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This is Pinter $10.G.5$

Let:

$a \in G$

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

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?

balddraz
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dharmatech
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    You should use \le rather than <=. – Angina Seng Aug 13 '19 at 02:52
  • @LordSharktheUnknown Thanks! Updated! – dharmatech Aug 13 '19 at 02:54
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    @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. – dharmatech Aug 13 '19 at 08:49
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    @José Carlos Santos Do you really think that it's a duplicate? – Michael Rozenberg Aug 13 '19 at 09:06
  • @MichaelRozenberg This basic result from a first course in abstract algebra has been referred to on our site hundreds if not thousands of times. If you are unhappy with my choice of duplicate, you are welcome to suggest a better one. It is your duty as a trusted user to participate in such maintenance. – Jyrki Lahtonen Aug 13 '19 at 09:22
  • @dharmatech Your argument never goes outside the cyclic subgroup generated by $a$, so that linked result covers everything. – Jyrki Lahtonen Aug 13 '19 at 09:23
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    @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. – Bill Dubuque Aug 14 '19 at 13:32

1 Answers1

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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

Bill Dubuque
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  • Do you agree with the duplicate flag above? – dharmatech Aug 13 '19 at 09:48
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    @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. – Bill Dubuque Aug 13 '19 at 12:54
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    @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 $a$. – Jyrki Lahtonen Aug 14 '19 at 06:18
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    @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). – Bill Dubuque Aug 14 '19 at 13:36
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    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. – Bill Dubuque Aug 14 '19 at 13:40
  • For the record, I'm not the downvoter. And, I take it that you think the target is not a close enough duplicate. While we are at it. How else can I try and learn how others think if not by speculating and hoping for a response? You seem to think that is a faux pas. – Jyrki Lahtonen Aug 14 '19 at 14:16
  • But, I do believe you, too, spend a lot of time searching. If only everybody did that :-( – Jyrki Lahtonen Aug 14 '19 at 14:17
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    @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 – Bill Dubuque Aug 14 '19 at 14:27
  • If someone is curious about how many duplicate votes they cast: There is a Top duplicate closevoters query on the SE Data Explorer. (The numbers are not per tag, though.) – Martin R Aug 14 '19 at 14:38
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    @MartinR Nor does that (all-time) search reflect temporal constraints (only more recently have I been spending much more time dealing with duplication, and it is that more recent timeframe that I refer to above). Nor do I peruse cherry-picking paradise (calculus / real analysis) or CRUDE closures, where most of the action occurs. – Bill Dubuque Aug 14 '19 at 15:05
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    @MartinR E.g. in the most recent elem-number-theory dupes I closed about the same number as the rest of the community (about 25 each), and Jyrki closed about 4, so my above claim is very close (counted quickly by eye / mentally so may not be exact, but surely close). – Bill Dubuque Aug 14 '19 at 15:31
  • @Jyrki See prior comment. – Bill Dubuque Aug 14 '19 at 15:31
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    FWIW I have cast 20 duplicate votes in August. I do not concentrate on elem-number-theory, and we have different requirements about what is a good enough fit for a duplicate. Does not matter much. Best of luck in dupe hunting. – Jyrki Lahtonen Aug 14 '19 at 16:26
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    And, I have noticed that you have been active in the dupe closure front lately. Sorry about the argument here getting out of hand. – Jyrki Lahtonen Aug 14 '19 at 17:33
  • @JyrkiLahtonen My intention with this post is to demonstrate the particular approach that Pinter recommends in his book. I'll try to make it clear in the future when I'm primarily interested in a particular approach (such as that recommended by the text). It's certainly interesting to note other solutions and pointers to those in comments are welcome. – dharmatech Aug 14 '19 at 17:41