A THEOREM ON GROUPS. 
151 
[659 
659] A THEOREM ON GROUPS. 151 
ient A-lU-l . A 2 TJ 2 . 
nmutative, A. 2 A X 
the symbols, 1 
3r to use 12, 21 
(A 2 m B 2 ) m UJJ 2 = 1 . A,B. 2 m ; A 1 m B 1 (A 2 m B 2 ) m UJJ 2 = A^B^B^, 
and 
12 A^B, (A 2 m B 2 ) m U 1 U 2 = A./ tl B 2 A 1 B 1 m , 
where of course the right-hand sides denote arrangements. Hence in the formula 2°, 
die substitution- 
the two substitutions operating on U-JJ 2 give each of them the same arrangement 
A 2 m B 2 A 1 B 1 m , that is, the two substitutions are equal. And similarly the other formulae 
1°, 3°, 4° may be proved. 
By interchanging A and B, in the formulae I obtain 
1, A, B, ... is 
i. the order of 
1°. A, A™ . B x B 2 m = A 1 B 1 (A 2 B. 2 ) m ; 
B x Br . A x Ar = B l A 1 (B 2 A 2 ) m = A X B X (A 2 B 2 ) m , 
which is 
= A x A 2 m . B x Bp; 
2° and 3°. A x A 2 m . 12 B x B. 2 m =12 A^B, (A 2 m B 2 ) m 
irime, we have 
t be composite, 
12B x B 2 m . A x A 2 m = 12B X A X (B 2 A 2 ) m = 12A X B X (A 2 B 2 ) m , 
which is not 
= A x A 2 m . 12B X B™ \ 
3° and 2°. 12A x A 2 m . B x B 2 m =12A 1 B x (A. 2 B.^ m -, 
course is that, 
any two sub- 
* case in virtue 
B x B 2 m . 12A x A 2 m = 12A 1 B 1 m (A 2 B 2 m ) m =12A 1 B 1 m A 2 m B 2 , 
which is not 
= 12A X A™ . B x B 2 m ; 
4°. 12A x A 2 m . 12B 1 B 2 m = A x m B x (A 2 m B 2 ) m ; 
12B x B 2 m . 12A 1 A 2 m = A x B x m (A,B.; n ) m = (A 1 m B 1 ) m A 2 m B 2 , 
which is not 
= 12^.^™. 12B 1 B i m . 
a substitution 
G 1} A 2 B 2 = C 2 , 
forms C 1 C 2 m , 
iem. 
That is, in the double group any two substitutions of the form A x A 2 m are commutative, 
but a substitution of this form is not in general commutative with a substitution of 
the form 12B 1 B 2 m , nor are two substitutions of the last-mentioned form 12A 1 A 2 m in 
general commutative with each other; hence the double group is not in general 
commutative. 
In the formula 4°, writing B = A, we have 
s obtained by 
(12A 1 A 2 m ) 2 = A x m+1 4 2 “ i+m = A x m+1 . A, m+1 ; 
hence, if A, is the least integer value such that 
% aa™, 
A. (to + 1) = 0 (mod. fi), 
gements. By 
n 2 = 1 (mod. /¿) 
the arrange- 
we have (12A. X Aj n ) 2K = 1, viz. in the double group the substitutions of the second row 
are each of them of an order not exceeding 2 A, the substitution 12 being of course 
of the order 2. In particular, if ??x = /x —1, then A = 1: and the substitutions of the 
second row are each of them of the order 2.