If $G$ is a finite group, show that there exists a positive integer $N$ such that $a^N = e$ for all $a \in G$.
I know that there are numerous threads in the forum about this topic and each and everyone of it (more or less) asks for a proof. However, I have a question regarding it, so, I am posting it. I was recently reading about group theory when I encountered this question.
I was given a definition about the order of an element: The order of an element $a$ in a group $G$ is the least positive integer $n$ such that $a^n=e$, where $e$ is the identity element in $G$. If no such $n$ exists, we call the element of infinite order. For example: If $G={1,\omega, \omega^2}$ under the usual multiplication as the binary operation forms a group. Now, here order of $1$ is $1$, the order of $\omega$ is $3$ and the order of $\omega^2$ is $3$. So each and every element in the group is of finite order. Now, if we consider the group $\mathbb{Z}$ under the usual addition as the binary operation then, $1\in\mathbb{Z}$ is of infinite order as the identity element $e=0$ and for any $n\in\mathbb{N}$, $$1^n=\underbrace{1+1+1+...+1}_{\textit{n times}}\neq 0.$$
So, $1$ is of infinite order and $1\in\mathbb{Z}$. Now, in the above question how is the above thing possible at all... I mean the element in the group may be of infinite order and then we may never have such an $n\in \mathbb{N}$. How is then the above thing true?
I am not quite getting it.