I have no clue what to do here !
Let $n>1$ be a natural number. the m-cycle $\sigma=(a_1,a_2,a_3,\dots,a_m)$ denotes a permutation in $S_n$ where $\sigma(a_1)=a_2,\sigma(a_2)=a_3,\dots, \text{ and }\sigma(a_m)=a_1$.
Prove by induction on $k \geq 1$ that if $k+i\equiv j (mod\text{ m})$, then $\sigma^k(a_i)=a_j$ whenever $1\leq i \leq m$ and $1\leq j \leq m$.
Pained Attempt :
Let $n>1$ be a natural number. the m-cycle $\sigma=(a_1,a_2,a_3,\dots,a_m)$ denotes a permutation in $S_n$ where $\sigma(a_1)=a_2,\sigma(a_2)=a_3,\dots, \text{ and }\sigma(a_m)=a_1$.
[Base Step]
For the base case,$k=1$, assume that $1+i\equiv j (mod\text{ m})$ and prove that $\sigma^1(a_i)=a_j$ whenever $1\leq i \leq m$ and $1\leq j \leq m$.
With $1+i\equiv j (mod\text{ m})$ then $i\equiv j-1 (mod\text{ m})$ iff $m|i-j+1$ or equivalently $i=mg+j-1$ for some $g \in \Bbb{Z}$. Therefore $\sigma^1(a_i)=\sigma(a_{mg+j-1}).....?$
Is this approach even right?? Thank you for your help