Q:

Prove the order of an element is the order of the group

If $g \in G$ and $|G| = n$, prove $g^n = e$.
I actually don't know if this is true or not, which is why I am trying to prove this. I know that the order of $g$ is $n$, but I don't really understand what the order of an element means. I know that when you apply the group's operation to that element by itself $n$ number of times, you get the identity element, but I don't know how that helps prove or disprove this.

A:

Consider the subgroup of $G$ generated by $g$. By Lagrange, the order of $\langle g\rangle$ divides $n = |G|$. But the order of this subgroup is precisely the order of $g$!
So if the order of $g$ is $k$, we have that: $k|n$, that is: $n = kt$ for some (positive) integer $t$.
Hence $g^n = g^{kt} = (g^k)^t = \dots$?

