CC BY
8
Journal of Siberian Federal University. Mathematics & Physics 2014, 7(2), 186-190
УДК 517.9
Hall's Polynomials of Finite Two-Generator Groups of Exponent Seven
Alexander A. Kuznetsov* Konstantin V. Safonov^
Institute of Computer Science and Telecommunications, Siberian State Aerospace University, Krasnoyarsky Rabochy, 31, Krasnoyarsk, 660014
Russia
Received 13.01.2014, received in revised form 18.02.2014, accepted 24.03.2014 Let Bk = B0(2, 7, k) be the largest two-generator finite group of exponent 7 and nilpotency class k. Hall's polynomials of Bk for k < 4 are calculated.
Keywords: periodic group, collection process, Hall's polynomials.
Let Bk — Bo (2,7, k) be the largest two-generator finite group of exponent 7 and nilpotency class k. In this class, the largest group is the group B28, which has the order 720416 [1]. For each Bk a power commutator presentation is obtained [1].
Let aXl ... and ay1 ... a^n be two arbitrary elements in the group Bk recorded in the commutator form. Then their product is equal
ax ... an • Qjy ... a П — a у ... a^ .
Powers zi are to be found based on the collection process (see [2,3]) which is implemented in the computer algebra systems GAP and MAGMA. Furthermore, there is an alternative method for calculating products of elements of the group, proposed by Hall (see [4]). Hall showed that zi are polynomial functions (over the field Z7 in this case), depending on the variables x1,..., xi, y1,..., yi, which is now called Hall's polynomials. According to [4]
Zi = Xi + yi + Pi(x 1, . . .,Xi-1,yi, ... ,yi-i).
Hall's polynomials are necessary in solving problems that require multiple products of the elements of the group. Study of the structure of the Cayley graph of some group is one of these problems [5,6]. The computational experiments carried out on the computer in two-generator groups of exponent five (see [7]) showed that the method of Hall's polynomials has an advantage over the traditional collection process. Therefore, there is a reason to believe that the use of polynomials would be preferable than the collection process in the study of Cayley graphs of Bk groups. It should also be noted that this method is easily software-implemented including multiprocessor computer systems.
Previously unknown Hall's polynomials of Bk are calculated within the framework of this paper. For k > 4 polynomials are calculated similarly but their output takes considerably more space so it makes impossible to verify the proof without use of computers.
The main result of this paper is
Theorem. Let aX1 ... a^T and «i1 • • • «n1 be two arbitrary elements of the group Bk recorded in
ix1 ... • «y1 • • • «n
the commutator form where k £ N u k ^ 4. Then their product is equal aX1 «Xn • «y1 =
* [email protected] [email protected] © Siberian Federal University. All rights reserved
a^1 ... a^, where z £ Z7, are Hall's polynomials given by formulas (1-2) for k = 1, (1-3) for k = 2, (1-5) for k = 3 and (1-8) for k = 4.
Z1 = X1 + yi, (1)
Z2 = X2 + y2, (2)
Z3 = X3 + y3 + x2yi, (3)
Z4 = X4 + y4 + 3x2yi + X3yi +4x2y2, (4)
Z5 = X5 + y5 + 3x2yi + X3 y2 + 4x2yi + X2yiy2, (5)
Z6 = X6 + y6 + 5x2 yi + 3x3yi + X4yi + 3x2y2 + 6x2 y3 + 4x3y2, (6)
2 2 2 2 2 Z7 = X7 + yr + 2x2yi + 2x2yi + X4y2 + X5 yi + 5x2yi + 5x2yi + 4x2yi y2 + 3x2yiy2 + X3yiy2, (7)
9 q Q Q Q
Zg = xg + yg + 5x2yi + 3x3y2 + x5y2 + 3x2yi + 6x2yi + 4x3 y2 + 4x2yiy2 + 4x2yiy2 + 6x2yiy2. (8)
1. Proof of the Theorem
Let's calculate the Hall's polynomials for the group B4. Dealing with this group we also obtain polynomials for groups Bi, B2 and B3 so no need to study separately these cases. When k < 4 commutators which has a weight is more than k are not considered because of they are definitionally equal to the group identity.
Using GAP we obtain a power commutator presentation of B4. Commutators of weight 1:
a2 _ generators of the group.
Commutators of weight 2:
«3 = [«2, ai]. Commutators of weight 3:
a4 = [a3,ai] = [«2, ai, ai],
a5 = [«3, «2] = [«2, ai, «2]. Commutators of weight 4:
«6 = [«4,«i] = [«2, «i, «i, «i], «7 = [«5,«i] = [«2, «i, «2, «i], «8 = [«5, «2] = [«2, «i, «2, «2].
List of defining relations R for commutators:
ar = 1(1 ^ i ^ 8), [«2, «i] = «3, [«3, «i] = «4, [«3, «2] = «5, [«4, «i] = «6, [«4, «2] = «7, [«4, «3] = 1, [«5, «i] = «7, [«5, «2] = «8, [«5, «3] = 1, [«5, «4] = 1, [«6, «i] = 1, [«6, «2] = 1, [«6, «3] = 1, [«6, «4] = 1, [«6, «5] = 1, [«r, «i] = 1, [«r, «2] = 1, [«r, «3] = 1, [«r, «4] = 1, [«r, «5] = 1, [«r, «6] = 1, [«8, «i] = 1, [«g, «2] = 1, [«g, «3] = 1, [«g, «4] = 1, [«g, «5] = 1, [«g, «6] = 1, [«g, «r] = 1.
Thus,
B4 = ( | R }.
Each element of the group is expressed uniquely as a normal commutator word:
Vg G B4 g = a?1 a?2 a?3 a?4a?5 a?6 a?7a?8, x4 G Z7.
Sometimes we will write g = (xi,..., x8).
In order to determine the functions zj first we need to calculate the products of ay a? for all 1 ^ i < j ^ 8, x, y = 1,2,3,4, 5, 6. For the pair (j, i) it is required to find the interpolation polynomial for each of the 8 commutators by the 36 values of the product (y, x). Let's start with the first pair a^ a ?:
a2ai =
14
a2ai = a2ai =
24
a2ai = 31
a2ai =
34 a2ai =
41
a2ai = 44
a2ai = 51
a2ai =
54
a2ai =
a^ai =
6 4 a2ai =
Let's write:
1,1,1, 0,0,0,0,0), 12 a2a1 = (2,1, 2,1,0,0,0,0), 13 a2a1 = (3,1, 3, 3, 0,1,0,0
4,1, 4, 6,0,4,0,0), 1 5 a2a1 = (5,1, 5, 3,0, 3,0,0), a2a1 = (6,1, 6,1, 0, 6,0,0
1, 2, 2, 0,1,0,0,0), 22 a2a1 = (2, 2, 4, 2, 2,0,1,0), 23 a2a1 = (3, 2, 6, 6, 3, 2, 3,0
4, 2,1, 5,4,1, 6,0), a2a1 = (5, 2, 3, 6, 5, 6, 3,0), a2a1 = (6, 2, 5, 2, 6, 5,1,0
1, 3, 3, 0, 3,0,0,1), 32 a2a1 = (2, 3, 6, 3, 6,0, 3, 2), 33 = (3, 3, 2, 2, 2, 3, 2, 3
4, 3, 5, 4, 5, 5, 4, 4), 3 5 a2a1 = (5, 3,1, 2,1, 2, 2, 5), afa! = (6, 3, 4, 3, 4,4, 3, 6
1, 4, 4, 0, 6,0,0,4), 4 2 a2a1 = (2,4,1, 4, 5,0, 6,1), 43 a2a1 = (3, 4, 5, 5, 4, 4, 4, 5
4, 4, 2, 3, 3, 2,1, 2), 45 a2a1 = (5, 4, 6, 5, 2, 5, 4, 6), 46 a2a1 = (6,4, 3, 4,1, 3, 6, 3
1, 5, 5, 0, 3,0,0, 3), a2a1 = (2, 5, 3, 5, 6,0, 3, 6), 53 a2a'1 = (3, 5,1,1, 2, 5, 2, 2
4, 5, 6, 2, 5, 6, 4, 5), a2af = (5, 5, 4,1,1,1, 2,1), a2a! = (6, 5, 2, 5, 4, 2, 3,4
1, 6, 6, 0,1,0,0, 6), a2a1 = (2, 6, 5, 6, 2,0,1, 5), a2af = (3, 6, 4, 4, 3, 6, 3,4
4, 6, 3,1,4, 3, 6, 3), a2af = (5, 6, 2, 4, 5,4, 3, 2), a2a! = (6, 6,1, 6, 6,1,1,1
ay a? = a? a2 a.^'2 (x'y) a^"' (x'y)
1 a2 a3
,/i1,2)(x,y)
where /r(i'2) (x, y) = 5^P=i 26=i xpyq are some polynomials over the field Z7. To find them
66
let's perform interpolation for each commutator r = 3,4,..., 8. To find /r(i'2)(
x, y) it is required to solve a system of linear equations over the given field:
6 6
EE^xV = zyx V x,y = 1, 2, 3,4, 5, 6,
p=1q=1
(9)
where zyx is a value of r-th commutator for the pair (y, x). This system will have 36 variables and consist of 36 equations.
8
Let's show how to find f ' ^(x, y) at the example of the 8-th commutator. For short, let's write ppq instead of . Substituting in (9) all values of z|x we receive:
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 MA /0\
2 2 2 2 2 2 4 4 4 4 4 4 1 1 1 1 1 1 2 2 2 2 2 2 4 4 4 4 4 4 1 1 1 1 1 1 ßl2 0
3 3 3 3 3 3 2 2 2 2 2 2 6 6 6 6 6 6 4 4 4 4 4 4 5 5 5 5 5 5 1 1 1 1 1 1 ßl3 0
4 4 4 4 4 4 2 2 2 2 2 2 1 1 1 1 1 1 4 4 4 4 4 4 2 2 2 2 2 2 1 1 1 1 1 1 ßl4 0
5 5 5 5 5 5 4 4 4 4 4 4 6 6 6 6 6 6 2 2 2 2 2 2 3 3 3 3 3 3 1 1 1 1 1 1 ßl5 0
6 6 6 6 6 6 1 1 1 1 1 1 6 6 6 6 6 6 1 1 1 1 1 1 6 6 6 6 6 6 1 1 1 1 1 1 ßl6 0
2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 2 4 1 ß21 0
4 1 2 4 1 2 1 2 4 1 2 4 2 4 1 2 4 1 4 1 2 4 1 2 1 2 4 1 2 4 2 4 1 2 4 1 ß22 0
6 5 3 6 5 3 4 1 2 4 1 2 5 3 6 5 3 6 1 2 4 1 2 4 3 6 5 3 6 5 2 4 1 2 4 1 ß23 0
1 2 4 1 2 4 4 1 2 4 1 2 2 4 1 2 4 1 1 2 4 1 2 4 4 1 2 4 1 2 2 4 1 2 4 1 ß24 0
3 6 5 3 6 5 1 2 4 1 2 4 5 3 6 5 3 6 4 1 2 4 1 2 6 5 3 6 5 3 2 4 1 2 4 1 ß25 0
5 3 6 5 3 6 2 4 1 2 4 1 5 3 6 5 3 6 2 4 1 2 4 1 5 3 6 5 3 6 2 4 1 2 4 1 ß26 0
3 2 6 4 5 1 3 2 6 4 5 1 3 2 6 4 5 1 3 2 6 4 5 1 3 2 6 4 5 1 3 2 6 4 5 1 ß3l 1
6 4 5 1 3 2 5 1 3 2 6 4 3 2 6 4 5 1 6 4 5 1 3 2 5 1 3 2 6 4 3 2 6 4 5 1 ß32 2
2 6 4 5 1 3 6 4 5 1 3 2 4 5 1 3 2 6 5 1 3 2 6 4 1 3 2 6 4 5 3 2 6 4 5 1 ß33 3
5 1 3 2 6 4 6 4 5 1 3 2 3 2 6 4 5 1 5 1 3 2 6 4 6 4 5 1 3 2 3 2 6 4 5 1 ß34 4
1 3 2 6 4 5 5 1 3 2 6 4 4 5 1 3 2 6 6 4 5 1 3 2 2 6 4 5 1 3 3 2 6 4 5 1 ß35 5
4 5 1 3 2 6 3 2 6 4 5 1 4 5 1 3 2 6 3 2 6 4 5 1 4 5 1 3 2 6 3 2 6 4 5 1 ß36 6
4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 4 2 1 ß4l 4
1 4 2 1 4 2 2 1 4 2 1 4 4 2 1 4 2 1 1 4 2 1 4 2 2 1 4 2 1 4 4 2 1 4 2 1 ß42 1
5 6 3 5 6 3 1 4 2 1 4 2 3 5 6 3 5 6 2 1 4 2 1 4 6 3 5 6 3 5 4 2 1 4 2 1 ß43 5
2 1 4 2 1 4 1 4 2 1 4 2 4 2 1 4 2 1 2 1 4 2 1 4 1 4 2 1 4 2 4 2 1 4 2 1 ß44 2
6 3 5 6 3 5 2 1 4 2 1 4 3 5 6 3 5 6 1 4 2 1 4 2 5 6 3 5 6 3 4 2 1 4 2 1 ß45 6
3 5 6 3 5 6 4 2 1 4 2 1 3 5 6 3 5 6 4 2 1 4 2 1 3 5 6 3 5 6 4 2 1 4 2 1 ß46 3
5 4 6 2 3 1 5 4 6 2 3 1 5 4 6 2 3 1 5 4 6 2 3 1 5 4 6 2 3 1 5 4 6 2 3 1 ß5l 3
3 1 5 4 6 2 6 2 3 1 5 4 5 4 6 2 3 1 3 1 5 4 6 2 6 2 3 1 5 4 5 4 6 2 3 1 ß52 6
1 5 4 6 2 3 3 1 5 4 6 2 2 3 1 5 4 6 6 2 3 1 5 4 4 6 2 3 1 5 5 4 6 2 3 1 ß53 2
6 2 3 1 5 4 3 1 5 4 6 2 5 4 6 2 3 1 6 2 3 1 5 4 3 1 5 4 6 2 5 4 6 2 3 1 ß54 5
4 6 2 3 1 5 6 2 3 1 5 4 2 3 1 5 4 6 3 1 5 4 6 2 1 5 4 6 2 3 5 4 6 2 3 1 ß55 1
2 3 1 5 4 6 5 4 6 2 3 1 2 3 1 5 4 6 5 4 6 2 3 1 2 3 1 5 4 6 5 4 6 2 3 1 ß56 4
6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 6 1 ß6l 6
5 2 5 2 5 2 3 4 3 4 3 4 6 1 6 1 6 1 5 2 5 2 5 2 3 4 3 4 3 4 6 1 6 1 6 1 ß62 5
4 3 4 3 4 3 5 2 5 2 5 2 1 6 1 6 1 6 3 4 3 4 3 4 2 5 2 5 2 5 6 1 6 1 6 1 ß63 4
3 4 3 4 3 4 5 2 5 2 5 2 6 1 6 1 6 1 3 4 3 4 3 4 5 2 5 2 5 2 6 1 6 1 6 1 ß64 3
2 5 2 5 2 5 3 4 3 4 3 4 1 6 1 6 1 6 5 2 5 2 5 2 4 3 4 3 4 3 6 1 6 1 6 1 ß65 2
1 6 1 6 1 6 6 1 6 1 6 1 1 6 1 6 1 6 6 1 6 1 6 1 1 6 1 6 1 6 6 1 6 1 6 1 W/ 1
The rank of the system matrix is equal to 36, therefore it has the only solution: pu = 5, P12 = 3, P13 = 6, all the remaining coefficients are equal to zero. Therefore,
/g1'2 (x, y) = 5xy + 3xy2 + 6xy3.
Similar calculations are applied for all polynomials fr ' ^x, y). Here they are: f(1'2)(x,y) = x,
f2(1'2)(x,y) = ^
f31,2)(x,y) = xy,
f,(1'2) (x, y) = 3xy + 4x2y,
f5(1'2)(x, y) = 3xy + 4xy2,
fg1'2) (x, y) = 5xy + 3x2y + 6x3y,
f71,2)(x, y) = 2xy + 5xy2 + 5x2y + 2x2y2,
f81,2)(x, y) = 5xy + 3xy2 + 6xy3.
Thus,
a22 aX = (x, y, xy, 3xy + 4x y, 3xy + 4xy , 5xy + 3x y + 6x y,
2xy + 5xy2 + 5x2y + 2x2y2, 5xy + 3xy2 + 6xy3).
Using this method let's calculate other noncommutative a| aX.
a3 aX = (x, 0, y, xy, 0, 3xy + 4x2y, 0, 0),
a3 a? = (0, x, y, 0, xy, 0, 0, 3xy + 4x2y), a| a? = (x, 0, 0, y, 0,xy, 0, 0), a| a? = (0, x, 0, y, 0, 0,xy, 0), ay a? = (x, 0, 0, 0, y, 0,xy, 0), ay a? = (0, x, 0, 0, y, 0, 0,xy).
Not listed pairs are commutative, i.e. a| a? = a? a|.
Thus, we have a complete set of relations for the implementation of the collection process in analytical form:
ay a? = a? ay aj+f^ aff^ . .. af^, 1 < i < j < 8. (10)
Using (10) we can calculate the product a?1 ... a?8 • ay1 ... a88 = a^1 ... a88. Following this procedure we will find all Zj (1-8).
The theorem is proved.
The investigation was supported by the Ministry of Education and Science of the Russian Federation (Project B 112/14).
References
[1] E.O'Brien, M.Vaughan Lee, The 2-Generator Restricted Burnside Group of Exponent 7, Int. J. Algebra Comput., 12(2002), 459-470.
[2] C.Sims, Computation with finitely presented groups, Cambridge, Cambridge University Press, 1994.
[3] D.Holt, B.Eick, E.O'Brien, Handbook of computational group theory, Boca Raton, Chapman & Hall/CRC Press, 2005.
[4] Ph.Hall, Nilpotent groups, Notes of lectures given at the Canadian Mathematical Congress 1957 Summer Seminar, in the collected works of Philip Hall, Oxford, Clarendon Press, 1988, 415-462.
[5] A.A.Kuznetsov, A.S.Kuznetsova, Computer modeling of finite two-generator groups of exponent five, Vestnik SibGAU, 45(2012), no. 5, 59-62 (in Russian).
[6] A.A.Kuznetsov, A.S.Kuznetsova, Relation betweengrowth functions in symmetric groups and tasks of combinatorial optimization, Vestnik SibGAU, 46(2012), no. 6, 57-62 (in Russian).
[7] A.A.Kuznetsov, A.S.Kuznetsova, Fast multiplication in finite two-generated groups of exponent five, Prikl. Diskr. Mat., 18(2013), no. 1, 110-116 (in Russian).
Полиномы Холла конечных двупорожденных групп периода семь
Александр А. Кузнецов Константин В. Сафонов
Пусть Бк = Б0(2, 7, k) — максимальная конечная двупорожденная бернсайдова группа периода
7 ступени нильпотентности k. В настоящей статье вычислены полиномы Холла для Бк при k < 4.
Ключевые слова: периодическая группа, собирательный процесс, полиномы Холла.
