226 Mr. G. A. Miller on the Groups of Isomorphisms 
subgroup corresponds to itself in every holomorphism of G. 
The invariant operator of order 2 also corresponds to itself. 
Hence G contains a characteristic subgroup of index 2, and 
each of the remaining operators may correspond to either of 
two operators after any holomorphism of this characteristic 
subgroup has been established. As the order of I is equal to 
that of G and as I contains a complete group which is simply 
isomorphic with the complete subgroup of index 2 in G, it 
follows that I is simply isomorphic with G. 
Theorem V. — The necessary and sufficient condition that an 
operator s of G transforms the operators of G according to an 
invariant operator under its group of cogredient isomorphisms 
is that the conjugates of s under G may be obtained by multi- 
plying s by invariant operators under G. The highest order 
of such an invariant operator is the order of the corresponding 
operator in the group of cogredient isomorphisms. 
This theorem follows directly from the isomorphism be- 
tween G and its group of cogredient isomorphisms. It is 
clear that a holomorphism corresponds to an invariant 
operator in the group of cogredient isomorphisms whenever 
the operators which correspond to themselves in the holo- 
morphism constitute an invariant subgroup, the corresponding 
divisions are invariant, and the remaining operators corre- 
spond to themselves multiplied by invariant operators. 
Similarly, it may be observed that a holomorphism corre- 
sponds to an invariant operator of I when the operators 
which correspond to themselves in this holomorphism consti- 
tute a characteristic subgroup, the corresponding divisions 
are characteristic, and the operators of G which are not in 
the given characteristic subgroup correspond to themselves- 
multiplied by characteristic operators of G. 
Theorem VI. — If a substitution-group of degree n contains 
a subgroup of degree n — 1 and involves no subgroup which is 
both of degree n and also of index n, then its group of iso- 
morphisms is simply isomorphic with a substitution-group of 
degree n which contains the given group invariantly. 
No two substitutions of G could transform the n subgroups 
of degree n— 1 in the same manner, since each substitution 
transforms these subgroups in the same manner as it trans- 
forms its elements*. If an operator of I were commu- 
tative with each one of these n subgroups it would also be 
commutative with every operator of G. As this is impossible, 
* Bulletin of the American Mathematical Society, vol. ii. (1896) 
p. 145. 
