Finite abelian groups with the same order statistics are isomorphic

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Statement

Suppose G and H are Finite abelian group (?)s that are order statistics-equivalent: the order statistics of G equal the order statistics of H. Then, G is isomorphic to H.

Facts used

  1. Structure theorem for finite abelian groups

Proof

Proof idea

We show that the invariants needed to describe a finite abelian group by the structure theorem are completely determined by the order statistics.

Proof details

By fact (1), any finite abelian group can be written as a direct product of cyclic groups of prime power order, with the number of copies of a cyclic group of prime power order independent of the choice of decomposition.

We claim that the order statistics of a finite group determine the number of times each cyclic group of prime power order occurs as a direct factor. First, for every prime p, consider the subgroup of elements whose order is a power of p (in other words, the p-Sylow subgroup). This is obtained by grouping together, in the direct product decomposition, all cyclic groups of order a power of p. Thus, we may restrict attention to this subgroup, so it suffices to consider the case that G is a finite abelian p-group.

For any k, let s_k be the number of cyclic group factors of order p^k, let s be the sum of the s_ks, and let t_k be the logarithm to base p of the number of elements of order dividing p^k. This makes sense since theelements of order dividing p^k form a subgroup. Also, t_k is clearly determined by the order statistics. We claim that the t_ks determine the s_ks.

An easy count shows that for all k:

t_k = \sum_{i < k} is_i + k\sum_{j \ge k} s_j.

We thus get, taking a first difference:

t_{k+1} - t_k = \sum_{j > k} s_j.

Taking a second difference yields:

2t_k - t_{k+1} - t_{k-1} = s_k.

This shows that the t_ks determine the s_ks, completing the proof.