Let $(G,\cdot, e)$ a finite group. Prove that for all $a\in G$, exists $n\in\mathbb{N}$ such as $a^n=e$.
Intuitively I understand that there should be equivalence departments according to the generator organ but I don't know to translate it to math, thanks.
The (ostensibly infinite) set $\{e,a,a^2,a^3,\dotsc\}$ is a subset of your finite group. Therefore some of elements must be equal (there is no injection from an infinite set to a finite set). Thus there is some $a^m=a^n$ with $m\neq n$ (assume WLOG that $m > n$). Therefore $a^{m-n}=e.$
Alternatively, if you've learned it, you can cite Lagrange's theorem.
Consider the sequence: $a, a^2, a^3,.....$, Each member belongs to G. Since G is finite, $a^p = a^q, 0<p < q$.
Hence $a^{p-q} = e$
Assume that for some $a$, that does not hold, then $a^{n}=e$ for all $n=1,2,...$, then the elements in $\{a_{n}\}$ are pairwise distinct, say, $a^{n}=a^{m}$ for some $m>n$, then $a^{m-n}=e$, contradiction.
