### The number of subgroups of given index in nondenumerable Abelian groups

W. R. Scott
1954 Proceedings of the American Mathematical Society
Let G be an Abelian group of order A >No-It 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]
more » ... 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. Lemma. Let H^O be an Abelian group, and let G= E^«> «£■\$, Ha=Hfor aliaES, o(S)= A ^i\$o. Then thereareat least 2A subgroups Ks of G such that G/K^H. Proof. (This proof is the same as Baer's, and is included only for the sake of completeness.) Identify Ha with H. Let ea be 0 or 1 for each a, aES. Let K be the set of elements of G such that ha a j^clo-Then it is easy to verify that AT is a subgroup of G and G=Hat +K. Ii ea = 0, then HaQK, but if e0 = l, then 7JaPiA = 0. Thus all of the K's are distinct, and the lemma is proved.