Groups in which the squares of the elements are a dihedral subgroup

Abstract

1. The dihedral subgroup is non-abelian. If a group G has the property that the squares of its operators constitute a given group then the direct product of G and any abelian group of order 2m and of type (1, 1, 1, * ) has the same property. Such direct products will always be excluded in what follows. It has been noted that a necessary and sufficient condition that there is at least one group G which has the property that the squares of its operators constitute a given non-abelian dihedral group H is that the order h of H is not divisible by 8 and that every odd prime number which divides h is congruent to unity modulo 4.t To determine the number of these groups when h is given and satisfies these conditions we let n represent the number of the different prime numbers which divide h. Since the cyclic subgroup K of index 2 contained in H is the direct product of its Sylow subgroups and all the groups of isomorphisms of these Sylow subgroups are cyclic it results that when h is twice an odd number there is one and only one group of order h 2n-2 which satisfies the condition that the squares of its operators constitute K. The given group of order hk 2n-2 is the largest group which has the property that the squares of its operators constitute K when h is twice an odd number, as will be assumed until the contrary is explicitly stated. It is merely the direct product of n-1 dihedral groups, each of order twice the power of an odd prime. The commutator subgroup of every G is K and the corresponding quotient group is the abelian group of type (2, 1, 1, * * ). This results directly from the fact that if two operators of G have squares which are contained in K then the square of their product is also contained therein since this square could not be an operator of order 2 in H in view of the fact that this product could not transform this cyclic subgroup according to an operator of order 4. It should be emphasized that an operator of G which transforms some of the operators of the cyclic subgroup of index 2 in H according to an operator of order 2 does not necessarily transform all the operators of this subgroup, besides the identity, according to such an operator, but that every operator