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MATHEMATICS: G. A. MILLER Proc. N. A. S. 
t2 as its constituent whose order is a power of q while S2 has ^2 ~ ^ as its corre- 
sponding constituent. It has, therefore, been proved that when m, n, k 
do not have a highest common factor greater than unity, and Si and S2 
are commutative, the largest group which they can generate subject to 
the condition ^i"" = ^2"^, (^152)^ = 1 is the C5^clic group of order (w + n)k. 
When the highest common factor of m, ^ is <i> 1 the Sylow subgroups 
of G whose orders are prime to d are seen to be cyclic just as in the pre- 
ceding case. The order of every operator in such a group must divide 
km km nk kn km 
(m -\- n)k/d, since Si^ = S2~^ Sind hence Si ^ = S2 = S2^ , Si ^ =Si'^. 
Let r be any prime divisor of d and let r* be the highest power of r 
which divides d while + * is the highest power of r which divides 
{m -\- n)k. From the preceding proof it follows that G involves an operator 
h of order / since such an operator generates a Sylow subgroup of the 
group determined by the equations 
53- = 5/, {s,s,f^' = 1. 
Suppose that 53 generates ts. If an operator which generates the Sylow 
subgroup whose order is a power of r contained in the group generated 
by Si is multiplied by an independent operator of order this product 
and Sz satisfy the conditions 
Sr = S2\ (5i5o)^ =1. • 
Hence it results that the Sylow subgroup of G whose order is a power of 
r involves two independent operators of orders / and r*, respectively. 
To complete a proof of the theorem under consideration it is, therefore, 
only necessary to establish the fact that r ^ ' is the order of a Sylow 
subgroup of G. 
When k is not divisible by r ^ ^ this fact is evident since the order of 
neither of the two operators Si, S2 is divisible by Z"*"^ and the order of at 
least one of them is divisible by /. If the order of the Sylow subgroup 
in question would exceed / ' the order of the product of these two 
operators would have to be divisible by f * ^ . When k is divisible by 
r*"^^ at least one of the two numbers m, n is not divisible by f ^ . 
Suppose that m is not divisible by \ Hence the order of the Sylow 
subgroup in question cannot exceed f times the order of the correspond- 
ing Sylow subgroup in the group generated by 52. This completes a proof 
of the theorem in italics announced in the second paragraph of this note. 
This theorem is useful in the study of the generalized groups of the 
regular polyhedrons as may be seen from the fact that if two commutative 
operators satisfy the equations Si^ = 52^, (^152)^ = 1, they generate the 
cyclic group of order 14 or one of its subgroups. This explains why the 
largest non-abelian group, which is generated by two operators satisfying 
these conditions, is a direct product of the group of order 7 and some 
other group (Miher, Blichfeldt, Dickson, Finite Groups, 1916, p. 157). 
