$Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$

Question

What can be said about $Ord_n(ab)$ when $a,b$ are positive integers both relatively prime to $n$ and $Ord_n(a)$ is not relatively prime to $Ord_n(b)$?

To start the proof I let $r=Ord_n(a)$, $s=Ord_n(b)$, and $t=Ord_n(ab)$. Since the orders of $a$ and $b$ are not relatively prime, I let $(r,s)=d$. Then I calculated the following:

$$(ab)^{rs/(r,s)}\equiv a^{rs/d}b^{rs/d} \equiv 1^{s/d}1^{r/d}\equiv 1 \mod n$$ and $$a^{rt}\equiv (ab)^{rt} \equiv a^{rt}b^{rt}\equiv 1^{t}(b^t)^r\equiv 1 \mod n $$

The first calculation shows that $t\bigg|\frac{rs}{d}$. The second calculation shows that $s|t$. A similar calculation would yield $r|t$. In general, its true that $rs\not|t$, but $\frac{rs}{(r,s)}\bigg|t$. Am I on the right track?

My reasoning is that $b^t \equiv 1$. Hence $t$ must be some multiple of $s$. — emka, Mar 28 '14 at 06:26
So the definition of order, for relatively prime $b$ and $n$, is the least positive integer $x$ such that $b^{x} \equiv 1 \mod n$. Multiples of $x$ will still make it one, but they won't be the least such. — emka, Mar 28 '14 at 06:30
yes, in your case $x=s$. But, how can you know $b^t\equiv 1$? — Jlamprong, Mar 28 '14 at 06:34
I don't think so. Yo only have $b^{dt}\equiv 1$. I haven't seen yet if $b^t\equiv 1$. Please see, my answer. — Jlamprong, Mar 28 '14 at 07:23
You "start the proof" before having made any claim. What do you think you will be able to establish? — Marc van Leeuwen, Mar 28 '14 at 07:27

score 2 · Accepted Answer · edited Mar 28 '14 at 07:39

2

Using your notation, let $p=\frac rd, \quad q=\frac sd$ then $(p,q)=1$, so there are integers $x$ and $y$ such that $px+qy=1$.

From your first equation, you have got $t|dpq$.

Since $a^tb^t\equiv 1$ then $a^{dt}=(a^{px+qy})^{dt}\equiv 1$ that implies $r|dt\iff p|t$. Similarly, $q|t$. Since $(p,q)=1$ then $pq|t$.

In conclusion, we have $\frac {rs}{d^2}\mid t\mid \frac{rs}d$. Another way of saying this is that $\frac{\mathop{\rm lcm}(r,s)}{\gcd(r,s)} \mid t \mid \mathop{\rm lcm}(r,s)$.

Remark: If $d=1$ then $t=rs$ which is consistent with the existing fact.

edited Mar 28 '14 at 07:39

Greg Martin

78,820

answered Mar 28 '14 at 07:21

Jlamprong

1,837

And if $r=s$, then $t$ can be any divisor of $r$ ... including $1$, which certainly happens when $a=b^{-1}$. – Greg Martin Mar 28 '14 at 07:40

$Ord_n(ab)$ when $(a,n)=(b,n)=1$ but $(Ord_n(a), Ord_n(b))\neq 1$

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