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The number of subgroups of given index in nondenumerable Abelian groups

1954
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Proceedings of the American Mathematical Society
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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]

doi:10.1090/s0002-9939-1954-0059904-6
fatcat:vf3dhkzh25dizhaaqk2qm6gz2a