COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.

240 Takayuki Tamura
The calculation 4X23 = 32, 32 -2 = 30 gives the number of non-isomorphic
G’s. If a dual automorphism exists, then it is in $j. Accordingly
if G is self-dual, it must be commutative.
We find the commutative G’s:
(1.2)~ = 2.1 = 1.2, 1.2 = 3 or 4.
We have 8 non-isomorphic commutative G’s, and the calculation 30- 8 =
22, 2292 = 11 gives the number of non-self-dual G’s, and we have a
total of 8 + 11 = 19 non-isomorphic and non-anti-isomorphic G’s.
(4) @ = [(I, 2x3, 41, (1, 3x2, 411.
I@ / = 4, Q is normal, it = 24/4 = 6, c = 1,
The number of groupoids G with $j c a(G) which appeared in the previous
cases is
14X3+2+2 = 46.
Let c( = (1, 2)(3, 4), /I = (1, 3)(2, 4). We have the ‘B-classes:
‘.I*;;\ j.4 1~2~ /,9 l.3A //?.2 ‘.4\ 2,’
3.3 3.4 3.1 3.2
The calculation 44-46 = 210, 210~6 = 35 gives the number of nonisomorphic
G’s. There is no commutative G in the 35 G’s, because
(1*2)a = 1.2
and we have no value 1.2.
Suppose y = (1,3) is a dual automorphism. We then have the ‘B-classes:
2.2
7-c
1.1 4.4
x/d
3.3
2.4
2-t
1.3 4.2
\Jd
3.1
The class of 22~4 = 16 groupoids contains 2 of those corresponding to
u, 2, 31, (1, 2,4)1, and so we calculate 16-2 = 14, 14t2 = 7, for the
number of self-dual, non-commutative G’s. Since the normalizer of [(l, 3),
(1, 2, 3, 4)] is itself, 8t4 = 2, and we calculate 35-7 = 28, 28~2 = 14
non-self-dual G’s, giving 14+7 = 21 for the total up to isomorphism and
dual-isomorphism.
Semigroups and groupoids
(5) @ = to, 2, 3, 4)l.
/.!!I = 4. The normalizer of $j is [(l, 3), (1,2,3,4)] of order 8, n = 8/4 = 2,
c = 2418 = 3.
The number of the groupoids corresponding to the groups which contain
Sj is
14+2 = 16.
Let tc = (1,2, 3,4). We have the ‘B-classes:
y2.2\ // ;‘i y3.2\ /2’4\
1.1 3.3 1.2 3-4 2.1 4.3 1.3 3-I
\ / \ / y / 1, p’
4.4 4.1 1-4 4.2
We calculate 44- 16 = 240, 240 -z- 2 = 120 for the number of non-isomorphicG’s.Wehave[(1,2,3,4)]~[(1,3),(1,2,3,4)].Supposey
=(1,3)isa
dual-automorphism. Then (1.2)a = (l-2)7, but there is no value l-2 which
satisfies this. Suppose some G is commutative. Then (1.3)~~ = 3.1 = 1.3,
but there is no 1.3 fixed by a2. Consequently there is no self-dual G in this
case, and we have only the
120 a 2 = 60 non-self-dual G’s.
(6) 43 = [Cl, 21, (3>4)1.
/Q 1 = 4. Its normalizer is [(l, 2), (1,3,2,4)] of order 8, n = 814 = 2,
c = 2418 = 3.
The number of the groupoids corresponding to the groups 2 $j is
14+2 = 16.
Let a = (1,2), B = (3,4). We have the %-classes:
We calculate 24~ 42 = 256, 256- 16 = 240, 240t2 = 120 for the number
of non-isomorphic G’s. We can prove that there is no self-dual G, so we
have
120 + 2 = 60 non-self-dual G’s.
(7) hj = cu, 2, 311.
/ $j j = 3. The normalizer is [(l, 2), (1, 3)] of order 6, n = 6/3 = 2,
c = 2416 = 4.
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