Prove that every element of a finite group has an order

Sometimes it is much clearer to argue the general case. Consider any $g \in G$, a finite group. Since $G$ is a group, we know that $g\cdot g = g^2 \in G$. Similarly, $g^n \in G$ for any $n$. So there is a sequence of elements,

$$g, g^2, g^3, g^4, g^5, \ldots, g^n, \ldots $$

in $G$. Now since $G$ is finite, there must be a pair of number $m \neq n$ such that $g^m = g^n$ (well, there are many of these, but that's irrelevant to this proof).

Can you finish the proof from this point? What does $g^m = g^n$ imply in a group?

Hope this helps!

Since the group is finite, it can not be the case that $g^n$ is different for all $n$. There must be $n_1$ and $n_2$ such that $g^{n_1} = g^{n_2}$ where $n_1 \neq n_2$ (unless $g = e$; ). Therefore, $g^{n_1 - n_2} = e$.

Yes, the order of an element is always less than or equal to the order of the group. In the proof above, assume $n_1$ and $n_2$ are all positive and that $n_1 < n_2$. Find the least such pair $n_1$ and $n_2$. If $n_1$ is greater than the order of the group, then that meant you saw at least $n_1$ different thing before seeing a repeat. But the group has only less than $n_1$ elements.

$ab=a$ happens only when $b=e$. Therefore $a^n$ is different for all $n<|G|$.

Because $ax=b$ has only one solution, there exists a finite $z$ with $a^z=e$, so all elements have a finite order.