# Proof on order of element of cyclic group must divide the order of the group.

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:

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

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.

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.

• 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. Jan 30 '18 at 6:51
• @Harry I think the using the $\operatorname{lcm}$ is a good approach here. Jan 30 '18 at 6:59
• Thanks, I will try to use $lcm$. Do you mind looking at my edited proof to see this approach work or not? Jan 30 '18 at 7:06
• 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 $d=gcd(n,k)$. Jan 30 '18 at 7:13
• @Harry All right. Of course, if you are allowed to use that theorem, everything becomes quite easy. Jan 30 '18 at 7:14

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 .