$a^n = e$ for $a \in G$ where $G$ is a group with finitely many elements. 
Let $G$ be a group with a finite number of elements. Show that for any $a \in G$, there exists an $n \in \mathbb{Z}^+$ such that $a^n = e$, where $e$ is the identity and $a^n = a * a * a \space ... *\space a$ where $*$ is a binary operation.

Being a new student to algebra, I find this question very counterintuitive. Let, for example, $G$ be the group of the positive rational numbers over multiplication. Thus we have the identity $e = 1$ and inverse $\frac{1}{a}$ for any number $a \in G$. I claim that there is no positive integer  $n$ where $2^n = 1$. How does one make sense of all this?
 A: If $a=e$, then the result is trivial. For $a\neq e$: Since the group is finite then $a^n$ can not be distinct elements for all $n$. Therefore, there exist $m_1$ and $m_2$ $\in \mathbb{Z}^+$ where $m_1\neq m_2$  such that $a^{m_1}=a^{m_2}$. Without loss of generality suppose $m_1>m_2$, so that $m_1-m_2 \in \mathbb{Z}^+$. Thus, $a^{m_1-m_2}=e$ which completes the proof. As an example in $\mathbb{Z}_4$: $0$ is the identity. $1+1+1+1=2+2=3+3+3+3=0$ in $\mathbb{Z_4}$.
A: Hint: How many distinct elements can the sequence
$$ a, a^2, a^3, a^4, \ldots$$
have?
A: Let $\exists a \in G$ such that $a^k \ne e$ $\forall k \in \mathbb{N}$ 
then we can write $k = k_1 - k_2$ where $k_1, k_2 \in \mathbb{Z}$
$\implies a^{k_1 - k_2} \ne e \implies a^{k_1} \ne a^{k_2} \forall k_1, k_2$ $\in \mathbb{Z}$
So all powers of $a$ (i.e. $a^1, a^2, a^3, a^4, \ldots$) are distinct.
$\implies G$ is an infinite group
Hence $\exists N \in \mathbb{N}$ satisfying $a^N = e $
A: By Lagrange's theorem,  the order of any element divides the order of the group. If $G$ is finite,  then any element $g\in G$ must have finite order, since $a\mid b\implies a\le b$.
