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From the first continued equation it follows that sx and s2 transform each other into their inverses and have a common square. From the second and third similar results follow with respect to s2, ss and sv sB respectively. Hence sv s2, sB are three operators such that each is transformed into its inverse by the other two. As any set of operators which fulfill the condition that each one is transformed into its inverse by all the others generates either the hamiltonian group of order 2 or the abelian group of this order* and of type (1, 1, 1, • • •), it follows that sv s2, s3 generate one of the following four groups : identity, the group of order 2, the four-group, or the quaternion group. That is, if sv s2, ss satisfy the three conditions imposed on them at the beginning of this paragraph, Q must be one of these four groups, and it is evident that these operators may be so chosen that G is any one of these four groups. If the given conditions are replaced by :

The groups generated by three operators each of which is the product of the other two