If $(ord_m(a), ord_m(b)) = 1$ prove that $ord_m(ab) = ord_m(a)*ord_m(b) $












2












$begingroup$



$DeclareMathOperatorord{ord}$Let $a, b$, and $m$ be positive integers such that $(a,m) = (b,m) = 1$. Assume that $(ord_m(a), ord_m(b)) = 1$. Prove that $ord_m(ab) = ord_m(a)*ord_m(b)$.




So I got $(ab)^{ord_m(a)*ord_m(b)} = 1 bmod m$



so $ord_m(ab) mid ord_m(a) * ord_m(b)$.



I am stuck on how to proceed from here though.






elementary-number-theory modular-arithmetic multiplicative-order







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  • $begingroup$
    Have you tried to show that $ord_m(a)mid ord_m(ab)$?
    $endgroup$
    – abiessu
    Apr 24 '14 at 16:48
















2












$begingroup$



$DeclareMathOperatorord{ord}$Let $a, b$, and $m$ be positive integers such that $(a,m) = (b,m) = 1$. Assume that $(ord_m(a), ord_m(b)) = 1$. Prove that $ord_m(ab) = ord_m(a)*ord_m(b)$.




So I got $(ab)^{ord_m(a)*ord_m(b)} = 1 bmod m$



so $ord_m(ab) mid ord_m(a) * ord_m(b)$.



I am stuck on how to proceed from here though.






elementary-number-theory modular-arithmetic multiplicative-order







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you tried to show that $ord_m(a)mid ord_m(ab)$?
    $endgroup$
    – abiessu
    Apr 24 '14 at 16:48














2












2








2


2



$begingroup$



$DeclareMathOperatorord{ord}$Let $a, b$, and $m$ be positive integers such that $(a,m) = (b,m) = 1$. Assume that $(ord_m(a), ord_m(b)) = 1$. Prove that $ord_m(ab) = ord_m(a)*ord_m(b)$.




So I got $(ab)^{ord_m(a)*ord_m(b)} = 1 bmod m$



so $ord_m(ab) mid ord_m(a) * ord_m(b)$.



I am stuck on how to proceed from here though.






elementary-number-theory modular-arithmetic multiplicative-order







share|cite|improve this question











$endgroup$





$DeclareMathOperatorord{ord}$Let $a, b$, and $m$ be positive integers such that $(a,m) = (b,m) = 1$. Assume that $(ord_m(a), ord_m(b)) = 1$. Prove that $ord_m(ab) = ord_m(a)*ord_m(b)$.




So I got $(ab)^{ord_m(a)*ord_m(b)} = 1 bmod m$



so $ord_m(ab) mid ord_m(a) * ord_m(b)$.



I am stuck on how to proceed from here though.






elementary-number-theory modular-arithmetic multiplicative-order




elementary-number-theory modular-arithmetic multiplicative-order






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Dec 21 '15 at 10:26









rabota

14.4k32783




14.4k32783










asked Apr 24 '14 at 16:44









terminix00terminix00

1569




1569












  • $begingroup$
    Have you tried to show that $ord_m(a)mid ord_m(ab)$?
    $endgroup$
    – abiessu
    Apr 24 '14 at 16:48


















  • $begingroup$
    Have you tried to show that $ord_m(a)mid ord_m(ab)$?
    $endgroup$
    – abiessu
    Apr 24 '14 at 16:48
















$begingroup$
Have you tried to show that $ord_m(a)mid ord_m(ab)$?
$endgroup$
– abiessu
Apr 24 '14 at 16:48




$begingroup$
Have you tried to show that $ord_m(a)mid ord_m(ab)$?
$endgroup$
– abiessu
Apr 24 '14 at 16:48










1 Answer
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$begingroup$

$DeclareMathOperatorord{ord}$Let $x=ord_m(a)$, $y=ord_m(b)$ and $z=ord_m(ab)$. You already found that $zmid xy$. It remains to prove that $xymid z$.



We have that
$$begin{align}1&equiv((ab)^z)^y\
&=a^{zy}(b^y)^z\
&equiv a^{zy}.end{align}$$
Therefore, $xmid zy$. As $gcd(x,y)=1$, it follows that $xmid z$.



Analogously we can show that $ymid z$. Because $gcd(x,y)=1$, this means that $xymid z$, which completes the proof. $square$






share|cite|improve this answer











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    $begingroup$

    $DeclareMathOperatorord{ord}$Let $x=ord_m(a)$, $y=ord_m(b)$ and $z=ord_m(ab)$. You already found that $zmid xy$. It remains to prove that $xymid z$.



    We have that
    $$begin{align}1&equiv((ab)^z)^y\
    &=a^{zy}(b^y)^z\
    &equiv a^{zy}.end{align}$$
    Therefore, $xmid zy$. As $gcd(x,y)=1$, it follows that $xmid z$.



    Analogously we can show that $ymid z$. Because $gcd(x,y)=1$, this means that $xymid z$, which completes the proof. $square$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      $DeclareMathOperatorord{ord}$Let $x=ord_m(a)$, $y=ord_m(b)$ and $z=ord_m(ab)$. You already found that $zmid xy$. It remains to prove that $xymid z$.



      We have that
      $$begin{align}1&equiv((ab)^z)^y\
      &=a^{zy}(b^y)^z\
      &equiv a^{zy}.end{align}$$
      Therefore, $xmid zy$. As $gcd(x,y)=1$, it follows that $xmid z$.



      Analogously we can show that $ymid z$. Because $gcd(x,y)=1$, this means that $xymid z$, which completes the proof. $square$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        $DeclareMathOperatorord{ord}$Let $x=ord_m(a)$, $y=ord_m(b)$ and $z=ord_m(ab)$. You already found that $zmid xy$. It remains to prove that $xymid z$.



        We have that
        $$begin{align}1&equiv((ab)^z)^y\
        &=a^{zy}(b^y)^z\
        &equiv a^{zy}.end{align}$$
        Therefore, $xmid zy$. As $gcd(x,y)=1$, it follows that $xmid z$.



        Analogously we can show that $ymid z$. Because $gcd(x,y)=1$, this means that $xymid z$, which completes the proof. $square$






        share|cite|improve this answer











        $endgroup$



        $DeclareMathOperatorord{ord}$Let $x=ord_m(a)$, $y=ord_m(b)$ and $z=ord_m(ab)$. You already found that $zmid xy$. It remains to prove that $xymid z$.



        We have that
        $$begin{align}1&equiv((ab)^z)^y\
        &=a^{zy}(b^y)^z\
        &equiv a^{zy}.end{align}$$
        Therefore, $xmid zy$. As $gcd(x,y)=1$, it follows that $xmid z$.



        Analogously we can show that $ymid z$. Because $gcd(x,y)=1$, this means that $xymid z$, which completes the proof. $square$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 29 at 7:43









        Martin Sleziak

        44.9k10122277




        44.9k10122277










        answered Apr 24 '14 at 17:49









        rabotarabota

        14.4k32783




        14.4k32783






























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