Facts of a finite cyclic group

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.
  • #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
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
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.
 
  • #4
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
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.
 
  • #6
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
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
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
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.
 
  • #10
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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What is a finite cyclic group?

A finite cyclic group is a mathematical structure that consists of a finite set of elements and an operation that combines any two elements to produce a third element in the set. The operation is associative, has an identity element, and every element has an inverse. Additionally, the elements in a finite cyclic group can be generated by repeatedly applying the operation to a single element, which is known as a generator.

How do you determine the order of a finite cyclic group?

The order of a finite cyclic group is equal to the number of elements in the group. This can be determined by counting the number of distinct elements that can be generated by repeatedly applying the operation to the generator element. Alternatively, the order can also be determined by finding the highest power of the generator that results in the identity element.

What is the structure of a finite cyclic group?

The structure of a finite cyclic group is determined by its order. If the order is a prime number, then the group is known as a cyclic group of prime order. If the order is a power of a prime number, then the group is known as a cyclic group of prime power order. In general, finite cyclic groups have a simple and predictable structure, which makes them useful in various areas of mathematics and science.

What are the properties of a finite cyclic group?

Finite cyclic groups have several important properties, including closure, associativity, identity, and inverse. Additionally, every element in a finite cyclic group can be expressed as a power of the generator, which allows for convenient calculations. Furthermore, these groups have cyclic subgroups and a cyclic subgroup of every order that divides the group's order.

What are some real-world applications of finite cyclic groups?

Finite cyclic groups have various applications in cryptography, coding theory, and group theory. In cryptography, they are used to create secure encryption methods and to generate pseudorandom numbers. In coding theory, they are used to construct error-correcting codes. In group theory, they are used to study the structure and properties of groups in general.

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