Finite abelian groups with the same order statistics are isomorphic

Statement
Suppose $$G$$ and $$H$$ are fact about::finite abelian groups 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) uses::Structure theorem for finite abelian groups

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 $$s_k$$s, 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 $$t_k$$s determine the $$s_k$$s.

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 $$t_k$$s determine the $$s_k$$s, completing the proof.