150 
A THEOREM ON GROUPS. 
[659 
may for shortness consider for instance A X A 2 as denoting the arrangement A 1 U 1 .A 2 U 2 . 
But observe that in this use of the symbols the A 1} A 2 are not commutative, A 2 A X 
would denote the different arrangement A 2 U 2 . A 1 U 1 : in this use of the symbols, 1 
would denote U l U 2 , and 12 would denote TJ 2 U l , but it would be clearer to use 12, 21 
as denoting U-JJ 2 and U 2 U 1 respectively. 
These explanations having been given, I remark that in every case the substitution- 
group 1, A, B,... gives the double group 
1, A X A % , B X B 2 ,... 
12, 12 A X A 2 , 12B x B 2 ,... 
as is at once seen to be true: but further when the original group 1, A, B, ... is 
commutative, then if m be any integer number, such that m 2 = 1 (mod. the order of 
the original group), we have also the double group 
1, A x A 2 ™, B x B 2 m ,... 
12, 12A x A 2 m , 12B x B 2 m ,... 
where of course if the order of the original group (=/a suppose) be prime, we have 
m = 1 or else m = —1 (mod. ¡x), say m= 1 or ¡i — 1; but if the order fx be composite, 
then the number of solutions may be greater. 
The condition in order to the existence of the double group of course is that, 
in the system of substitutions just written down, the combination of any two sub 
stitutions may give a substitution of the system. And this is in fact the case in virtue 
of the formulae 
1°. A 1 Aj M . B x B 2 m = A X B X (A 2 B 2 f\ 
2°. A : A 2 m . 12B x B 2 m = 12A 1 m B 1 (A 2 m B 2 ) m , 
3°. 12A x A 2 m . B x B 2 m = 12 (A.B,) (A 2 B 2 ) m , 
4°. 12A 1 A % m . 12BjBf* = A 1 m B 1 (A 2 m B 2 ) m , 
inasmuch as 1, A, B,... being a group, AB and A m B are each of them a substitution 
of the group, = G suppose; we have of course in like manner A 1 B 1 = G 1> A 2 B 2 = C 2 , 
etc., and the right-hand sides of the four formulae are thus of the forms CjC 2 m , 
126 , 1 C 2 m , 12C' 1 C 2 w , G^C^ 1 respectively, viz. these are substitutions of the system. 
To prove for instance the formula 2°, considering the arrangements obtained by 
operating upon U-JJ 2 , we have 
B x B 2 m U 1 U 2 = B 1 B 2 m , 12B 1 Bj n U 1 U 2 = B 2 B™, A x A 2 m 12B 1 B 2 m U 1 U 2 = A 2 m B 2 A x B x m , 
where of course the expressions on the right-hand side denote arrangements. By 
reason that the original group is commutative (A m B) m is =A m2 B ,n or since m 2 = 1 (mod. ¡i) 
this is =AB m \ hence also (A 2 m B 2 ) m = A 2 B 2 m : hence, considering as before the arrange 
ments obtained by operating on U x U 2i we have