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Previously shown:

Let $m\in\mathbb{N}$ and $a\in\mathbb{Z}$ s.t. $(a,m)=1$. Then the order $d$ of $a$ modulo $m$ exists and $d\mid\phi(m)$.

Moreover, whenever $a^k\equiv 1\pmod{m}$, one has $d\mid k$

Trying to show: Suppose that $a$ has order $h$ modulo $m$. Then $a^k$ has order $h/(h,k)$ modulo $m$

Proof: From above, one has $(a^k)^j\equiv 1\pmod{m}$ iff $h\mid kj$.

But, $h\mid kj\Leftrightarrow h/(h,k)\mid (k/(h,k))j\Leftrightarrow h/(h,k)\mid j$

Thus the least positive inter $j$ s.t. $(a^k)^j\equiv 1\pmod{m}$ is $j=h/(h,k)$

Point of contention: Surely here we have $(h,k)=k$, so we have shown that the order of $a^k$ has order $h/k$ as opposed to $h/(h,k)$

Bart Michels
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Sam Houston
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    Why should we have $\gcd(h,k)=k$? Why should we have $h$ a multiple of $k$? If $a$ has, say, order 3, I can certainly ask what the order of $a^2$ is, for example. – Gerry Myerson May 01 '15 at 09:33
  • Surely we assumed $(h,k)=k$ in the step $h/(h,k)\mid (k/(h,k))j\Leftrightarrow h/(h,k)\mid j$? – Sam Houston May 01 '15 at 09:35
  • If the goal is to find a formula for the order of $a^k$ when the order of $a$ is $h$, then you'd want a formula that works for all $k$, not just for those that are divisors of $h$. If you don't assume $h$ is a multiple of $k$, but somehow that pops out in your proof, then there's something wrong with the proof. – Gerry Myerson May 01 '15 at 09:52
  • Thanks that makes more sense, but, the proof is from a reputable source and I don't understand why in the step above is $h/(h,k)\mid (k/(h,k))j\Leftrightarrow h/(h,k)\mid j$? – Sam Houston May 01 '15 at 09:56
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    $h/(h,k)$ and $k/(h,k)$ are relatively prime. If $r$ and $s$ are relatively prime, and $r$ divides $st$, then $r$ divides $t$. – Gerry Myerson May 01 '15 at 09:58
  • Totally understand now, thanks so much, if you want to give a brief answer below, I'll give you the correct answer and +1 – Sam Houston May 01 '15 at 10:01
  • Let us continue this discussion in chat. – Sam Houston May 01 '15 at 10:04

2 Answers2

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One can also prove it without Euclid's Lemma, instead using the gcd Distributive Law

$$ h\mid kj\iff h\mid kj,hj\iff h\mid \underbrace{(kj,hj)}_{\Large (k,h)\,j}\iff h/(k,h)\mid j\qquad $$

Community
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Bill Dubuque
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As requested by OP, in the comments:

$h/(h,k)$ and $k/(h,k)$ are relatively prime. If $r$ and $s$ are relatively prime, and $r$ divides $st$, then $r$ divides $t$. Therefore, if $h/(h,k)$ divides $(k/(h,k))j$, then it divides $j$.

Gerry Myerson
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