# Finite abelian groups with the same order statistics are isomorphic

## 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 $s_k$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.