Prove that a cyclic group $(G,\cdot)$ has order $n$ with $n\in\omega_+$ iff for any $x\in G$ s.t. $G=\langle x\rangle$ the equality $x^n=1$ holds.

Question

First of all I point out that to follow the order of a group is only its cardinality!

So I am trying to prove that a cyclic group $(G,\cdot)$ has order $n$ with $n\in\omega_+$ if and only for any $x\in G$ such that the equality $$ G=\langle x\rangle $$ then even the equality $$ x^n=1 $$ holds.

First of all, let's we prove that if there exist $x\in G$ and $n\in\omega_+$ such that $$ x^n=1 $$ then $G$ has at least $2n$ elements provided $x$ generates $G$. Well, if $m\in\omega$ is such that $$ n<m $$ then there exists $h\in m$ such that $$ m=n+h $$ so that the equality $$ \tag{1}\label{1}x^h=1\cdot x^h=x^n\cdot x^h=x^{n+h}=x^m $$ holds. So \eqref{1} proves that for any $m\in\omega\setminus n$ there exists $h\in m$ such that $$ x^h=x^m $$ so that putting $$ h_0:=\min\{h\in m:x^h=x^m\} $$ then by recurrence we can put $$ h_{i+1}:=\min\{h\in h_i:x^h=x^{h_i}\} $$ for any $i\in\omega$ such that $h_i\in\omega\setminus n$. Now given $m\in\omega\setminus n$ the set $m\setminus n$ is finite so that if by definition for any $i\in\omega$ the inequality $$ h_{i+1}<h_i $$ holds then the recursive process cannot be iterated infinitely and what above proved ensures that it can be iterated exactly as long as $h_i$ is $n+1$ so that for any $m\in\omega\setminus n$ there exist $h\le n$ such that $$ x^m=x^h $$ which proves that $G$ has order at most $2n$ since the equality $$ x^{-m}:=(x^{-1})^m=(x^m)^{-1}=(x^h)^{-1}=(x^{-1})^h=x^{-h} $$ holds.

After all, if for any $h,k\in\omega$ distinct even $x^h$ and $x^k$ was it then $G$ would be infinite so that by contrapposition if $G$ is finite then there exist $h,k\in\omega$ distinct such that $x^h$ and $x^k$ are equal. So, assuming without loss of generality that the inequality $$ k-h>0 $$ holds, then we observe that $$ x^{k-h}=x^k\cdot x^{-h}=x^k\cdot(x^{-1})^h=x^k\cdot(x^h)^{-1}=x^k\cdot(x^k)^{-1}=1 $$ concluding there exist $n\in\omega_+$ such that $x^n$ is equal to $1$.

So first of all, I ask if I correctly proved the first part and then I ask to complete the second part showing that $n$ can be actually the order of $G$; finally, I ask if it is actually possible to prove that a cyclic group $(G,\cdot)$ has order $n\in\omega$ if and only if the equality $$ G=\{x^0,\dots,x^{n-1}\} $$ holds, provided $x$ generates $G$. So could someone help me, please?

N.B.

I Point out I found this question: initially, I thought it is could be solve my doubts but unfortunately I can say this because I do not really understand the notation so that I thought to put a more accurately question where I asked to solve the problem using my background which I hope is revealed by the notation I used above. For sake of completeness, I put here an image of my textbook (in Italian, sorry) to make it clearer what I'm trying to prove.

Answer 1

There is a fair amount of stuff that is wrong here, as well as parts that are overly complicated and overdone.

First, the statement seems to be:

A cyclic group $G$ has order $n$ if and only if for every $x\in G$ such that $\langle x \rangle = G$, we have $x^n=1$.

This is false. If $G$ is cyclic of order $k$, and $k\mid n$, then the for every $x\in G$ we have $x^n=1$, so in particular it holds for every generator of $G$. For a trivial counterexample, take $G$ cyclic of order $2$, $G=\{1,x\}$. Then for every even number $n$ we have $x^n=1$; if the quoted claim above were true, this would imply that $G$ is cyclic of order every positive even number, which is of course absurd.

What is true is:

A cyclic group $G$ has order $n$ if and only for every $x$ such that $G=\langle x\rangle$, $n$ is the smallest positive integer such that $x^n=1$.

Continuing on:

[L]et's we prove[sic] that if there exists $x\in G$ and $n\in \omega_+$ such that $x^n=1$, then $G$ has at least $2n$ element provided that $x$ generates $G$.

Again, this is just plain wrong. If $G=\langle x\rangle$, and $x^n=1$, then $G$ has at most $n$ elements, not at least $2n$ elements.

(What one could prove is that if $x^n=1$, and $n$ is the smallest positive integer such that $x^k=1$, and $x$ does not generated $G$, then $G$ has at least $2n$ elements; this would follow because $\langle x\rangle$ would be a subgroup of order $n$, and by Lagrange's Theorem the order of $G$ would be a multiple of $n$; the assumption that $x$ does not generate $G$ will imply that the order is not equal to $n$, and so it would be at least $2n$. But while this statement is true, it seems to be totally irrelevant to what you claim to be trying to prove, so I cannot fathom why we would be trying to prove it in this context...)

If $m$ [...] is such that $n\lt m$[,] $m=n+h$, so that the equality $x^h=1\cdot x^h=x^n\cdot x^h=x^{n+h}=x^m$ holds; [this proves that for every $m\gt n$ there eixsts $h\lt m$ such that] $x^h = x^m$.

This is true, though convoluted. Simpler to just write $x^m = x^{n+h} = x^nx^h = x^h$.

So if we put $h_0=\min\{h\lt m\mid x^h = x^m\}$ [and define] $h_{i+1}=\min\{h\lt h_i\mid x^h=x^{h_i}\}$,

This is wrong. It is wrong because the defined set is empty, so it has no minimum.

To see this, note that by definition, $h_0$ is the least positive integer less than $m$ such that $x^h=x^m$. If $h$ is a positive integer and $h\lt h_0$, then by the minimality of $h_0$ we must have $x^h\neq x^m$. But that means that $x^h\neq x^m = x^{h_0}$. Thus, the set $$\{ h\lt h_0\mid x^h = x^{h_0}\}$$ is empty, has no minimum, and $h_1$ is not properly defined. Neither is $h_{i+1}$ for any nonnegative integer $i$.

This observation completely obviates the next three paragraphs, which just say that we cannot "repeat this process" indefinitely. In fact, we cannot even do it twice.

So for every $m\gt n$ there exists $h\leq n$ such that $x^h=x^m$.

This is true, but rather than the confused argument given, a simple application of the division algorithm shows this: given $m\gt n$, write $m=qn+r$ with $0\leq r\lt n$. Then $$x^m = x^{qn+r} = (x^n)^qx^r = (1)^qx^r = x^r,$$ thus $x^m = x^r$ with $0\leq r\lt n$.

This proves that $G$ has order at most $2n$[...]

ehr... you said you were going to prove that $G$ has order at least $2n$: "$G$ has at least $2n$ elements". But now you are telling me you were proving it has at most $2n$ elements?

The argument in fact shows that $x$ has at most $n$ elements: while it is true that $x^{-m}=x^{-h}$, it is also true that because you can find $h$ with $0\leq h\lt n$ such that $x^m=x^h$, then $x^mx^{n-h} = x^hx^{n-h} = x^{n}=1$. If $h\gt 0$, then that means that $x^{-m}$ is also of the form $x^r$ with $0\leq r\lt n$; and if $h=0$, then $x^{-m}=x^m=1$ already. So in fact, you have at most $n$ elements, which is a much better bound that at most $2n$ elements.

The next part is the argument that if $G$ is finite and $x\in G$, then there exists $n\gt 0$ such that $x^n=1$. This is true, and the argument is correct.

This is all that is presented.

The above neither proves that if $G$ is cyclic of order $n$ then for every $x\in G$ with $\langle x\rangle = G$ we have $x^n=1$, nor does it prove that if for all $x\in G$ such that $\langle x\rangle = G$ we must have $x^n=1$, then it must be the case that $G$ has order $n$. Why not?

The first part proves that if $G$ has an element $x$ with $\langle x\rangle= G$ and $x^n=1$, then $G$ has at most $2n$ elements; this does not show that the group has order $n$; it shows that the group has order at most $2n$, a very different proposition. Even improving the argument as noted above, it only proves that $G$ has order at most $n$, it does not prove that $G$ has order $n$. This is to be expected, since the given statement, namely

For every $x\in G$ such that $\langle x\rangle=G$, we have $x^n=1$

only tells us that the order of $G$ divides $n$, it does not tells that the order is exactly $n$. So we cannot hope to actually prove the claimed result with this statement.

And the second part of the argument tells us that if $G$ is finite and $x\in G$, then there must exist a positive integer $n$ such that $x^n=1$; it does not tell us, however, that $G$ is cyclic, or that $G$ has order exactly $n$. Now, it is true that if $G$ is cyclic of order $n$, and $x$ is such that $\langle x\rangle = G$, then $x^n=1$. In fact, it is true that if $G$ is any finite group, cyclic or not, and $x\in G$, then $x^n=1$.

So the arguments proffered do not establish either of the implications; and one of the implications is false in any case.

I ask, it possible to prove that a cyclic group $G$ has order $n$ if and only if the equality $G=\{1,x,\ldots,x^{n-1}\}$ holds, provided $x$ generates $G$?

No, you cannot prove that for a very simple reason: that statement is false. It is false because there is nothing in that statement requiring the set $\{1,x,\ldots,x^{n-1}\}$ to have cardinality $n$; it just lists $n$ elements, but a set may be listed with the same element multiple times without changing what the set is. How we write a set does not determine its cardinality. The set $\{1, 2-1, 3-2, 5-4\}$ has cardinality $1$, not four, even though I listed it with four elements. It is true that if $G$ is cyclic of order $n$ and $x$ generates $G$, then $G=\{1,x,x^2,\ldots,x^{n-1}\}$. It is not true that if $G=\langle x\rangle$ and $G=\{1,x,x^2,\ldots,x^{n-1}\}$, then $G$ has order $n$; that statement only tells us that $G$ has order at most $n$, not that it has order exactly $n$.