322 The Rev. Thos. P. Kirkman on 



The second is clearly true; because, 9^ being in G^^, 



GdGj- = G<f , 

 and this identity can be placed under any S. 



The former has to be proved in spite of the inequality 



a true negation, because 0^- is not in G^^, and no group is 

 identical with its derangement by a substitution not in the 

 group. 



But this required demonstration of the first is easy, if it 

 be granted that when two substitutions to be multiplied 

 together are placed under barred S, and the product is 

 equated to the tactical result under barred 2, that equation, 

 if the bars are at once suddenly removed, must be algebrai- 

 cally true. 



Let H be any substitution of the group G^, and let 0^ be 

 any substitution of the index-group which is not in G^, also 

 let barred 2 be written over both H and 0^. 



We have to perform, without regard to exponents, by the 

 rule for sinister multiplication {M.M, art. 3, p. 275) the 

 operation 



H-C= ^. 

 Looking now at the clusters under the repeated indices of 

 S, we see that, by the rule, the final cluster on the right of 

 H is deposited in K in the places and order of the final 

 cluster of 0,-, as well as, by the form of I<+i (art. 6), under the 

 final cluster of unity. But what change, if the bars be at 

 the same moment removed, has been made in the algebraical 

 value of that final cluster, in unity, in Qi or in H, by its 

 reading in K ? None ; the cluster is in all four the same 

 product of powers. The like is true of the ^th cluster, 

 whatever it be. We have nothing before us but 



where the left member stands only for what the derange- 

 ment H.O, becomes under 2. 



