Why Does the Proof Assume ##|x| = n## in Finite Cyclic Groups?

[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?

 

[fresh_42]

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 $n$.

 

[Mr Davis 97]

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.

 

[fresh_42]

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.

 

[Mr Davis 97]

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.

 

[fresh_42]

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

 

[Mr Davis 97]

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 $|H| = \infty$"?

 

[fresh_42]

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.

 

[member 587159]

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.

 

[fresh_42]

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.