# Facts of a finite cyclic group

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• Mr Davis 97
In summary, the author is trying to prove that an element of finite order always generates a cyclic group of the same order. The first step in this proof is to "let" an element of order n generate a group of order n. This is done by assuming that an element of order n can generate a group of order n. If this assumption is true, then the order of the element is n.

#### Mr Davis 97

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?

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##.

fresh_42 said:
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.

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.

fresh_42 said:
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.

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
$$x^a=x^b \,(n > a > b \geq 0)\, \Longrightarrow x^{a-b}=1 \Longrightarrow n\,\mid\, (a-b) \Longrightarrow a \geq a-b =nq \geq n \,\,\lightning$$

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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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.

• Mr Davis 97
fresh_42 said:
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.

Math_QED said:
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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