Two finite abelian groups with the same number of elements of any order are isomorphic

Suppose $G, H$ are finite abelian groups with the same number of elements of any given order. Show they are isomorphic.

Since finite abelian groups are isomorphic if and only if their Sylow subgroups are, we may restrict our attention to the case where $G, H$, are $p$-groups, but I can't quite make it past there...

Solution 1:

You don't say how much structure you have proven for abelian groups, so I will not assume much. If you do know some structure theorems, please let us know. (E.g., there is a theorem that if $G$ is abelian, and $a$ is an element of $G$ of maximal order, then there is a subgroup $H$ of $G$ such that $G = H\oplus \langle a\rangle$; do you know that?) Anyway...

Let $G$ and $H$ be finite abelian $p$-groups that have the same number of elements of each order. We want to prove that $G$ and $H$ are isomorphic.

Let $p^n$ be the largest order of an element of $G$ (and of $H$). If $n=1$, then $G$ and $H$ are elementary abelian $p$-groups; so they are vector spaces over $\mathbf{F}_p$, and since they have the same number of elements, they are isomorphic (same dimension).

Assume the result holds for abelian groups whose largest orders are $p^k$, and let $G$ and $H$ be groups satisfying our hypothesis and in which the largest elements have oder $p^{k+1}$.

Show that $pG$ and $pH$ have the same number of elements of each order, and that the elements of largest orders have order $p^k$. Apply induction to conclude $pG\cong pH$. Now see if you can leverage that to get $G\cong H$. If you need more help with those steps, please ask through comments.

Solution 2:

Here's a "low-tech" approach, which yields an explicit formula of elementary divisors using the information of the number of elements of each order.

First we count the number of elements of each order in $\mathbb Z/p^n\mathbb Z$. If $m=d_n\cdots d_1$ is the $p$-adic expansion of $m\in\mathbb{Z}/p^n\mathbb{Z}$, then its order is equal to $p^{m+1-l}$, where $l$ is the smallest index for which $d_i\neq0$. So the number of elements in $\mathbb Z/p^n\mathbb Z$ whose order is at most $p^k$ is equal to $p^k$, where $k\leq n$.

Now let $G$ be an abelian $p$-group, and let $N_k$ denote the number of elements of order $p^k$ in $G$. The fundamental theorem of finite abelian group asserts that there is a unique sequence of nonnegative integers $\lambda_1,\cdots,\lambda_m$ such that $G\simeq\bigoplus_{i=1}^m(\mathbb Z/p^{i}\mathbb Z)^{\lambda_i}$. Since an element in $\bigoplus_{i=1}^m(\mathbb Z/p^{i}\mathbb Z)^{\lambda_i}$ has order at most $p^k$ if and only if each of its component has order at most $p^k$, it follows from the result in the preceding paragraph that $$ N_0+\cdots +N_k=(\prod_{i\leq k}(p^i)^{\lambda_i})(\prod_{i> k}(p^{k})^{\lambda_i}) $$ for $k\geq0$. Thus $$\begin{eqnarray} N_0+\cdots+N_{k+1}&=&(\prod_{i\leq k+1}(p^i)^{\lambda_i})(\prod_{i> k+1}(p^{k+1})^{\lambda_i})\\ &=&(\prod_{i\leq k}(p^i)^{\lambda_i})(\prod_{i> k}(p^{k})^{\lambda_i})p^{\lambda_{k+1}\ +\cdots+\lambda_m}\\ &=&(N_0+\cdots+N_k)p^{\lambda_{k+1}\ +\cdots+\lambda_m} \end{eqnarray}$$ for $0\leq k<m$. We can use the above system of equations the express $\lambda_i$ in terms of $N_j$, which was to be proved.