Suppose $G$ is abelian and finite of order $n$ and $x\in G$. Prove that the order $x$ divides $n$.

Question

Suppose $G$ is abelian and finite of order $n$ and $x\in G$. Prove that the order $x$ divides $n$.

My solution:

Denote $m:=\text{the order of }x.$

Suppose the order of $x$ does not divide $n$. Then

$$\exists 0<b<m , a\in \mathbb{Z} \space n=am+b.$$

Also

$$x^n=x^{am+b}=(a^m)^ax^b=x^b=e.$$

Contradiction since $m$ is the order of $x$.

Is my proof correct? Why it was important to note that $G$ is an abelian group?

Answer 1

Yes, your proof is correct. I would state the conclusion once $x^b=e$ is shown though.

This result is known as Lagrange's Theorem. Having $G$ abelian is not necessary.