If $\beta$ is the transposition $\beta = (1, 4)$, compute both $\beta\alpha$ and $\beta\alpha\beta^{−1}$ and compute orders. Let $\alpha$ be the permutation in the symmetric group $S_9$ defined by
$$\alpha =\left(\begin{matrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ 5 & 1 & 7 & 6 & 3 & 9 & 2 & 8 & 4\end{matrix}\right)$$

If $\beta$ is the transposition $\beta = (1, 4)$, compute both $\beta\alpha$ and $\beta\alpha\beta^{−1}$ and give a list of those positive integers which are the orders of the elements in the
  symmetric group $S_9$. For each such integer, also give one permutation having that
  order.

So I have found out that $\alpha = (1,5,3,7,2)(4,6,9)$ as a product of two disjoint cycles. However, I am not sure how to use the transposition of $\beta$ in this case to compute the composition of $\beta\alpha$ and $\beta\alpha\beta^{−1}$. 
Also for the second part of the question, wouldn't the orders of the elements in the first cycle $(1,5,3,7,2)$ all be 5 since thats how many elements it takes to cycle? and similarly for the second cycle $(4,6,9)$ they would all be order 3? 
I know this is a few few parts but I'm having some clarification issues, thank you for any and all help.
 A: In the symmetric group $S_n$, the group operation is the composition of permutations. The permutation $\beta\alpha$ is the permutation $\alpha$ followed by $\beta$. For example, $(\beta\alpha)(1)=\beta(\alpha(1))=\beta(5)=5$.
To answer your second question, $1,2,...n$ are not elements of $S_n$. The elements are $\textit{permutations}$ on $1,2,...n$.
You are correct that the order of a permutation that is a single cycle is the length of that cycle. More generally, if a permutation is a product of multiple disjoint cycles, its order is the least common multiply of the lengths of those cycles.
A: Permutation means bijective function on Set to itself .Any cycle say $\beta$ as given (14) means $\beta(1)=4,\beta(4)=1$ Similarly hold for any cycle define in  such way. Now $\beta\alpha$ and $\beta\alpha\beta^{-1}$ can be calculated as just function composition .$\beta\alpha$=(15372469) and  $\beta\alpha\beta^{-1}$=(169)(53724) .We can also see now this Permutation can form group under composition. So any of its element has some order means $\beta^{n}$=Identity  that is $\beta$ $\beta$ $\beta$ $\beta$ .....$\beta$ (n times)=Identity  .This can also be obtained using cycle size But for that disjoint cycle are must .Even if they are not then we have to take composition until unless we are left with disjoint cycle.As $\beta\alpha$ is single 8 cycle Its order is 8.If you see $\beta\alpha\beta^{-1}$ has same cycle type as original permutation this is because both are conjugate elements . Here we have 2 disjoint cycle of order 3 and 5 so we can calulate order of combine permutation as lcm of both. that 15.Coming to your last question .In $S_9$ or any symmetric group order of any elements related to cycle type .SO we can even predite possible order in that group(Stronger than Lagrange theorem! Here In Symmetric Group we guarantees these order element are existing in that group)  We have to just count cycle type in that group like (1)(2),(2)(2),(2)(2)(2),(2)(2)(2)(2),(2)(2)(2)(3).(3),(3)(3),(3)(3)(3),(3)(4),(3)(4)(2)...
And SO On this is Painful listing but we count directly this using Partition .P(9)=30. From each cycle listed above we can obtain size of elements possible .So 1,2,3,4,5,6,7,8,9,10,12,....So on 
