In a Abstract Algebra textbook of "Dummit and Foote" , There is a Group theory problem at 1.32 . The Problem goes like this :
"If $x$ is an element of finite order $n$ in $G$, prove that the elements $1$ , $x$ , $x$$2$ , $x$$3$ , $x$$4$ , ….. , $x$$(n-1)$ are all distinct.Deduce that $|x|$ $\le$ $|G|$. "
My Attempt: Lets say there is two numbers $a$ and $b$ and they follow this inequality:
$0$ $\le$ $a$ $\lt$ $b$ $\le$ $(n-1)$
Lets assume that:
$x$$a$ = $x$$b$
$\rightarrow$ $x$$(b-a)$ $=$ $1$
$\rightarrow$ $x$$(b-a)$ $=$ $x$$n$
So , $(b-a) = n$
On the other hand , From the inequality , I can deduce that $(b-a)$ $\lt$ $n$ ,Which is a contradiction .So , $x$$a$ $\neq$ $x$$b$
On the other hand , if $|x|$ $>$ $|G|$ , Then $x$ can generate more distinct elements than the Group even has , which is never possible .So , $|x|$ $\le$ $|G|$.
Confusion: First of all , The question is unclear for me if I choose the order of $x$ to be $0$ or $1$.Because if $|x|$ is $0$ or $1$ , then , $x = e$ and all powers of $e$ are the same ($e$). The problem doesn't tell me what values of $n$ I am restricted to.Can someone tell me what that is?
Also , My attempt seems to be very hand wavy .Is there a way to write the proof with Formal logic in a rigorous manner and where are the major flaws of my proof?. (I am asking this because in the future , I want to learn how to write proper proofs and learn Formal logic).
Best Answer
As you want to prove that distinct powers correspond to distinct elements, you aim to get a contradiction by assuming $\exists a,b, 0\le a<b\le n-1$, such that $x^a=x^b$. This is indeed the case, because this latter would mean that $\exists l(:=b-a), 0<l\le n-1$, such that $x^l=1$, against the minimality of $n$ (by definition of order of an element).
Now, by closure axiom, $\langle x\rangle :=\{1,x,\dots,x^{n-1}\}\subseteq G$, whence $|x|:=|\langle x\rangle|\le |G|$.