Let $A$ and $B$ be finite abelian groups.
Suppose that for every natural number $m$, the number of elements of order $m$ in $A$ is equal to the number of elements of order $m$ in $B$.
Prove that $A$ and $B$ are isomorphic.
Idea
Given that these groups are finite, I think you have to use the primary decomposition theorem somehow.
Best Answer
Here's how I would look at it. Suppose you are given the order $g$ of the group which has canonical factorization $p_1^{\alpha_1} p_2^{\alpha_n} \dots p_n^{\alpha_n}$. By the fundamental theorem of finite group your group is direct product of $n$ abelian $p$-groups, which are in turn direct products of cyclic subgroups of different orders. We want to determine for each prime which $p$-group is being used.
To do this look at the elements whose order has only $p$ as a prime divisor. Those are the elements which are of the form $(1,1,1,g,1,1\dots 1)$ These elements form a subgroup which is isomorphic to the $p$-subgroup.
So we would like to discern which subgroup it is by knowing the elements of order power of $p$.
To prove we can do this we need to prove that if
$\mathbb Z_{p^{a_1}}\times\mathbb Z_{p^{a_2}}\dots \mathbb Z_{p^{a_r}}$ and $\mathbb Z_{p^{b_1}}\times\mathbb Z_{p^{b_2}}\dots \mathbb Z_{p^{b_s}}$ are groups of order $p^n$ with different exponents they have a different count of orders.
To see this note that the order of an element in a direct product $g=(g_1,g_2\dots g_n)$ is the least common multiple of all of the orders, in a $p$-group it is the highest order.
So go order the factors from least to greatest. and suppose they differ for the first time at factor $k$, where the first group has a larger exponent (call it $p^x$) Then that group has more elements of order $p^x$.