Exercise 1.3.11 ($(1\ 2\ \ldots m)^i$ is a $m$-cycle iff $\mathrm{g.c.d.}(m,i) = 1$)

Proof. Let us denote $a\wedge b$ the g.c.d. of $a$ and $b$.

(a)

• Suppose that $d=i\wedge m>1$.

  If $d=m$, then $m\mid i$, thus ${\sigma }^i=e$ is not a $m$-cycle. We may suppose now that $d<m$.

  Then there are integers $k$ and $\delta$ such that $i=\mathit{dk}$ and $m=\mathit{d\delta }$, so $1<\delta <m$. Moreover

  $${({\sigma }^i)}^{\delta }(1)={\sigma }^{\mathit{dk\delta }}(1)={\sigma }^{\mathit{mk}}(1)=1,$$

  where $1<\delta <m$. This shows that ${\sigma }^i$ is not a $m$-cycle.

• Suppose that $i\wedge m=1$.

  Since $\sigma$ is a cycle, for all integers $j$, ${\sigma }^j(1)=1⟺m\mid j$ (see Ex. 10). Then for all integers $k$, using $i\wedge m=1$,

  $${({\sigma }^i)}^k(1)=1⟺{\sigma }^{\mathit{ik}}(1)=1⟺m\mid \mathit{ik}⟺m\mid k.$$

  Therefore ${\sigma }^i(1),{({\sigma }^i)}^2(1),{({\sigma }^i)}^3(1),\dots ,{({\sigma }^i)}^{m-1}(1)$ are distinct from $1$, and ${({\sigma }^i)}^m(1)=1$. Moreover $1,{\sigma }^i(1),{({\sigma }^i)}^2(1),{({\sigma }^i)}^3(1),\dots ,{({\sigma }^i)}^{m-1}(1)$ are all distinct, otherwise ${({\sigma }^i)}^j(1)={({\sigma }^i)}^l(1)$ where $0\le j<l\le m-1$. The image by ${\sigma }^{-\mathit{ij}}$ gives $1={({\sigma }^i)}^{l-j}(1)$, where $0<l-j<m$, which is impossible.

  This shows that the orbit ${\mathcal{}O}_{{\sigma }^i}(1)$ de $1$ for ${\sigma }^i$ is

  $${\mathcal{}O}_{{\sigma }^i}=\{1,{\sigma }^i(1),{({\sigma }^i)}^2(1),{({\sigma }^i)}^3(1),\dots ,{({\sigma }^i)}^{m-1}(1)\}=[[1,m]],$$

  We prove now that

  $$\begin{array}{lll}\hfill {\sigma }^i=(1,({\sigma }^i)(1),{({\sigma }^i)}^2(1),{({\sigma }^i)}^3(1),\dots ,{({\sigma }^i)}^{m-1}(1)).& & \hfill \text{(1)}\end{array}$$

  Indeed,

  • if $j\notin {\mathcal{}O}_{{\sigma }^i}$, then $j>m$, so $j$ is fixed for $\sigma$, and ${\sigma }^i(j)=j$.

  • if $j\notin {\mathcal{}O}_{{\sigma }^i}$, then $j={({\sigma }^i)}^k(1)$ for some integer $k$, $0\le k\le m-1$.

    $${\sigma }^i[{({\sigma }^i)}^k(1)]={\sigma }^{i(k+1)}(1),$$

    in particular if $k=m-1$, then ${\sigma }^i[{({\sigma }^i)}^k](1)=1$. This shows (1), so ${\sigma }^i$ is a $m$- cycle.

  In conclusion, ${\sigma }^i$ is a $m$-cycle if and only if $i$ is prime to $m$.