Facts of a finite cyclic group

  • #1
Mr Davis 97
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Problem: If ##H = \langle x \rangle## and ##|H| = n##, then ##x^n=1## and ##1,x,x^2,\dots, x^{n-1}## are all the distinct elements of ##H##.

This is just a proposition in my book with a proof following it. What I don't get is the very beginning of the proof: "Let ##|x| = n##. The elements ##1,x,x^2,\dots, x^{n-1}## are all distinct elements because..."

Isn't the fact that ##|x| = n## part of what we wanted to prove? Why does the proof just "let" this be the case?
 

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  • #2
fresh_42
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How do you want to prove this? There are elements of any finite order. The question is where you want to start from. I read this above as: Given an element ##x## of finite order ##n##, then ##x## generates of cyclic group of order ##n##.
 
  • #3
Mr Davis 97
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How do you want to prove this? There are elements of any finite order. The question is where you want to start from. I read this above as: Given an element ##x## of finite order ##n##, then ##x## generates of cyclic group of order ##n##.
I feel like it reads: given a group of order ##n## generated by ##x##, prove that the order of ##x## is ##n## and ##1,x,x^2,\dots, x^{n-1}## are all the distinct elements of ##H##. If this is the case I feel like one must prove that the order of ##x## is ##n##, not just let it be the case.
 
  • #4
fresh_42
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Yeah, that's possible, too. In this case ##|H|=n## is given and ##\operatorname{ord}x =n## has to be shown. Hard to tell not seeing the book. Both ways are possible and make equal sense - and are equally easy to show.
 
  • #5
Mr Davis 97
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Yeah, that's possible, too. In this case ##|H|=n## is given and ##\operatorname{ord}x =n## has to be shown. Hard to tell not seeing the book. Both ways are possible and make equal sense - and are equally easy to show.
The only way I see to prove that ##|x| = n## is this: if ##|x|## were larger, ##|H| > n## and if ##|x|## were smaller, ##|H| < n##. Thus ##|x| = n##. But I feel like this depends on the distinctness of each element ##x^k## where ##0 \le k < n##, which is what must be proved it seems.
 
  • #6
fresh_42
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Not quite. For ##|x|=m## you are right that this implies ##|H|=: n \geq m##. But why is it equal? There could theoretically be other elements which fill up the gap from ##m## to ##n##. The distinctness follows from the definition of the order as minimal number, because
$$x^a=x^b \,(n > a > b \geq 0)\, \Longrightarrow x^{a-b}=1 \Longrightarrow n\,\mid\, (a-b) \Longrightarrow a \geq a-b =nq \geq n \,\,\lightning $$
 
  • #7
Mr Davis 97
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https://imgur.com/a/4jZbtfl

So the proof of 1) starts with "Let ##|x| = n##" and the proof of 2) starts with "Next suppose ##|x| = \infty##".

But shouldn't the hypothesis of 1) be "Let ##|H| = n##" and the hypothesis of 2) be "Next suppose ##|H| = \infty##"?
 
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  • #8
fresh_42
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I don't know what exactly is written in the book.

If you want to prove, that the generator in a cyclic group of finite order is of the same order, then start with ##|H|=n## and ##|x|=m## and show ##n=m##.
If you want to prove, that an element of finite order always generates a cyclic group of the same order, then start with ##|x|=n## and show ##|H|=n##.

An infinite order can be ruled out. In the first case, because the group is finite, in the second per assumption.
 
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  • #9
Not quite. For ##|x|=m## you are right that this implies ##|H|=: n \geq m##. But why is it equal? There could theoretically be other elements which fill up the gap from ##m## to ##n##. The distinctness follows from the definition of the order as minimal number, because
$$x^a=x^b \,(n > a > b \geq 0)\, \Longrightarrow x^{a-b}=1 \Longrightarrow n\,\mid\, (a-b) \Longrightarrow a \geq a-b =nq \geq n \,\,\lightning $$

The contradiction symbol (lightning) isn't formatting on my phone. Is this standard Latex or do you need a package for that? Can be useful.
 
  • #10
fresh_42
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The contradiction symbol (lightning) isn't formatting on my phone. Is this standard Latex or do you need a package for that? Can be useful.
I thought the same. It's library isn't loaded here either. I just kept it as a silent protest that it would be a good idea to have one. It's so convenient compared to "... and this is a contradiction to our original assumption."
\usepackage{ stmaryrd }
Btw., funny name "St Mary Road" to look for a lightning.
 
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