We are working with $S_{10}$, the group of permutations of $\{1,...,20\}$, and $\sigma=(a_1 ,...,a_{10})$ and $\pi=(b_1 , ..., b_3)$. We wish to find the order of $\sigma^2$, and $\sigma^3$.
Can't we just start by stating the order of $\sigma$ is just $10$ because the cardinality of $\sigma$ is $10$.
Isn't $\sigma^2 = (a_1 , ..., a_{10})(a_1 , ... ,a_{10})$, and since there are repeats all throughout the multiplication the order of $\sigma^2$ is also $10$.
If not then what am I missing? Wouldn't the case for $\sigma^3$ just be the same thing as well.
True it is that since $\sigma$ is a $10$-cycle its order is $10$. But then $\sigma^2$ has order $5$. Meanwhile, $\sigma^3$ has order $10$ again, because $\rm{gcd}(3,10)=1$.
General fact about cyclic groups, if $g$ has order $n$, then $g^k$ has order $n/\rm{gcd}(n,k)$.
Finally, what does $\pi$ have to do with it?