About Cycles And Permutations Comsider a $m$ cycle $$\phi =(1, 2, ........., m)$$. Let $k < m$ be any positive integer. Prove that ${\phi}^k$ is again a $m$ cycle if and only if $$\gcd(m, k)=1$$.
$$$$Let us assume that $1 \neq d=(m, k)$ and ${\phi}^k$ is a $m$ cycle. Now after applying $\phi$, $k$ times, we get $${\phi}^k=(1, k+1, r_1, r_2, .........., r_{m-2}),$$  where $r_1$ is such that $$(k+1)+k \equiv r_1 \mod m$$, and for all $2 \leq i \leq m-2$, $r_i$ is defined such that $$r_{i-1}+k \equiv r_i \mod m$$, also each $r_i\leq m $. Now as $d$ divides both $m$ and $k$ so we get $$r_i \equiv 1  \mod d$$ for every $1 \leq i \leq m-2$. Now as the cycle $$(1, k+1, r_1, ........, r_{m-2})$$ is a $m$ cycle hence all of $r_is$ are different and also different from $k+1$. Hence $k$ and all the $r_i-1$ for $1 \leq i \leq m-2$ are $m-1$ distinct multiples of $d$ and all are less than $m$. But as $d\mid m$ and $d>1$ hence we cannot have $m-1$ distinct multiples of $d$ all of which are less than $m$, hence a contradiction.
$$$$Now suppose that $(m, k)=1$ and there is a cycle $$(1, k+1, r_1, r_2, ...... r_{n-2})$$ where $r_is$ are defined in similar manner and $n<m$. Now we see that $$(k+1)+k \equiv r_1 \mod m$$, $$r_1 +k \equiv r_2 \mod m , \\\vdots\\\vdots$$ $$r_{n-3}+k \equiv r_{n-2}  \mod  m,$$ $$ r_{n-2} +k \equiv 1  \mod m.$$ Adding all these we get $$m|2k + k(n-2)$$. Now as $(m, k)=1$ hence we get $m\mid n$, but as $n<m$, it is a contradiction. Hence we get the required result.
$$$$Is My Proof Correct??
 A: Well done! Your proof is correct although there are other much shorter ways of completing the proof. For example:
Let ${\phi}^k=\psi $ and note that  ${\phi}^m=I$, the identity.
If  $ d=(m, k)\ne 1$, then $ {\psi}^{m/d}=({\phi}^{mk/d})=I^{k/d}=I$ and so $\psi$ is not an $m$-cycle.
If interested, you might like to work out a similar proof for the case when $(m, k)=1$.
A: Alternatively you could use your $m$-cycle to define a cyclic group of order $m$:
$$\langle (1,2,3,...,m) \rangle = \langle \phi\rangle$$
It turns out that $\langle \phi^k \rangle = \langle \phi^{(k,m)} \rangle$ and the generators of the cyclic group are precisely when $(k,m)=1$ i.e. $\phi^k$ is an $m$-cycle when $(k,m)=1$.
Proof:
Since $(k,m)|k$, $\phi^k \in \langle \phi^{(k,m)}\rangle$ so $\langle \phi^k \rangle \subseteq \langle \phi^{(k,m)} \rangle$.
Similarly $(k,m)= ky+mx$ for some $x,y \in \mathbb{Z}$, $\phi^{(k,m)}= \phi^{ky+mx}=\phi^{ky}\phi^{mx}=\phi^{ky}$ since $\phi^{m}=(1)$. Thus $\phi^{(k,m)} \in \langle \phi^k \rangle$. $\langle \phi^{(k,m)} \rangle \subseteq \langle \phi^k \rangle$.
