,j ), with all positions described in Lemma 5 equal to zero. To find a final form of a linearized system, we write the last one as: (E + Z )x (1) = x (1)(E + a1 T1 + a2 . . . )(E + a2 T2 + a2 . . . )· 1 2
2 2 · (E + b1 X1 + b2 X1 /2 + . . . ) . . . (E + c6 X-6 + c2 X-6 /2 + . . . )(E + Z ), 1 6

where X is a corresponding element of the Lie algebra in its adjoint representation, T1 = diag [1, 1, 0, 0, 1, 1, 2, 2, 3, 3, 3, 3, 0, 0]; T2 = diag [0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 0, 0]. Finally we have Z x (1) - x (1)(Z + a1 T1 + a2 T2 + b1 X1 + · · · + c6 X6 ) = 0. This equation can be written for every (naturally, with another aj , bj , cj ), and can be written only for generating roots: for 1 , 2 , -1 , -2 . The number of free variables is not changed.


15

We have four equations: Z x1 (1) - x1 (1)(X + a1,1 T1 + a2,1 T2 + +b1,1 X1 + · · · + b6,1 X6 + c1,1 X-1 + Z x (1) - x (1)(X + a1,2 T1 + a2,2 T2 + 2 2 +b1,2 X1 + · · · + b6,2 X6 + c1,2 X-1 + X x-1 (1) - x-1 (1)(X + a1,3 T1 + a2,3 T2 + +b1,3 X1 + · · · + b6,3 X6 + c1,3 X-1 + X x (1) - x (1)(X + a T + a T + -2 -2 1,4 1 2,4 2 +b1,4 X1 + · · · + b6,4 X6 + c1,4 X-1 +

· · · + c6,1 X-6 ) = 0; c6,2 X-6 ) = 0; · · · + c6,3 X-6 ) = 0; · · · + c6,4 X-6 ) = 0.

The matrix Z = (zi,j ) has zero elements on the positions z4,6 , z4,12 , z8,8 , z10,8 , z10,10 , z10,12 , 12,4 , z12,6 , z12,8 , z12,10 , z12,12 , z12,14 , z14,6 , z14,8 , z14,12 . From the matrix of the second condition we have: the position (3, 6): z3,4 = 0; the position (3, 7): z3,9 = 0; the position (3, 11): c5,1 = 0; the position (10, 6): z10,4 = 0; the position (10, 7): z10,9 = 0; the position (10, 12): b2,1 = 0; the position (11, 6): z11,4 = 0; the position (11, 13): z11,1 = 0; the position (2, 6): z2,4 = 0; the position (2, 7): z2,9 = 0; the position (3, 8): z3,6 = 0; the position (3, 13): z3,1 = 0; the position (5, 6): z5,4 = 0; the position (6, 9): z8,9 = 0; the position (11, 8): z11,6 = 0; the position (11, 2): z11,13 = 0; the position (12, 8): c3,1 = 0; the position (12, 6): c4,1 = 0; the position (12, 10): c2,1 = 0; the position (14, 11): c6,1 = 0; the position (14, 12): b6,1 = 0; the position (10, 13): z10,1 = 0; the position (12, 7): z12,9 = 0; the position (11, 8): z11,6 = 0; the position (10, 2): z10,13 = 0; the position (13, 12): z2,12 = 0; the position (13, 11): z2,11 = 0; the position (12, 5): z12,7 = 0; the position (6, 1): z8,1 = 0; the position (14, 7): z14,9 = 0; the position (12, 13): z12,1 = 0; the position (14, 13): z14,1 = 0; the position (14, 5): z14,7 = 0; the position (12, 2): z12,13 = 0; the position (1, 9): z13,9 = 0; the position (4, 9): z6,9 = 0; the position (14, 2): z14,13 = 0; the position (2, 5): z2,7 = 0; the position (1, 12): z13,12 = 0; the position (10, 5): z10,7 = 0; the position (9, 4): z7,4 = 0. From the matrix of the second condition it follows: the positions (9, 6): z9,2 = 0; (13, 13): z13,3 = 0; (12, 13): z12,3 = 0; (12, 14): 6,2 = 0; (14, 11): z4,11 = 0; (12, 9): z12,11 = 0; (13, 5): 3,2 = 0; (11, 10): z9,10 = 0; (8, 9): z8,11 = 0; (11, 8): z9,8 = 0; (10, 9): z10,11 = 0; (6, 9): z6,11 = 0; (2, 3): z6,3 = 0; (4, 1): z4,5 = 0; (10, 1): z10,5 = 0; (10, 5): z12,5 = 0; (7, 12): z7,10 = 0; (5, 10): z1,10 = 0; (7, 5): b1,2 = 0; (12, 10): c2,2 = 0; (5, 3): z1,3 = 0; (13, 7): c4,2 = 0; (13, 8): b4,2 = 0; (10, 8): c1,2 = 0; (12, 4): c5,2 = 0; (7, 2): b5,2 = 0; (1, 6): z1,2 = 0; (4, 12): z4,10 = 0; (5, 11): z1,11 = 0; (13, 9): z13,11 = 0; (4, 6): z4,2 = 0; (7, 9): z7,11 = 0; (10, 6): z10,2 = 0; (14, 8): z4,8 = 0; (4, 13): z4,3 = 0; (7, 13): z7,3 = 0; (8, 13): z8,3 = 0; (9, 13): z9,3 = 0; (10, 13): z10,3 = 0; (6, 1): z6,5 = 0; (8, 1): z8,5 = 0; (9, 1): z9,5 = 0; (13, 1): z13,5 = 0; (10, 2): z12,2 = 0; (6, 4): z6,14 = 0; (7, 4): z7,14 = 0; (8, 4): z8,14 = 0; (10, 4): z10,14 = 0; (13, 4): z13,14 = 0; (4, 5): z14,5 = 0; (8, 6): z8,2 = 0; (2, 7): z6,7 = 0; (5, 7): z1,7 = 0; (14, 7): z4,7 = 0; (3, 7): z13,7 = 0; (3, 8): z13,8 = 0; (8, 8) (10, 10): a1,2 = 0 a2,2 = 0; (1, 1): z1,5 = 0; (2, 2): z6,2 = 0; (3, 3): z14,3 = 0; (4, 4): z4,14 = 0; (9, 9): z9,11 = 0; (10, 12): b2,2 = 0; (3, 11): z14,11 = 0; (2, 10): z6,10 = 0; (2, 14): z2,3 = 0; (2, 13): z6,13 = 0; (14, 13): z4,13 = 0; (14, 1): z4,1 = 0; (14, 9): z4,9 = 0; (14, 4): z14,14 = 0; (3, 9): z3,11 = 0; (3, 14): z13,13 = 0; (3, 13): z3,3 = 0. z


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Again return to the first condition. Now the elements z1,9 , z2,5 , z6,1 , z8,7 , z8,13 , z5,6 , z5,11 become to be zeros. If we consider the condition 3, we have that from all coefficients ai , bi , ci only b2,3 , b3,3 , b4,3 , c1,3 , a can be not zeros. The following coefficients of Z are zeros: z5,3 , z5,10 , z1,12 , z7,8 , z11,2 , z11,10 , z11,8 , z14,2 . Now in the second condition the position (14, 6) gives b3,2 = 0, therefore z13,2 , z6,8 , z1,14 , z7,5 , z1,13 , z8,10 , z9,6 , z9,7 , z9,12 are zeros. Again from the third condition a1,3 = 0, z7,12 = 0. Finally, come to the last, fourth condition. Since the roots 2 and -2 are conjugate with the element w2 , it is clear that in the fourth condition from all ai , bi , ci only b5,4 can be not zero. Therefore the following elements of Z are zeros: z1,4 , z5,7 , z5,9 , z5,12 , z5,1 , z5,13 , z5,14 , z2,1 , z2,6 , z2,14 , z3,2 , z3,12 , z3,14 , z3,5 , z3,7 , z3,8 , z3,13 , z13,1 , z13,6 , z6,4 , z2,8 , z2,13 , z7,1 , z7,6 , z7,9 , z8,4 , z8,6 , z8,12 , z9,4 , z10,6 , z11,7 , z11,9 , z11,12 , z11,14 , z3,10 , z13,4 , z14,4 , z5,8 . From the first equation now z2,10 = z6,12 = z1,6 = 0, from the second one z1,8 = z5,2 = z7,2 = z13,10 = z14,10 = z11,3 = z9,13 = z9,14 = 0, from the third one z9,1 = z11,5 = z7,13 = 0. Again from the second condition it follows z6,6 = z2,2 , z5,5 = z1,1 , z11,11 = z9,9 . From the first condition z4,4 = z1,1 = z2,2 = z7,7 = z9,9 = 0. Therefore Z = 0, what we need. Theorem 2 is proved. From Theorems 1 and 2 it directly follows the main theorem of this paper: Theorem 3. Suppose that G(R) and G(S ) are Cheval ley groups of type G2 , R, S are local rings with 1/2 and 1/3. Then every isomorphism between G(R) and G(S ) is standard, i. e., it is the composition of a ring isomorphism and an inner automorphism. References
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