# Power of a r-cycle

Given $$n \ge 3$$ and $$2 \le r \le n$$, let's define the permutation $$\rho \in S_n$$ (the symmetric group of degree $$n$$) by:

• $$\rho(k)=k+1$$, for $$k \in I_<:=\lbrace 1,...,r-1 \rbrace$$
• $$\rho(r)=1$$
• $$\rho(k)=k$$, for $$k\in I_>:=\lbrace r+1,...,n \rbrace$$

I've thought to prove that $$\rho^r=\iota_{S_n}$$ -the identical permutation- in the following manner (reductio ad absurdum). Could you double check if the proof is correct, please?

Proof. Let's suppose that $$\rho^r \ne \iota_{S_n}$$; then, $$\exists \tilde k \in I:=\lbrace 1,...,n \rbrace$$, such that $$\rho^r(\tilde k)=\rho(\rho^{r-1}(\tilde k)) \ne \tilde k$$. By the very same definition of $$\rho$$, it follows that $$\rho^{r-1}(\tilde k) \in I_<^*:=\lbrace 1,...,r-1 \rbrace \cup \lbrace r \rbrace$$. Now, let's assume that $$\rho^j(\tilde k) \in I_<^*$$ for some $$0 \le j \le r-2$$ (inductive hypothesis); then $$\rho^{j-1}(\tilde k)=\rho^{-1}(\rho^j(\tilde k)) \in I_<^*$$, because the restriction $$\rho_{|I_<^*}$$ is a bijection of $$I_<^*$$ into itself; so, finally, $$\rho^j(\tilde k) \in I_<^*$$ $$\forall j=0,\dots,r-1$$. This means that $$\tilde k+j \in I_<^*$$, $$\forall j=0,\dots,r-1$$, which in its turn implies $$\tilde k=1$$: contradiction, because by definition of $$\rho$$, it is $$\rho^r(1)=1$$.

• "This means that $\tilde k+j \in I_<^*$, $\forall j=0,\dots,r-1$," Why? If you have intermediate claims, it makes sense to delimit them (and their proofs) clearly, particularly when you are trying to formalize the proof. – darij grinberg Nov 20 '18 at 4:04
• @darij grinberg - You're right, that conclusion is undue, and frankly I'm stuck there. Any chance to conclude on that way? – Luca Nov 20 '18 at 14:55
• A quick way is by proving (by induction) that $\rho^i \left(k\right) \in \left\{1,2,\ldots,r\right\}$ and $\rho^i \left(k\right) \equiv k+i \mod r$ for each $k \in \left\{1,2,\ldots,r\right\}$ – darij grinberg Nov 20 '18 at 16:05