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I am proving the following problem. I finished proving it but do not know is it sufficient enough.

Problem: Prove that the order of an element in a cyclic group $G$ must divide the order of the group.

My original proof:

Let the order of $G$ be $n$, and the generator of $G$ be $a$, then $a^n=e$. Then, for each element $b\in G, \exists k, m \in \mathbb{Z}$ s.t. $|b|=k$ and $b=a^m$. It must follow that:

$b^{k}=(a^{m})^{k}$ $\implies$ $e=a^{mk}$ $\implies$$a^n=a^{mk}$ $\implies$ $n=mk$.

Since $m\in \mathbb{Z}$, then $k|n$.

My edited proof:

Let the order of $G$ be $n$, and the generator of $G$ be $a$, then $a^n=e$. Then, for each element $b\in G, \exists k, m \in \mathbb{Z}$ s.t. $|b|=k$ and $b=a^m$. It must follow that:

$k=\frac{n}{gcd(n,m)} \implies n=(k)gcd(n,m)$

Since $gcd(n,m)\in \mathbb{N}$, then $k|n$.

My question is that I have seen lots of people used Division Algorithm to prove this, but it is only when the element is the generator, here it is any element. Yet, I don't know my prove is sufficient or not. Any suggestion is appreciated.

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Your proof is wrong because you state that $a^n=a^{mk}\implies n=mk$. Why do you think that this is true?

However, $a^n=a^{mk}=e\implies n\mid mk$, but this is not enough to deduce that $k\mid n$, which is what you want to prove.

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  • $\begingroup$ That makes sense, I knew something was wrong with my proof. Thank you. Also, can I use the $gcd$ to approach this? or is there any other way to do this. $\endgroup$ – Harry Jan 30 '18 at 6:51
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    $\begingroup$ @Harry I think the using the $\operatorname{lcm}$ is a good approach here. $\endgroup$ – José Carlos Santos Jan 30 '18 at 6:59
  • $\begingroup$ Thanks, I will try to use $lcm$. Do you mind looking at my edited proof to see this approach work or not? $\endgroup$ – Harry Jan 30 '18 at 7:06
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    $\begingroup$ It's from the theorem states that if $G$ is cyclic and has a generator $a$ with order $n$, then for any element $b \in G, b=a^{k}$. The order of $b$ is $n/d$ where $d=gcd(n,k)$. $\endgroup$ – Harry Jan 30 '18 at 7:13
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    $\begingroup$ @Harry All right. Of course, if you are allowed to use that theorem, everything becomes quite easy. $\endgroup$ – José Carlos Santos Jan 30 '18 at 7:14
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According to your hypothesis that $|G|=n \in {\mathbb N}~$ . Therefore , $G$ is a finite group.

Now for each $a\in G~$ ,$~\langle a\rangle$ forms a subgroup of the group $G$ .Then we see on account of the Lagrange theorem that $|\langle a\rangle |~|~|G|,$ that is ,$|\langle a\rangle |$ is a divisor of $~|G|~.$

Keep in mind that this proof holds for all group $G$ , and hence of course the cyclic one , then our conclusion follows .

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