On several theorems of operation groups
G. A. Miller · Bulletin of the American Mathematical Society · 1896
we proved the theorem, "Every group (G) whose order is divisible by p*,p being any prime number, contains a commutative group (G x ) of order p 9 ."The following proof of this theorem is much simpler and can readily be extended to apply to more general theorems.G contains a subgroup (G') of order p a , a >3.-G' contains a subgroup of order p whose substitutions* are commutative to all the substitutions of G'.fWith respect to this subgroup G' is isomorphic to a group ( G/) of order p*" 1 .G/ contains a subgroup of order p whose substitutions are commutative to all the substitutions of G/.With respect to this subgroup G/ is isomorphic to a group (G 2 ') of order p a ~2.Hence we may suppose the substitutions of G' so arranged that the first p^(/5= 0,1, 2, 3, •-, a -1) constitute a self-conjugate (invariant) subgroup of G' and that each of its p sets of p^~x substitutions, in order, is transformed into itself by all the substitutions of G'.J If we suppose p = 2 each of the p sets contain p substitutions.The substitutions of p -1 of these sets must be transformed, by all the substitutions of G', according to the cyclical group of order p or according to identity.Those in the first set are known to be transformed according to identity.Hence each of these p 2 substitutions must be commutative to at least ^a~~1 substitutions of G' and the the first p 4 * in the given arrangement must contain a commutative group of order p*.This proves the given theorem.In general, the first p$~l substitutions in the given arrangement are transformed by G' according to a group (H) of order p 0 .A substitution which is commutative to all the substitutions in the second set of p&~2 is commutative to each of the given pP~x substitutions.Hence IT is simply isomorphic to a group whose degree cannot exceed p&~2 and the maximum value (M) of 0 is given by the formula § * The operations are throughout represented by means of substitutions.