251 
The substitution 13« 25 '67034 is 
1537284609«-«943867520 l=:a 073684259 1 = 0 C , 
whose circular factors are 2 0 9 5 6 8, 4 3 7, a 1 . 
It can be demonstrated, either a priori, or by simple in- 
spection of the above table of the 55 triads, that none of the 
twelve just written transposes a consecutive pair of the circle 
2 0 9 5 6 8, i.e., none has a transposition which fractures 
that circle into one of five and another of one. And as each 
of the twelve has one element of each of the three circles 
undisturbed, it must have for one of its four transpositions a 
non-consecutive pair of the circle 2 0 9 5 6 8, while the other 
three transpositions will of necessity be junctures of the four 
resulting circles into one circle of eleven. 
Theor. A transposition of two letters of any circle always 
fractures that circle into two : a transposition of elements of 
tioo circles always unites those circles into one. 
From this follow many important theorems on substitu- 
tions. Hence we deduce easily, as above for substitutions of 
the seventh order, that there are 12T0;’ substitutions of the 
eleventh order, represented by 12 • 2 * 55 • 6 triplets A(BC) 6 , 
each substitution by at least six different triplets having the 
same A. Hence r= 1 or r=ll, of which the former only is 
the true value ; and there are 120 substitutions of the 
eleventh order. We prove readily that no quadruplet A B C D 
is irreducible; hence no quintuplet, &c. And the 55 substi- 
tutions of the second order, 1 5 7, 5 71, &c., form with their 
products a group of 
6 6 -4 5 +55 -2 6 + 55-2 3 + 55- 1 2 + 12- 10„+ lj = 11 * 10 * 6 
substitutions. This group is maximum, and has 9 , 8 - 7 , 5 , 4 , 3 , 2 
equivalents. 
We thus find that, in the discussion and construction of 
these hitherto difficult groups, we can dispense with con- 
gruences, and with the still more formidable apparatus of 
imaginary subindices, which MM. Betti and Mathieu have 
so skilfully handled. 
