The concept of associate for finite abelian groups

Question

Let $G$ be a finite abelian group of order $n$, and let $a$ and $b$ be elements of $G$. If $a$ generates the same subgroup as $b$, must there be an integer $i$ prime to $n$ such that $ia=b$?

I can prove this when $G$ is cyclic, but not in general.

Answer 1

I should have thought more before posting... The answer is yes and the proof goes like this: Suppose first that $G$ is a $p$-group. By hypothesis there are integers $i$ and $j$ such that $ia=b$ and $jb=a$. We must show that it is possible to choose $i$ and $j$ prime to $p$. But the latter two equations imply that $ijb=b$, hence $(ij-1)b=0$. We may assume that $b\ne0$, otherwise there is nothing to prove. But then the order of $b$ divides $ij-1$. Since the order of $b$ is divisible by $p$, it follows that $i$ and $j$ are prime to $p$, which gives the required conclusion.

Finally, an arbitrary $G$ can be decomposed into a product of $p$-groups for distinct primes $p$. The result we want then follows readily from the Chinese Remainder Theorem.