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For this proof, is it simply that $\operatorname{ord}(a)=km$ says be definition that $a^{km}=e$, so $\operatorname{ord}(a^k)=x$, so $a^{kx}=e$, giving $m=x$?

Moz
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    $\operatorname{ord}(a)=km$ means that by definition $a^{km}=e$, you are right about that. We can rewrite $a^{km}=\left(a^k\right)^m$ which is also equal to $e$. This means that the order of $a^k$ is $m$. – NoetherianCheese Feb 27 '16 at 05:31
  • Well, you know that it is at most $m$, then you need to show that it is at least $m$ (which is just as easy). –  Feb 27 '16 at 05:38
  • For reference, this is Pinter 10.D.3. – dharmatech Aug 08 '19 at 04:03

3 Answers3

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If $\:{\rm ord}(a) = KM\:$ then $\,(\color{#0a0}{a^{ K}})^{\color{#c00}{J}}\! = 1\!\iff KM\mid KJ\iff M\mid\color{#c00} J,\ $ thus $\,{\rm ord}(\color{#0a0}{a^{K}})= M$

Bill Dubuque
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  • I'm reviewing this proof and am looking at your approach. I'm trying to understand what the theorem you link to is saying. What does ϕ(n) mean? At first I read the theorem as: ord(a) mod n | ord(n) or ord(a mod n) | ord(n) but not sure those even make sense. Yay, math and all the different notations. :-) – dharmatech Aug 05 '19 at 05:08
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    @dharmatech In that question $\varphi$ denotes Euler's totient function., thus $\large , a^{\large \varphi}\equiv 1,\Rightarrow, o(a)\mid\varphi.,$ Note: in that answer $,\rm n,$ does not denote the modulus (there is no need to mention the modulus there). – Bill Dubuque Aug 05 '19 at 13:38
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Let $k$ and $m$ be positive integers. From $\text{ord}(a)=km$, we can see that $a^{km}=(a^k)^m=e$.

It follows that $a^k$ has order less than or equal to $m$. But we must show that $a^k$ has order exactly $m$.

Suppose to the contrary that $a^k$ has order $l\lt m$. Then $a^{kl}=e$. Since $kl\lt km$, this contradicts the fact that $a$ has order $km$.

André Nicolas
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    Why do you say "less than or equal to?" Why isn't the equality immediate since I know that $(a^k)^m=e$? – Moz Feb 27 '16 at 05:38
  • Think of $\mathbb Z_5$. I know that $(\bar 4)^{50} = \overline {200} = \bar 0$ but that does not mean $o(\bar 4) = 50$, just that the order is at most $50$ (and more strongly, that it divides $50$). –  Feb 27 '16 at 05:40
  • I'm sorry, I haven't seen that notation before. I'd appreciate an explanation. – Moz Feb 27 '16 at 05:41
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    ;Example: Look at the group whose elements are $1,2,3,4$ under multiplication modulo $5$. Note that $2^{12}=1$. But $12$ is not the order of $2$, since the smallest positive $s$ such that $2^s=1$ is $s=4$. When we say that an element $a$ has order $q$, we are saying two things: (i) $a^q=e$ and (ii) there is no positive $s\lt q$ such that $a^s=e$. – André Nicolas Feb 27 '16 at 05:44
  • That makes perfect sense! – Moz Feb 27 '16 at 05:45
  • The order of an element $b$ of finite order is by definition the smallest positive integer $q$ such that $b^q=e$. So to show $b$ has order $q$, we must show two things, (i) $b^q=e$ and (ii) nothing smaller than $q$ "works." – André Nicolas Feb 27 '16 at 05:48
  • In your proof, should you have $a^{kl}=e$ and since $kl<km$, we have a contradiction? – Moz Feb 27 '16 at 05:53
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    @Moz: Thank you, fixed. You really do understand it! – André Nicolas Feb 27 '16 at 05:56
  • Thanks very much for this! Very simple and direct solution. – dharmatech Aug 08 '19 at 03:00
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It's well-known that for any element (of finite order) of a group, $\mid a^l\mid =m/(m,l),$ where $m=\mid a\mid. $ Here's a proof.

So, we get $\mid a\mid=km\implies \mid a^k\mid=km/(km,k)=km/k=m.$

calc ll
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