2
$\begingroup$

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.

  • 0
    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 $A\times B$.2011-08-17

1 Answers 1

9

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.

  • 0
    No problem! I w$a$sn't going to $a$nswer, 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 :-)2011-08-17