3 
Vol. 6, 1920 MATHEMATICS: G. A. MILLER 71 
hut are not restricted otherwise they generate the direct product of two cyclic 
groups of orders {m + n)k/d and d, respectively, d being the highest common 
factor of m, n, k. In particular, a necessary and sufficient condition that 
the group G generated by Si and 52 is cyclic is that m, n, k have no com- 
mon factor greater than unity. For instance, the cyclic group of order 
75 is determined by two commutative operators which satisfy the equations 
si' = 52^ isiS2y = 1. 
This group is also determined by ^i^*^ = 52^^, (5 1^2)^ = 1, etc. 
It may first be noted that G is always cyclic when m and n are relatively 
prime. In fact, when this condition is satisfied and p is any prime number 
which divides m then the Sylow subgroup of order contained in the 
group generated by Si includes the Sylow subgroup of order p^ contained 
in the group generated by 52, and a> ^ when k is divisible by p. When 
k is not divisible by p then a = (3 = 0, Hence it results that the Sylow 
subgroups of G whose orders are not prime to both of the numbers m 
and n are cyclic whenever m and n are relatively prime. The other 
Sylow subgroups of G must also be cyclic since they are generated by 
Si". This proves that G is cyclic whenever m and n are relatively prime. 
When m and n are not relatively prime let di represent their highest 
common factor. Just as before it follows that the Sylow subgroups 
of G whose orders are prime to di are cyclic, and those whose orders are 
not prime to di must also be cyclic when k is prime to di since the generator 
of such a Sylow subgroup in the group generated by Si must be the inverse 
of such a subgroup in the group generated by ^2. It has, therefore, been 
proved that a sufficient condition that G be cyclic is that the three numbers 
m, n, k have no common factor greater than unity. 
To prove that the order of G is (m -\- n)k when this condition is satisfied 
it is only necessary to prove that the orders of Si and ^2 divide this number 
and that G contains at least one operator whose order is equal to this 
number. The former of these facts results directly from the following 
equations : 
-k km _ -km _ kn k(m + n) _ _ k{m + n) 
S\ — S2 , Si — S2 — S2 , S2 — I — Si 
To establish the latter fact it may be noted that when a>0, is the 
highest power of p which divides k, and p^ is the highest power of p which 
divides m -\- n, then we let ti represent a constituent of Si of order 
and ^1 ~ ^ + the corresponding constituent of ^2, where / is so chosen 
that ^i" = ti-"^ + As n is prime to p, the product nl represents all 
the residues prime to p mod p", and hence it follows that G involves the 
cyclic subgroup of order ^ ^. 
It remains only to prove that when is the highest power of the prime 
number q which divides m -\- n, q being prime to k, G involves a cyclic 
subgroup of order g^. Let t2 be a generator of such a group. As t2^ + = 
1 it follows that t2^ = (^2~^)'^ and hence Si may be supposed to have 
