( 355 ) 



We in the first place remember that the S •'>. 2), // = 9 deduced 

 from (S(4,3), // = 10 is regular commutative (Moore), i.e. that its 

 group possesses a regular commutative subgroup of the type : 



1 2 3 4 5 6 7 8 9 



2 :; I 5 6 4 8 9 7 



3 L 2 6 4 5 9 7 8 



4 5 6 7 8 ( .» 1 2 3 



5 6 4 8 9 7 2 3 I 



6 4 5 9 7 8 3 1 2 



7 8 9 1 2 3 4 5 (I 



8 9 7 2 3 1 5 6 4 



9 7 8 3 I 2 6 4 5 



Indeed, an 5(3,2), n = 9 appears if uc submit the triplets 

 L 2 3 : 14 7 ; 15 9 ; 1(1 8 

 lo all the substitutions of this group. 



If we now add to each of' these twelve triplets a zero and if we 

 submit the quadruplets 



1 2 4 5 and 1 2 6 9 

 to the substitutions of the group, then \\\v 1 2 -f- 'J 8 = 30 quadruplets 

 of the 8(4,3), n = 10 are formed. 



Secondly we observe, that the system is cyclic and appeal's among 

 others by submitting - the quadruplets 



1-237 ; 1245 ; 1 3 5 8 

 to the cycle (1 2 345 6 7 8 9 0). 



C. If we choose out of an 8(5,4), n = ll, that is aCf.(ll, ,66 § ), 

 a quintuple 1 2 3 4 5, then all triplets of it must appear still three 

 times, all pairs moreover still two times, the single letters after- 

 wards three times, with which the 10 3 -f- 10 ■ 2 -f 5 . 3 = 65 

 remaining quintuplets of the system are exhausted. The single letters 

 are completed with quadruplets out of 6, 7, 8, 9, 0, a, which may 

 have at most three letters in common and which form together a 

 Cf. (6 10 , 15 4 ), consisting of five 6 a , 3 4 as appeared in />. Of this but 

 one type exists 1 ), so that e.g. the 8(5,4 commences with: 



1 8 9 a 2 7 9 0./ 3780a 47 8 9 a 5 7 8 9 



1 (5 7 <t 2 6 8 9 n 3 <; s \\ 4 <; S ,/ 5 t; 9 „ 

 1 8 7 <s 9 2 6 7 8 3 (I 7 9,/ 4 (5 7 9 5 b" 7 H a 



') Deduced from the well-known system of live three-divisions of six elements 

 oi Sebukt. 



