COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

242 Takayuki Tamura
The number of groupoids corresponding to the groups bigger than @ is
Let r. = (1, 2, 3).
30+2+2 = 34.
We calculate 4j = 1024, 1024- 34 = 990, 990+2 = 495 for the number
of non-isomorphic G’s.
The number of the groupoids which have p = (1,2) as a dual automor-
phism is
2"x4 = 32
since1.1 = 1 or4, 2.3 = 1 or4, 3.2 = 1 or4.
Among the 1024 groupoids, there are 64 commutative ones. Eight of the
64 correspond to [(I, 2), (1,3)], and so 32 - 8 = 24 is the number of noncommutative
G’s which have (1,2) as a dual-automorphism. Two of these
24 G’s correspond to [(I, 2,3), (1,2,4)], leaving
24-2 = 22.
The number of commutative G’s is (64- 8) -+- 2 = 28 (up to isomorphism).
The number of non-commutative self-dual G’s is
22 t 2 = 11 (up to isomorphism).
To count the total number up to isomorphism and dual-isomorphism, vve
calculate
495- (28 + 11) = 456, 456 $2 = 228, 228 f39 = 267 for this number.
(8) h3 = u, 2x3,4)1.
1 Q / = 2, the normalizer is [(l, 2), (I, 3, 2, 4)] of order 8, n = 8/2 = 4,
c = 24/8 = 3.
The number of G’s with ‘%(G)x.@ is
240+240+210+42+2+2 = 736.
Under CE = (1, 2)(3, 4) we have the %-classes:
1.4-2 2 3 3x-4 4 $.2-.!- t2 ’ 3 42-d ?
14dL23 4:-a32 I 32-i 4 3jlt.P +i >
The calculation 48- 736 = 64,800, 64,800t 4 = 16,200 gives the number
of non-isomorphic G’s.
Semigroups and groupoids
Considering the self-dual G’S, we have
tU, 2X3, 411 c Ccl, 2),(3, 411,
u 2x3, 411 c [CL 3, 2, 411,
[(l, 2x3, 411 c [(I, 2)(3, 41, (1, 3x2, 4)i.
If ,!? = (1, 2) is a dual-automorphism, we have the &classes:
..+pr 334 a -cc 3; ; :
B,’ \\--
.3, I’ \ ‘:
,:
\ 4’ 31
I \
k\
a , ,f/p “‘> p
#2_321 3 4-a-4 3 24
B
12,33=3ar4 34=1,x2 I! = lOi2
Here we have 24~42 = 256 non-isomorphic G’s.
Ify = (1,3)(2,4), is a dual-automorphism, (1.3)~ = 1.3, but no element
is fixed by y. This case is impossible, therefore we see that there is no commutative
G.
If 6 = (1,3,2,4) is a dual-automorphism, under 6, we have the B-classes:
3.3
8
/
\“, /g3\ f3\ y3z\i,
“\ /2.2 1 2\ /2.’ +3\ /,2.4 3.1\, //2
4.4 3 4 !.4 4-l
We have 44 = 256 non-isomorphic G’s in this case. The two cases contain 16
groupoids in common among which 14 correspond to [( 1,2)(3,4), (I, 3)(2,4)1
and2to [(1,2, 3),(1,2,4)1, and we calculate 256- 16 = 240,240 X 2 = 480,
48Oi4 = 120 self-dual non-commutative G’s, and further calculation
gives 16,200-120 = 16,080, 16,080 + 2 = 8040, 8040f120 = 8160 for
the number up to isomorphism and dual-isomorphism.
(9) Q = [O, 211.
j.‘jj j = 2, the normalizer is [(l, 2), (3, 4)] of order 4, n = 4/2 = 2,
c = 2414 = 6.
The number of G’s with %(G)>$j is
60+240-l-14+2 = 316.
Under a = (1, 2) we have the %-classes:
243