# Prove that the $\mathrm{ord}_{mn}a = [\mathrm{ord}_ma, \mathrm{ord}_na]$ where $a, m, n$ are relatively prime

Problem

Prove that the $\mathrm{ord}_{mn}a = [\mathrm{ord}_ma, \mathrm{ord}_na]$ where $a, m, n$ are relatively prime.

My attempt was,
Let $x = \mathrm{ord}_ma$, $y = \mathrm{ord}_na$, and $z = \mathrm{ord}_{mn}a$. By definition of order, we have $a^x \equiv 1 \pmod{m}$ and $a^y \equiv 1 \pmod{n}$ which implies $(a^{x})^y \equiv 1 \pmod{m}$, and $(a^y)^x \equiv 1 \pmod{n}$. Hence, $a^{xy} \equiv 1 \pmod{m}$ and $a^{xy} \equiv 1 \pmod{n}$. Furthermore, we have $a^{xy}, m, n$ are pairwise relatively prime. Therefore, $a^{xy} \equiv 1 \pmod{mn}$. So if $z$ is the order of $a$ modulo $mn$, then $z$ is the smallest integer such that $a^z \equiv 1 \pmod{mn}$. On the other hand, $z|(xy)$ because $a^{xy} \equiv 1 \pmod{mn}$, which implies $xy = z \cdot k$ for some integers $k$. Next, to satisfy the smallest value that divides $xy$, $k$ must be the greatest common divisor of $x, y$. Hence $z = \dfrac{xy}{(x,y)} = [x,y] \text{ } \square$.

Am I in the right track? Any feedback would be greatly appreciated.

Thank you

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That's exactly right. Well done.

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Great thanks once again! You were also right that this was my homework. However, I won't copy other people' solutions and submit as my work. My rule of thumb is, grade is important, but learning something is much more important. I don't want to regret when I'm getting older, because at that time I won't be able to think like this anymore. Furthermore, understanding new things make me happy, even though it often takes a lot of my time. – Chan Apr 25 '11 at 6:02
@Chan: That's a good way to go about this. I look forward to hearing from you again. – mixedmath Apr 25 '11 at 6:05
Thank you. – Chan Apr 25 '11 at 6:10

Your solution shows good understanding of the situation, but I think it could be considered technically incomplete. You write "to satisfy the smallest value that divides $xy$." It is clear that what you have in mind is that you are looking for the smallest (positive) number which is a multiple of $x$ and of $y$, and divides $xy$. And you do not explicitly say why you are looking for that smallest multiple, although you certainly could explain if asked. The fact that $x$ and $y$ are the orders of $a$ modulo $m$, $n$ respectively is not used, and needs to be ($a^x \equiv 1 \pmod{m}$ and $a^y \equiv 1 \pmod{n}$ are used, but not the minimality of $x$ and $y$).

The following is a variant of your solution that (in the second part) makes explicit what you had left implicit. In the first part, I will not use the product $xy$, because of personal preference.

Each of $x$ and $y$ divides $[x,y]$. Let $[x,y]=xu=yv$. Then $a^{[x,y]}=(a^x)^u$. But since $a$ has order $x$ modulo $m$, it follows that $a^x\equiv 1 \pmod{m}$, and hence $a^{[x,y]}\equiv 1 \pmod{m}$. In the same way, we can show that $a^{[x,y]}\equiv 1 \pmod{n}$. Since $m$ and $n$ are relatively prime, it follows that $a^{[x,y]}\equiv 1 \pmod{mn}$.

Thus $z$, the order of $a$ modulo $mn$, must divide $[x,y]$. We will show that in fact $z=[x,y]$. Since $a^z \equiv 1 \pmod{mn}$, it follows that $a^z \equiv 1 \pmod{m}$. So the order of $a$ modulo $m$, namely $x$, divides $z$. Similarly, $y$ divides $z$. It follows that $[x,y]$ divides $z$, and hence $[x,y]=z$.

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@user6312: It was like you read my mind!Amazing! In fact, before posting, I was stuck with "You write "to satisfy the smallest ... divides xy.", then suddenly I came up with that. Although I didn't know how to say it explicitly, I knew it must be true, otherwise the proof was not true. Thanks for pointing that out. By the way, there are not many people can read other people minds this way, especially very few smart people want to hear and understand other thoughts. I'm totally convinced, everything you said is just verbatim! – Chan Apr 25 '11 at 23:15
@user6312: I wish I could give at least 5 votes for your answers! – Chan Apr 25 '11 at 23:20
@Chan: Thank you, but I do not need points, one cannot buy wine with points. – André Nicolas Apr 25 '11 at 23:28
@user6312: Just a way to show my appreciation! By the way, wine is not healthy :) – Chan Apr 25 '11 at 23:33

HINT $\$ This is obvious using CRT (Chinese Remainder Theorem). Since $\rm\ (m,n) = 1\$ we know $\rm\ \mathbb Z/mn\ \cong \mathbb Z/m \times \mathbb Z/n\: ,\:$ where $\rm\ a_{mn}\: \mapsto\: (a_m,\:a_n)\:,\:$ with obvious order $\rm\ lcm(o(a_m),o(a_n))\:,\:$ viz. $$\rm\ (1,1)\ =\ (a_m,\:a_n)^k\: =\ (a_m^k,\:a_n^k)\ \iff\ o(a_m),\:o(a_n)\ |\ k\ \iff\ lcm(o(a_m),\:o(a_n))\ |\ k$$

Note especially how this method efficiently simultaneously proves both directions of the proof by exploiting the universal bidirectional $(\iff)$ definition of $\rm lcm,$ namely $\rm\ a,b\ |\ c\ \iff\ [a,b]\ |\ c\:.\:$

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