Will $n$ for $A^n=\mathbb{I}$, where $A$ is any finite operation on a finite group and $\mathbb{I}$ is the identity operator, always be finite?

Question

Will $n$ for $A^n=\mathbb{I}$, where $A$ is any finite operation on a finite group and $\mathbb{I}$ is the identity operator, always be finite?

Consider for instance a finite sequence of moves (operations) on a Rubik's cube. How can we show that if we just carry out the same sequence of moves enough (but finitely many) times, we will always end up in the starting configuration again?

(I suspect that this is an elementary result in Group theory, but I haven't gotten around to studying it yet, so please bear with me.)

Answer 1

Since there are only finitely many states, you'll come across a state twice eventually, so there must be $k > m$ such that $A^k = A^m$. Define $n := k-m$. Now $A^n = A^{k-m} = A^kA^{-m} = A^mA^{-m} = I$.