COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

164 Marshall Hall Jr.
b=
(OO)(Ol, 02, 09, 16, 23, 20, 30, 17)(03, 35, 13, 29, 31, 24, 25, 11)
(04, 26, 27, 07)(05, 21, 14, 10)(06, 19, 32, 36)
(08, 18, 28, 22, 15, 12, 33, 34)(37)(38, 92, 67, 77, 99, 89,49, 84)
(39, 59, 82,46, 88, 54, 52, 68)(40, 55,47, 81,75, 95, 61, 78)
C=
(41,44, 63, 58, 72, 65, 50, 51)(42, 76, 53, 93, 86, 80, 79, 57)
(43, 85,48, 70, 96, 83, 66, 94)(45, 97)(56, 87)
(60, 71, 91, 69, 90, 73, 64,74)(62, 98)
(00)(01)(02, 28,21)(03,23, 18)(04,25, 22)(05,27, 19)(06, 29, 16)
(07,24, 20)(08,26, 17)(09, 13, 14)(10, 15, 11)(12)(30, 36, 34)(31)
(32, 33, 35)(37, 50, 63)(38,45, 60)(39,47, 64)(40, 49, 61)(41,44, 58)
(42,46, 62)(43,48, 59)(51, 65, 72)(52, 67, 76)(53, 69, 73)(54, 71, 77)
(55, 66,74)(56, 68, 78)(57, 70, 75)(79, 89,96)(80, 91, 93)(81, 86,97)
(82, 88,94)(83, 90, 98)(84, 92, 95)(85, 87, 99)
d=
(00)(01)(02)(03,05)(04,25)(06)(07, 20)(08, 17)(12)(13)(09,14)(10)(11,15)
(16,29)(18,27)(19, 23)(21,28)(22)(24)(26)(30)(31)(32,35)(33)(34,36)
(37, 44)(38, 88)(39, 64)(40, 90)(41, 50)(42, 95)(43, 97)(45, 82)(46, 92)
(47)(48, 86)(49, 83)(79, 89)(58, 63)(59, 81)(60,94)(61, 98)(62, 84)
(80, 85)(87, 93)(91, 99)(96)(51, 72)(52)(53, 68)(54, 71)(55, 66)(56, 69)
(57)(65)(67, 76)(70, 75)(73, 78)(74)(77)
(6.7)
InH = GM, the normalizer of a 7-group is N,(7) = {a, c) of order 21.
In G the normalizer N&7) of a ‘I-group is of order 42, and S(7) is its own
centralizer in G. Thus in G there is an involution t such that N&7) =
(a, c, t) and such that tat = a-l. Of the two classes of involutions
in G, one fixes 20 of the 100 letters, the other none. Since (a) has 2 fixed
letters and 14 7-cycles, an involution I( fixing 20 letters must fix two or
more letters in one of the 7-cycles and for such an involution we cannot
have uau = a-l. Hence the involution t in iVG(7) must move all letters.
We may choose a conjugate of t in N,(7) so that ct = tc is of order 6.
We shall determine this permutation t.
Since tat = a-l, and t moves all 100 letters, t interchanges the two
letters 00 and 01 fixed by a. Since tc = ct and c fixes exactly the four letters
00, 01, 12, and 31, it follows that t also interchanges 12 and 31. Hence, as
I
tat = a-l,
Thus
Simple groups of order less than one million 165
a = (09, 10, 11, 12, 13, 14, 15)(30, 31, 32, 33, 34, 35, 36),
a-1 = (34, 33, 32, 31, 30, 36, 35)(13, 12, 11, 10,09, 15, 14). W-8)
t = (00,01)(09, 34)(10, 33)(11, 32)(12, 31)(13, 30)(14, 36)(15, 35). (6.9)
At this stage an element of luck enters in. The element tb2t has a relatively
large number of its values determined:
. .
tb2t = 00, 01, 13, 30, 32, 35, 31, . . .
34,01, 00, 12, 15, 10, 09, .
We now form an element fixing 00:
*
(6.10)
tb2a4tb2t = 00, 01, 13, 30, 31,32,35, . . .
00,34,01,10,09,15,12 ,... -
(6.11)
This is sufficient to determine the permutation completely. It is a3bsada3.
Thus t must satisfy
tb2a4tb2t = a3b6ada3.
In permutation form this begins
(6.12)
b2 ( 30,01, 19,05 24
01,09,36, 14 11 ’
14 11
t ( 01,09,36, 00,34, 14,36,21, 32 1 ’
(6.13)
The element t interchanging 00 and 01 normalizes the group (a, d) of
order 168 fixing 00 and 01. Its orbits are
w,
PI*
(09, .. .) 15},
. . . . . .
(30, . . ., 36},
(02, . . ., 08, 16, . . . , 22, 23, . . ., 29},
(51, . . ., 57, 65, . . ., 71, 72, . . ., 78},
(6.14)
(37, .. .) 43, 44, . . . , 50, 58, . . ., 64, 79, . . ., 85, 86, , 92, 93, , 99).