I want to show that $x=\operatorname{ord}(a) \bmod m$ and $y=\operatorname{ord}(b)\bmod m$ then $\operatorname{ord}(ab)\equiv xy \pmod{\phi(m)}$

Question

Given:$\DeclareMathOperator{\ord}{ord}$

$\phi(m)$ is Euler's totient function

$\ord(a)$ is the least solution of $a^t\equiv 1\pmod{m}$

$\operatorname{gcd}(x,y)=1$

$x=\ord(a)$

$y=\ord(b)$

Show $\ord(ab)\equiv xy \pmod{\phi(m)}$

I haven't gotten to group theory yet, so I don't know what the "subgroup $$" is. — Anna Naden, Sep 25 '20 at 13:52
That's why I didn't close the question as a duplicate. But you can pretty much ignore the group theory language. The subgroup $$ just means the collection of powers of $x$. In your case, $$ would be ${1,a,a^2, \cdots, a^{x-1}}$. Calling it a subgroup just means that it is closed under multiplication and that there are inverses. — lulu, Sep 25 '20 at 14:20
To stress: in the accepted answer to the question I linked to, the writer just needs to claim that, in your case, the two sets ${1, a, \cdots, a^{x-1}}\pmod m$ and ${1, b, \cdots, b^{y-1}}\pmod m$ only intersect at $1$. You can prove that statement without referring to group theory at all. — lulu, Sep 25 '20 at 14:22
I was able to follow the argument in the accepted answer and translate it into the language of number theory without groups, mostly (see below). However, a question remains. — Anna Naden, Sep 25 '20 at 16:09

Anna Naden · Answer 1 · 2020-09-25T17:29:02.887

0

Lemma: If $a^r\equiv b^s\pmod{m}$

and $r,s<\phi$.

Then since $(a^r)^{ord(a)}\equiv 1 \pmod{m}$,

$ord(a^r)|x$

Similiary $ord(b^s)|y$.

Since $gcd(x,y)=1$, $ord(a^r)=1$ and

Then $a^r \equiv 1$

$a^{ord(ab)}\equiv b^{-ord(ab)}$

By the lemma, $a^{ord(ab)}\equiv 1\equiv b^{-ord(ab)}$

$x|ord(ab)$ and $y|ord(ab)$

Since $gcd(x,y)=1$, $xy=lcm(x,y)|ord(ab)$

But since $(ab)^{xy}\equiv 1$, $ord(ab)|xy$,

So we have $xy=ord(ab)$

This begs a question: can we have $xy>\phi$, requiring us to say that $xy\equiv ord(ab) \pmod{\phi}$ (as in the given problem) rather than saying that they are exactly equal? But $xy<=ord(ab)<=\phi$

edited Sep 25 '20 at 17:29

answered Sep 25 '20 at 16:13

Anna Naden

426

So, you've shown that those two lists of powers only intersect at $1$, good! Now note that, if order$(ab)=k$, we have $a^k\equiv b^{-k}$ which means that $x,y$ both divide $k$ (by virtue of what you have shown). But since $\gcd(x,y)=1$ this means that $xy,|,k$. Since the order of $ab$ is clearly $≤xy$ we are done. – lulu Sep 25 '20 at 16:19

I want to show that $x=\operatorname{ord}(a) \bmod m$ and $y=\operatorname{ord}(b)\bmod m$ then $\operatorname{ord}(ab)\equiv xy \pmod{\phi(m)}$

1 Answers1