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.

  • 2
    $\begingroup$ Are you familiar with a theorem describing the structure of finitely generated abelian groups? Invariant factors? $\endgroup$ – Jyrki Lahtonen Aug 17 '11 at 5:31
  • $\begingroup$ 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? $\endgroup$ – RHP Aug 17 '11 at 5:39
  • $\begingroup$ 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 $A\times B$. $\endgroup$ – Jyrki Lahtonen Aug 17 '11 at 5:55

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.

  • $\begingroup$ +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 :-) $\endgroup$ – Jyrki Lahtonen Aug 17 '11 at 18:28
  • $\begingroup$ @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. $\endgroup$ – Pierre-Yves Gaillard Aug 17 '11 at 18:46
  • $\begingroup$ 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 :-) $\endgroup$ – Jyrki Lahtonen Aug 17 '11 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.