COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA. COMPUTATIONAL PROBLEMS IN ABSTRACT ALGEBRA.
234 Takayuki Tamura For S, In the following table [(I, 2, 3)] is the permutation group generated by a 3-cycle (1,2,3). [(I, 2)] .1s one generated by a 2-cycle or substitution (I, 2). & 1 1 1 [Cl, 2, 311 1 2 4 -__~ Kl, 2)l 3 1 8 - - - {e> 1 6 116 ____ Total 129 TABLE 2. Groupoids of Order 3 L Self-dual, Non-selfnon- dual, comm, up to iso, up to iso up to dua 0 1 0 4 0 35 9 1556 9 1596 Total up to is0 ‘9 up to duz 11 2 8 43 1681 1734 - I - -. - Semi- Total groups up to is0 up to iso, up to dual 3 1 12 0 78 5 3237 12 3330 18 By Theorem 1, if 8 = &, we have two isomorphically, dual-isomorphically distinct groupoids : Case $j = [(I, 2, 3)]. Let E = (1, 2, 3). %-classes : + 11 122 i denotes the multiplication table i 12 Semigroups and groupoids 235 Since there is no restriction to choosing ((1 . l)e, (l-2)6, (2.1)6}, we have 27 groupoids G such that Cal S WG). However, the set of the 27 groupoids contains the 3 groupoids in which SI is the automorphism group; the number of isomorphically distinct groupoids G for $j = [E] is +(27- 3) = 12 where n = 2 in Table 2. If G has dual-automorphisms, b = (1, 2) must be a dual-automorphism. In this case, since (3*3)/l = 3.3, (l-2)8 = 1.2, (2*1)8 = 2.1, we must have (3.3)6 = (1*2)8 = (2*l)f3 = 3. Therefore if @ S. ‘%(G) and if (1, 2) is a dual-automorphism of G, then G is while this G is already obtained for LQ = Ss, and (1, 2) is an automorphism. In the present case there is no self-dual, non-commutative groupoid. There are formally 9 commutative groupoids; but excluding one we have nonisomorphic commutative groupoids : f(9-1) = 4 and 12-4 = 8 non-self-dual G’s, 8e-2 = 4 isomorphically distinct non-self-dual G’s. Therefore we have 4+4 = 8 non-isomorphic, non-dual-isomorphic G’s. Case $j = [(l, 2)]. Let CI = (1, 2). There are 5 &classes among which a class consists of only 3.3. Clearly (3.3) 8 = 3. The number of non-isomorphic G’s is 81-3 = 78. If G is commutative then (1.2)~ = 1.2, hence (l-2)0 = 3. The number of non-isomorphic commutative G’s is 9-l = 8. The number of non-self-dual G’s is 78-8 = 70, and hence the number of those, up to isomorphism, is 70+2 = 35. Therefore the number of non-isomorphic, non-dual-isomorphic G’s is 35+8 = 43. We remark that there is no non-commutative self-dual groupoid for sj because SI has no subgroup of order 4.
236 Takayuki Tamura Case ,Q = {e). First, we find the number of self-dual, non-commutative groupoids G for {e}. Let j? = (1, 2) be a dual-automorphism of G. Then we can easily see that and hence (1*2)/I = (1*2), (2.1)8 = (2,1), (3*3)/I = (3.3) (l.l)/!? = (2.2) (1.3)/I = (3*2), (3-l)/!? = (2.3), (1.2)8 = (2*1)8 = (3*3)8 = 3. The 27 G’s contain the 9 G’s which appeared in the previous cases. We have that the number of isomorphically distinct G’s is $(27-9) = 9 since we recall that the normalizer of [(I, 2)] is itself. The number y = 116 of all non-isomorphic commutative groupoids whose automorphism group is {e} is the solution of 6y+4X2+8X3+1 = 3s. The number x = 3237 of all non-isomorphic groupoids corresponding to {e} is the solution of 6x+78X3+12X2+3 = 3g. The number of non-self-dual G’s, up to isomorphism and dual-isomorphism, is +(3237-(116f9)) = 1556. The total number of G’s up to isomorphism and dual-isomorphism is the sum 116+9+1556 = 1681. For [(l, 2, 3)], let CC = (1, 2, 3): where (x, y, z) is I /x y 2 zu xu yx yd 2s xa? I I (1, 2, I), (2, 2, I), (2, 3, 2), (2, 1, 3) commutative, (1, 1, 2), (2, 1, l), (2, 1, 2), (2, 3, 1) non-self-dual. For [(l, 2)], let fl = (1, 2): where (x, z) is (1, 11, (1, 3), (2, 11, (2, 2), (2, 3), (3, 11, (3, 2), (3, 3). Non-self-dual : where (x, y, z, u) is Semigroups and groupoids commutative XY z YB XB ZB u u/9 3 (1,1,1,1),(1,1,3,1),(1,3,2,3),(2,1,2,2),(2,3,1,2),(3,1,1,3),(3,1,3,2), (1,1,1,2),(1,1,3,2),(2,1,1,1),(2,1,2,3),(2,3,1,3),(3,1,2,1),(3,1,3,3), (1,1,2,1),(1,1,3,3),(2,1,1,2),(2,1,3,1),(2,3,2,3),(3,1,2,2),(3,3,1,2), (1,1,2,2),(1,3,1,2),(2,1,1,3),(2,1,3,2),(3,1,1,1),(3,1,2,3),(3,3,1,3), (1,1,2,3),(1,3,1,3),(2,1,2,1),(2,1,3,3),(3,1,1,2),(3,1,3,1),(3,3,2,3). For {e}: where (x, y, z) is (1, 1, 2), (2, 1, 2), (3, 1, 2), (1, 1, 3), (2, 1, 3), (3, 1, 3), (1, 2, 3), (2, 2, 3), (3, 2, 3). x3 Y 3 XB ZP z YB 3 237
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236 Takayuki Tamura<br />
Case ,Q = {e). First, we find the number of self-dual, non-commutative<br />
groupoids G for {e}. Let j? = (1, 2) be a dual-automorphism of G. Then<br />
we can easily see that<br />
and hence<br />
(1*2)/I = (1*2), (2.1)8 = (2,1), (3*3)/I = (3.3)<br />
(l.l)/!? = (2.2) (1.3)/I = (3*2), (3-l)/!? = (2.3),<br />
(1.2)8 = (2*1)8 = (3*3)8 = 3.<br />
The 27 G’s contain the 9 G’s which appeared in the previous cases. We<br />
have that the number of isomorphically distinct G’s is<br />
$(27-9) = 9<br />
since we recall that the normalizer of [(I, 2)] is itself.<br />
The number y = 116 of all non-isomorphic commutative groupoids<br />
whose automorphism group is {e} is the solution of<br />
6y+4X2+8X3+1 = 3s.<br />
The number x = 3237 of all non-isomorphic groupoids corresponding<br />
to {e} is the solution of<br />
6x+78X3+12X2+3 = 3g.<br />
The number of non-self-dual G’s, up to isomorphism and dual-isomorphism,<br />
is<br />
+(3237-(116f9)) = 1556.<br />
The total number of G’s up to isomorphism and dual-isomorphism is the<br />
sum<br />
116+9+1556 = 1681.<br />
For [(l, 2, 3)], let CC = (1, 2, 3):<br />
where (x, y, z) is<br />
I<br />
/x y 2<br />
zu xu yx<br />
yd 2s xa?<br />
I I<br />
(1, 2, I), (2, 2, I), (2, 3, 2), (2, 1, 3) commutative,<br />
(1, 1, 2), (2, 1, l), (2, 1, 2), (2, 3, 1) non-self-dual.<br />
For [(l, 2)], let fl = (1, 2):<br />
where (x, z) is<br />
(1, 11, (1, 3), (2, 11, (2, 2),<br />
(2, 3), (3, 11, (3, 2), (3, 3).<br />
Non-self-dual :<br />
where (x, y, z, u) is<br />
Semigroups and groupoids<br />
commutative<br />
XY z<br />
YB XB ZB<br />
u u/9 3<br />
(1,1,1,1),(1,1,3,1),(1,3,2,3),(2,1,2,2),(2,3,1,2),(3,1,1,3),(3,1,3,2),<br />
(1,1,1,2),(1,1,3,2),(2,1,1,1),(2,1,2,3),(2,3,1,3),(3,1,2,1),(3,1,3,3),<br />
(1,1,2,1),(1,1,3,3),(2,1,1,2),(2,1,3,1),(2,3,2,3),(3,1,2,2),(3,3,1,2),<br />
(1,1,2,2),(1,3,1,2),(2,1,1,3),(2,1,3,2),(3,1,1,1),(3,1,2,3),(3,3,1,3),<br />
(1,1,2,3),(1,3,1,3),(2,1,2,1),(2,1,3,3),(3,1,1,2),(3,1,3,1),(3,3,2,3).<br />
For {e}:<br />
where (x, y, z) is<br />
(1, 1, 2), (2, 1, 2), (3, 1, 2),<br />
(1, 1, 3), (2, 1, 3), (3, 1, 3),<br />
(1, 2, 3), (2, 2, 3), (3, 2, 3).<br />
x3 Y<br />
3 XB ZP<br />
z YB 3<br />
237
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