This is Pinter $10.G.5$


$a \in G$

$\text{ord}(a) = n$

Prove: $\text{ord}(a^m) = \frac{\text{lcm}(m,n)}{m}$

Use $10.G.3$ and $10.G.4$ to prove this.

Here is $10.G.3$:

Let $l$ be the least common multiple of $m$ and $n$. Let $l/m = k$. Explain why $(a^m)^k = e$.

Here is $10.G.4$:

Prove: If $(a^m)^t = e$, then $n$ is a factor of $mt$. (Thus, $mt$ is a common multiple of $m$ and $n$.)

Conclude that: $l = mk \leq mt$

OK, let's begin.

By $10.G.3$:

$$ (a^m)^{\text{lcm}(m,n)/m} = e \tag{5} $$

If $\text{lcm(m,n)}/m$ is the lowest number such that (5) is true, then:

$$ \text{ord}(a^m) = \text{lcm}(m,n)/m \tag{9} $$

Let's assume that there is a number

$$ q < \text{lcm}(m,n)/m \tag{7} $$

such that:

$$ (a^m)^q = e \tag{6} $$

By $10.G.4$ with (6):

$$ n \ \big|\ mq $$

$$ l \lt mq $$

$$ \text{lcm}(m,n) \leq mq $$

Isolate q:

$$ \text{lcm}(m,n)/m \leq q \tag{8} $$

(8) contradicts assumption (7).

So (9) must be true.

That's how I approached it. If anyone notices any issues, I'd be happy to know about them.

Even if it is considered correct, do you feel there is a better way?

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    $\begingroup$ You should use \le rather than <=. $\endgroup$ – Angina Seng Aug 13 '19 at 2:52
  • $\begingroup$ @LordSharktheUnknown Thanks! Updated! $\endgroup$ – dharmatech Aug 13 '19 at 2:54
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    $\begingroup$ @JyrkiLahtonen Can you explain how this is a duplicate of the question you linked to? There is no requirement in this question that G be cyclic. Whereas the question you linked to is for cyclic groups. $\endgroup$ – dharmatech Aug 13 '19 at 8:49
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    $\begingroup$ @José Carlos Santos Do you really think that it's a duplicate? $\endgroup$ – Michael Rozenberg Aug 13 '19 at 9:06
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    $\begingroup$ @YuiToCheng Once again not a proper dupe target becauyse - among other things - the OP is working with LCMs not GCDs, and is working from specific lemmas. Please be more careful in choosing proper dupe targets. $\endgroup$ – Bill Dubuque Aug 14 '19 at 13:32

Yes, $\,(a^{m})^{k} = 1\!\! \overset{(1)\!\!}\iff n\mid mk \iff m,n\mid mk\!\! \overset{(2)\!\!}\iff {\rm lcm}(m,n)\mid mk\iff {\large{\frac{{\rm lcm}(m,n)}m\,\mid}}\, k$

$\!(1)$ and the (omitted) final conclusion is by Corollary' here, and $(2)$ is the LCM Universal Property

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    $\begingroup$ @dharmatech No, it's certainly not a dupe of the proposed target (though it may be a dupe of some other question). As such I voted to reopen. $\endgroup$ – Bill Dubuque Aug 13 '19 at 12:54
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    $\begingroup$ @BillDubuque Do you really think this question has not been handled on our site? The OP's complaint about my choice of dupe was mostly because "their version is not only about cyclic groups". Yet they only work within the cyclic group generated by $a$. $\endgroup$ – Jyrki Lahtonen Aug 14 '19 at 6:18
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    $\begingroup$ @Jyrki Once again I politely request that your refrain from making insulting wild guesses about what other users think. If you you wish to close it as a dupe then it is your job to find a good dupe target - not mine. I already waste too much time doing so (far more than you do). $\endgroup$ – Bill Dubuque Aug 14 '19 at 13:36
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    $\begingroup$ Btw, the non-mathematical based downvote (presumably having to do with the above remark) shows just how pitiful this site has become. It is sad that politics overrides mathematics and pedagogy. This will be the death of the site. $\endgroup$ – Bill Dubuque Aug 14 '19 at 13:40
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    $\begingroup$ @Jyr I probably close more dupes in my tags than everyone else combined. I searched for one here and did not find a target that I thought was good given the OP's context (which I know well having answered some of their prior questions). It is frustrating when someone comes along and proposes a very poor dupe target and the robovoters close it on that. Even more so when these actions end up leading to undeserved downvotes likely based on errant speculation about the motivation for answering the question - all of which are likely due to your misjudgements about others - both past and present $\endgroup$ – Bill Dubuque Aug 14 '19 at 14:27

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