Very poor proof of order of elements in a permutation group 
In the group $S_{10}$ there is an element of order $30$. Also prove that no
  element has order $11$.

(i)
$(1,2,3,4,5)(6,7,8)(9,10) = 5*3*2= 30$
How do they view this as an element i wanna look at this as a function or a group how is this an element?
(ii)
$11$ is a prime number thus to get $11$, you must have $11*1$ assume this is true and its in our group $S_{10}$ then $(1,2,3,4,5,6,7,8,9,10,11)$ must be in $S$, $S$ has only $10$ elements so this is absurd. 
This seems really cheesy is there a good way to prove this?
 A: The first one is correct.
For the second part, use Lagrange Theorem which says the order of element divides the order of group. Or, you can simply write out all possible cycle combinations in $S_{10}$ and see what are the possible orders.
A: You are correct, in a sense, in your second part: but you'd need to elaborate:
a permutation of order $11$ is of prime order, hence, there are no factors of $11$ to have a product of disjoint cycles of order, such that the $\text{lcm}\,$ of the lengths of those cycles equals $11$, save for a cycle of length $11$. A cycle of length $11$ has order $11$, but being a single cycle requires that there are $11$ distinct permuted elements, and $S_{10}$ is the set of permutations of only $10$ elements. So there can be no permutation which is a cycle of length $11$, or a product of two or more cycles each of length $11$. This would indeed be absurd.
Also, your computation of the order of the first permutation is correct, and it is correct because $30=\text{lcm}\,(5, 3, 2)$. That is, $30$ is the least common multiple of the lengths of the the disjoint cycles. 
In contrast, the permutation $\alpha = (12)(34)(567)(8\,9\,10)$ is the product of two cycles of length $2$ and two cycles of length $3$, so the order of $\alpha = \text{lcm}\,(2, 2, 3, 3) = 2 * 3 = 6$.
