$\operatorname{ord}(ab)=\operatorname{ord}(a)\operatorname{ord}(b)$ in $U(\mathbb{Z}_n)$

Asked Sep 06 '16 at 22:44

Active Sep 07 '16 at 13:55

Viewed 112 times

Let $(U(\mathbb{Z}_n), \cdot)$ the group of the units of $\mathbb{Z}_n$, and $a,b\in U(\mathbb{Z}_n)$. If $\operatorname{ord}(a)=r$, $\operatorname{ord}(b)=s$, $\operatorname{ord}(ab)=m$, $\gcd(r,s)=1$ show that $m=rs$.

Proof:

$ab^{rs}=1 \implies m\mid rs \implies rs \ge m$
$\forall π\in \mathbb{N^*}:(ab)^π=1 \implies (ab)^{π\ r}=1 \implies b^{π\ r}=1 \implies s|πr \implies s|π$. Working similar, $r|π$. So, $rs|π, \forallπ\in \mathbb{N^*} \implies rs|m \implies m\ge rs$.

Is this proof true?

edited Sep 07 '16 at 13:55

asked Sep 06 '16 at 22:44

Chris

3,026

@Crostul You are right, forgot that gcd is 1. Thank you. – Chris Sep 06 '16 at 22:50
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Your proof doesn't use the hypothesis $\gcd (r,s)=1$. Since this is a necessary hypothesis, something is missing. – Crostul Sep 06 '16 at 22:53
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The second line makes no sense. – Q-Zh Sep 06 '16 at 22:54
No. It only proves that $ord(a,b) \mid ord(a) ord(b)$ – Steven Alexis Gregory Sep 06 '16 at 22:55
I thought that it would be a good idea to say that from devision algorithm, we must have $υ=0$ because $m$ is the smallest positive integer such that $ab^m=1$... – Chris Sep 06 '16 at 22:56
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@Chris why can't $\pi=0$ and $v=m$? – stewbasic Sep 06 '16 at 23:25
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Also could you put parentheses around $ab$? What you have written now should be interpreted as $a(b^m)$. – stewbasic Sep 06 '16 at 23:26
@stewbasic Thank you for this comment. I corrected it. Any idea for tha second dot? – Chris Sep 06 '16 at 23:33
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@Chris The implication $(ab)^v=1\Rightarrow v=0$ is incorrect. You have that $m$ is the smallest positive integer such that $(ab)^m=1$, but you don't have that $v<m$, only that $v<rs$. – stewbasic Sep 06 '16 at 23:44
I tried to correct my proof. Have a look please. Thank you all of you. – Chris Sep 07 '16 at 13:57

$\operatorname{ord}(ab)=\operatorname{ord}(a)\operatorname{ord}(b)$ in $U(\mathbb{Z}_n)$

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