The Number of Subgroups of given Index in Nondenumerable Abelian Groups

Let G be an Abelian group of order A >NoIt has been shown [4, Theorem 9 ] that there exist 2A subgroups of G of order A, and that the intersection of all such subgroups is 0. In this paper, this result is improved to the following: If b$0^B^A and ^4>N0, then an Abelian group of order A has 2A subgroups of index B, and the intersection of all such subgroups is 0. In addition, it is shown that there is a set of 2A subgroups Ha of index B such that G/Ha=G/Ha' for all a, a'. Baer [l, p. 124] showed that if G is an Abelian £-group which is the direct sum of A cyclic groups of bounded order, then G has 2A subgroups of index p (here A may equal ^0). The proof in the present paper is accomplished by extending Baer's result in an obvious manner to a wider class of ^-groups, and then reducing all other cases to this one. We shall use + and E to denote direct sums, and o(S) to denote the number of elements in S.