Consider the two sets of M numbers given by $W^k$ , $ 0\leq k\leq M-1 $ and $ W^{kL} $, $ 0\leq k\leq M-1 $ where $ W = e^{-2j\pi/M} $. Show that these two sets are identical.
It is true for the small numbers. But how to prove it truth?(Without induction) What theorem is behind in this phenomenon? Thanks in advance!
Let me outline the idea leaving the proofs as an exercise. Let $$U = \{W^k : 0 \leq k \leq M - 1\},$$ where $W = e^{2\pi i /M}$. Let $$U^L = \{z^L : z \in U\}.$$
If you are not familiar with the language of groups, just recall that:
If $L$ and $M$ are relatively prime, then there exist $r, s \in \mathbf Z$ such that $$Lr + Ms = 1$$
Conclude that:
$W^k \in U^L$, for every $0 \leq k \leq M - 1$.
Hint: $W^k= W^{kLr + kMs} = e^{2\pi i (kLr + kMs)/M} = \dots$
Therefore,
$U = U^L$
Hint: count how many elements each set has
If you are familiar with the language of groups, regard $U$ as a subgroup of $\mathbf C^\times$.
$U$ is isomorphic to $\mathbf Z /M\mathbf Z$
Hint: consider the map $\varphi: \mathbf Z \rightarrow U$ given by $\varphi(k) = W^k$.
Under the above isomorphism, the set $$U^L = \{z^L : z \in U\}$$ is identified with $$L\mathbf Z/M \mathbf Z = \{ Lr + M \mathbf Z : r \in \mathbf Z\}.$$
If $L$ and $M$ are relatively prime, then there exists $r \in \mathbf Z$ such that $$Lr \equiv 1 \mod M$$
Hint: there exist integers $r$ and $s$ such that $Lr + Ms = 1$.
Conclude that
If $L$ and $M$ are relatively prime, then $$L\mathbf Z/M \mathbf Z = \mathbf Z/M \mathbf Z.$$ Hence, $U^L = U$.