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...

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    $\begingroup$ Are you allowed to use the classification theorem? $\endgroup$ – Bill Cook Oct 16 '11 at 1:34
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    $\begingroup$ Start at order p^n, the size of the group. Count the number of elements of that size, and see what that tells you about the structure. Then look at p^n-1. Keep going. $\endgroup$ – Noah Snyder Oct 16 '11 at 1:36
  • $\begingroup$ Yes. But that is essentially it: that abelian groups are isomorphic to their sylow subgroups. I can't reduce it any further because the decomposition isn't unique until we establish they are isomorphic. $\endgroup$ – JeremyKun Oct 16 '11 at 1:36
  • $\begingroup$ @JeremyKun Can you please share the proof using the parenthetical argument? $\endgroup$ – Sam Dec 15 '14 at 22:22
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    $\begingroup$ By induction on the order of $|G|$, for $|G|=1$ it is trivial ($G$ must be the trivial group). For larger $G$ take an element of maximal order, split it off as $G = H \oplus \langle a \rangle$ and apply induction on $H$ since $|H| < |G|$. Eventually you get a direct sum of cyclic groups, and the isomorphism is just piece by piece. $\endgroup$ – JeremyKun Dec 16 '14 at 2:45

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.

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    $\begingroup$ The parenthetical theorem made for an easier argument, and in fact I did know it. :) $\endgroup$ – JeremyKun Oct 16 '11 at 2:53
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    $\begingroup$ @Bean: Ah, well: then proceed by induction on $|G|$ instead. (-: $\endgroup$ – Arturo Magidin Oct 16 '11 at 2:55
  • $\begingroup$ @ArturoMagidin: Do you have a reference for the theorem you mention? $\endgroup$ – Seirios Aug 22 '13 at 15:15
  • $\begingroup$ I finally found a proof here: csus.edu/indiv/e/elcek/M210A/FiniteAbelian.pdf $\endgroup$ – Seirios Aug 23 '13 at 9:52
  • $\begingroup$ @ArturoMagidin Would you mind sharing the proof that uses the parenthetical argument? $\endgroup$ – Sam Dec 15 '14 at 21:39

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.

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