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 ?

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

  • $\begingroup$ Many thanks man for your help ! $\endgroup$ – Zakaria Oussaad Oct 26 '16 at 13:39
  • $\begingroup$ I did something new and i don't have if i have the right to, can you please comment on this. I thought : For a prime p let A-p be the p-adique valuation of a if v_p(a)>=v_p(b) and A_p=0 for other cases, similarly let B_p be the p-adique valuation of b if v_p(b)>v_p(a) and B_p=0 for other cases. I choose a'=product of primes(p^A_p) and b'=product of prime(p^B_p). we have then gcd(a',b')=1 and a', b' respectively divide a and b. And what's cool is this a'b'=product of primes(p^max(v_p(a),v_p(b))) so a'b'=lcm(a,b) , Now i'm free to consider x′=x^(a/a') and y'=y^(b/b'). $\endgroup$ – Zakaria Oussaad Oct 26 '16 at 13:55
  • $\begingroup$ I'm unfortunately not familiar with the $p$-adic theory at all. Hopefully someone else can help you. $\endgroup$ – Nitin Oct 26 '16 at 14:03

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