A Cubic Curve Connected with Two Triangles

/ = /i$i. Since (to)i = 5ii is in (s), and ni is prime to the order of Si, Si is a power of s. Thus h, as well as t, corresponds to t in the isomorphism of G with G/(s) ; but h and /' are of the same order tii. Every element of G/(s) whose order is a divisor of m/mi corresponds to an element of G whose order is a divisor of m. It follows that t', and hence every element of G/(s) whose order divides n, is commutative with every element whose order divides m /mi. Hence G/(s), being of order <tnn, contains an invariant subgroup of order n. The corresponding subgroup of Gy being of order min<mn, also contains an invariant subgroup of order n. Thus in all cases G contains a subgroup of order n. Similarly G contains a subgroup of order m. G is evidently the direct product of these two subgroups.