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How can I show that two finite abelian groups are isomorphic, knowing only that both groups have the same number of elements of any given order?

I feel like there should be a nice way to show this without having to actually label elements of either group, but I am at a loss. In fact I have few ideas at all on how to begin and would appreciate any/all help.

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Are you familiar with a theorem describing the structure of finitely generated abelian groups? Invariant factors? –  Jyrki Lahtonen Aug 17 '11 at 5:31
    
I am. A finite abelian groups is isomorphic to a direct product of cyclic groups of prime power order. So I can use this on both of my groups and get two decompositions, but how can I use the order condition to say that these factors are the same? –  RHP Aug 17 '11 at 5:39
    
Good. Start counting the number of elements of a prime power order. For all positive integers $k\le n$ the group $\mathbf{Z}/p^n\mathbf{Z}$ has $p^k$ elements of order that is a factor of $p^k$. If $a$ is of order $m$ in group $A$ and $b$ is of order $n$ in group $B$, then $(a,b)$ is of order $lcm(m,n)$ in the direct product $A\times B$. –  Jyrki Lahtonen Aug 17 '11 at 5:55
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up vote 7 down vote accepted

This argument was suggested by Jyrki Lahtonen in the comments. (I'm sure Jyrki would have expressed it in a much nicer way.)

Abelian groups will be written multiplicatively. For each positive integer $n$ let $C_n$ be a cyclic group of order $n$. For any finite abelian group $A$ and any positive integer $k$, let $f_k(A)$ be the number of elements of $A$ whose order divides $k$ (or, if you prefer, the number of $k$th roots of $1$ in $A$). Then we have $f_k(A\times B)=f_k(A)f_k(B)$. Moreover, $f_k(C_n)$ is equal to $k$ if $k$ divides $n$, and to $1$ otherwise.

Our two finite abelian groups, $A$ and $B$ say, clearly satisfy $f_k(A)=f_k(B)$ for all $k$. By the classification theorem, we have $$A\simeq C_{m(1)}\times\cdots\times C_{m(r)},\quad B\simeq C_{n(1)}\times\cdots\times C_{n(s)}$$ with $m(1)\ |\ \cdots\ |\ m(r)$ and $n(1)\ |\ \cdots\ |\ n(s)$ (where $|$ means "divides"). We have $m(r)=n(s)$ (take $k=m(r)$ and $k=n(s)$), and an obvious induction completes the proof.

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+1: Well done. I suggested studying elements of a prime power order, because the OP expressed that's the version of the structure theorem s/he is familiar with. Comes to much the same thing as one may (and with that version perhaps should) study it one prime at a time. I stopped giving hints for the simple reason that I had to commute :-) –  Jyrki Lahtonen Aug 17 '11 at 18:28
    
@Jyrki: Thanks! I completely agree. I didn’t know if you were planning to answer the question. I could have asked you, but I didn’t want to bother you with this. I just wanted to avoid that the question remain unanswered (even if it wasn’t clear that the OP was still around). I don’t know what the etiquette is in this kind of situation. –  Pierre-Yves Gaillard Aug 17 '11 at 18:46
    
No problem! I wasn't going to answer, as I hoped that the OP might get started on a solution, but it is difficult to tell here (different posters have different styles in this respect). You gave the OP ample time, so it was for the best to give a solution. I had forgotten about this one :-) –  Jyrki Lahtonen Aug 17 '11 at 19:37
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