If $G$ is a finite group, show that there exists a positive integer $N$ such that $a^N = e$ for all $a \in G$. 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.
 A: Any element $g$ of a finite group generates a cyclic subgroup, $\langle g\rangle $, which must also be finite, of order $n=\lvert g\rvert $.
By Lagrange's theorem if $N=\lvert G\rvert $, then $n\mid N$.  Hence $N$ does the trick.
The smallest such number is called the exponent of the group.
A: Alternative approach:
The problem, as stated in the title, can be solved with virtually no knowledge of Group theory, using only the fact that the group has a finite number of elements.
Take any element $a$ in the group.
Consider the infinite sequence:
$a^1, a^2, a^3, \cdots.$
Since the group is finite, there must exist two positive integers $i$ and $j$ such that $i < j$ and $a^i = a^j.$
This implies that $a^{(j-i)} = e.$
Further, for any element $(a)$ in the group, if $k$ is a positive integer such that $a^k = e,$ then any multiple of $k$, of form $(r \times k) ~: ~r \in \Bbb{Z^+}$ will be such that $a^{(r \times k)}$ must equal $e^r = e.$
So, to each element $a$ in the finite group, there exists at least one positive integer $N(a)$ such that $a^{[N(a)]}$ is equal to $(e)$.
Further, for each such element $(a)$, any multiple $[r \times N(a)]$ will also be such that $a^{[r \times N(a)]} = e.$
So, given that the complete list of distinct elements in the finite group is represented by the list
$a_1, a_2, \cdots, a_s$ in the finite group $G$, you can form the sequence of numbers
$$N(a_1), N(a_2), \cdots, N(a_s). \tag1 $$
By all of the discussion in this answer, simply set $N$ equal to the $~\color{red}{\text{least common multiple}}~$ (or any common multiple) of the positive integers referred to in (1) above.
One explicit example would be
$$N = \prod_{i=1}^s N(a_i).$$
