This exercise asks to provide elements of order 10, 20, and 30 in $S_{10}$. Thinking that the order of a permutation $\sigma$ is the least common multiple of the length of the disjoint cycles whose product is $\sigma$, I went for the following:
$\sigma_1=(a_1a_2)(b_1b_2b_3b_4b_5)$ of order 10
$\sigma_2=(a_1a_2a_3a_4)(b_1b_2b_3b_4b_5)$ of order 20
$\sigma_2=(a_1a_2a_3)(b_1b_2b_3b_4b_5)(c_1c_2)$ of order 30
The question also asked if there could be a permutation of order 40. This is where I got stuck, mainly because I could not find any combination of cycle lengths which would give 40 as least common multiple. Because the order of $S_{10}=3628800$, and 40 divides that, I figured there would have to be an element of order 40, and much larger orders as well, which it seems to be impossible to obtain considering a permutation written as disjoint cycles. Help in finding the flaw in my reasoning would be appreciated.
Every permutation (of finitely many elements) can be written as a product of disjoint cycles.
$40$ is divisible by $8=2^3$ and the only way for a least common multiple to be divisible by a prime power is if one of the operands to the lcm is divisible by that same prime power.
So in order to make a permuation of order 40, we need at least one cycle whose length is a multiple of 8. There's not room for that in $S_{10}$ while also getting a factor of 5 from somewhere.
Your assumption that just because 40 divides the order of the group there must be an element of order 40 is not true.
For example, consider the gorup $C_2\times C_2\times C_2$. Its order is 8, but there is no element of order 4 -- all elements except the identity have order 2.
In fact your reasoning would mean that every group $G$ has an element of order $|G|$ (because $|G|$ divides itself!) and so is cyclic. This is obviously not true.