# The concept of associate for finite abelian groups

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.

• Where did you encounter this problem? – Shaun Dec 10 '18 at 8:14
• Hint: Consider the subgroup generated by $a$ (or $b$, as this is the same subgroup). You have shown that the claim holds in this subgroup. How could it not hold in the bigger group? – Tobias Kildetoft Dec 10 '18 at 8:22
• @Shawn. The problem arises in the case that the decomposition of $G$ into a product of cyclic $p$-groups contains two or more factors of order a power of $p$. In the case that $G$ is cyclic this can't happen, and one proves the assertion by first handling the case that $n$ is a prime power (which is easy) and then applying the Chinese Remainder Theorem. – falang Dec 10 '18 at 8:26
• @Tobias: The bigger group may have order divisible by primes that do not divide the order of $a$. – falang Dec 10 '18 at 8:29
• @Shaun: Where did I encounter the problem? I noticed that it was true for $G$ finite cyclic and wondered if it was true for arbitrary products of finite cyclic groups. – falang Dec 10 '18 at 8:39

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.