Since s2, s2 transform each other into their inverses, they are either commutative or they generate the quaternion group ; in the former case either both of them are of order two and they generate the four group, or both of them are the identity. The hypothesis that one is of order 2 while the other is the identity leads to an absurdity in view of the given equations. Hence the orders of sI, s2 must have one of the following pairs of values: 1, 1; 1, 2 ; 2, 2; 4, 4; 8, 8. In the first two cases sj, s2 would generate the identity and the group of order two respectively. In the third cases they would generate the dihedral group. As these cases are practically trivial, we shall confine our attention in what follows to the groups generated by s1 s2 when their common order is either 4 or 8. The main results of this paragraph may be expressed as follows: It two operators, neither of which is the identity, are such that each is transformed into its inverse by the square of the other, they must have the same order, and this common order is 2, 4, or 8. We begin with the case when 8s is of order 8. The quaternion group sa, s'2 tenerated by 82, Se is invariant under the the group (G) generated by sj, s2. This fact results from the following equations: