THE GROUPS OF ORDER pm WHICH CONTAIN EXACTLY p CYCLIC SUBGROUPS OF ORDER

If a group ( G ) of order pm contains only one subgroup of order pa, a > 0, it is known to be cyclic unless both p = 2 and at = l.f In this special case there are two possible groups whenever m > 2. The number of cyclic subgroups of order pa in G is divisible by p whenever G is non-cyclic and p > 2. J In the present paper we shall consider the possible types of G when it is assumed that there are just p cyclic subgroups of order pa in G. That is, we shall consider the totality of groups of order pm which satisfy the condition that each group contains exactly p cyclic subgroups of order p". It is evident that a < m. Since the total number of subgroups of order pa in G is of the form 1 + kp, it follows that a > 1. For all values of a greater than unity there is at least one group of order pm which contains exactly p cyclic subgroups of order pa, viz., the abelian group of type (m — 1,1). When p is odd there is a non-abelian group which is conformai with this abelian group. It will be proved that these two groups are the only groups of order p"1, p> > 3, which contain exactly p cyclic subgroups of order pa. These two groups exist also when p = 3 or 2 and m > 3, but they are not the only groups which contain p cyclic subgpoups of order p", p = 2. When at = 2 and m = 4 there is another group of order 3TM which contains just 3 cyclic subgroups of order 9. Let the p cyclic subgroups of order p" be represented by _P,, P2, ■ ■ ■, P . Each of these transforms every other one into itself. The group generated by any two of them contains all the others, since it cannot be cyclic. Let sx, s2, be generators of Px, P2, respectively. From the fact that s~'s2sx = sj and s~1sls2= s\, it follows that the commutator subgroup of the group (K) generated by sx, s2, is composed of operators which are invariant under this group. The order of this commutator subgroup cannot exceed p since spx = s"f. This