If $\alpha, \beta$ are cycles s.t $\alpha^k(i)=\beta^k(i)$ then $\alpha=\beta$

Question

If $\alpha, \beta$ are cycles such that there is $i$ which is passed by both of them (as is $\alpha(i)\neq i$) such that $\alpha^k(i)=\beta^k(i)$ for all $k>0$ then $\alpha=\beta$

let $\alpha =(i_1,i_2,...,i_r), \beta=(j_1,j_2,...,j_s)$

$i_2=\alpha(i)=\beta(i)=j_2$

$i_3=\alpha^2(i)=\beta^2(i)=j_3$

For all $k$: $i_{k+1}=\alpha^k(i)=b^k(i)=j_{k+1}$ Then $\alpha=\beta$

It seems that there is a subtle point that I do not get, maybe an example will help my "digest" the proof

Answer 1

By definition of cycle, both $\alpha$ and $ \beta$ have one orbit only of cardinality $>1$, and by your condition their restrictions to it do coincide. For the points out of the orbit, it must be $i=\alpha(i)=\beta(i)$. So $\alpha(i)=\beta(i)$ for all $i$'s.