# Order of elements in a finite Abelian group

Suppose that $G$ is a finite abelian group and that $x, y \in G$ are of orders $a$ and $b$ respectively.

I'm trying to show that there exist two elements $x'$ and $y'$ of orders $a'$ and $b'$ such that

1. $a'b'=lcm(a,b)$ and

2. $gcd(a',b')=1$.

Well the first thing that glaces to my mind is this $x'=x^{\frac{a}{a'}}$ and $y'=y^{\frac{b}{b'}}$. So to make sure it's the answer, I've got to firstly show that $a$ and $b$ are respectively divisible by $a'$ and $b'$, but for (1) and (2) i have no clue. Please can you help me figure this out ?

• If $lcm\{a',b'\}=1$ then $a'$ and $b'$ divide $1$, so that $a'=b'=1$ since $a'$ and $b'$ are positive integers. – Firepi Oct 26 '16 at 2:41
• Oh sorry i just edited it, i reversed notations ^^ Thanks ! – Zakaria Oussaad Oct 26 '16 at 2:46

First note that if $gcd(a,b) = 1$, then you're done and $x = x'$, $y = y'$. If not, let $gcd(a,b) = k$, and choose $x' = x^k$,$y' = y$. The orders of $x'$ and $y'$ respectively are $\frac{a}{k}$ and $b$.
Since $lcm(a,b) gcd(a,b) = ab$ we have $lcm(a,b) = \frac{a}{k} b = a' b'$ as desired. We also know $gcd(a',b') = 1$, since $k = gcd(a,b)$ was divided from $a$.
• I'm unfortunately not familiar with the $p$-adic theory at all. Hopefully someone else can help you. – Nitin Oct 26 '16 at 14:03