We have $H_1 = \{e,(1234),(13)(42),(1432)\}$
As I understand it, "right cosets" are defined as: $H_1x:x\in S\setminus H$
The answer is given as:
$\{\operatorname{id},(1 2 3 4),(1 3)(2 4),(1 4 3 2)\},$
$\{(1 2),(1 3 4),(1 4 2 3),(2 4 3)\},$
$\{(1 3),(1 4)(2 3),(2 4),(1 2)(3 4)\},$
$\{(1 4),(2 3 4),(1 2 4 3),(1 3 2)\},$
$\{(2 3),(1 2 4),(1 3 4 2),(1 4 3)\},$
$\{(3 4),(1 2 3),(1 3 2 4),(1 4 2)\}$.
What I don't understand is how the first $x$ in each right coset given was chosen. Surely we have to go through all $x$ which is in $S$ but not in $H$? For instance I calculated the set:
$\{(123),(1324),(1342),(143)\}$ - is this not a right coset?
Okay after some trial and error and further reading, I discovered what my problem was:
Firstly, we know that in general $|S_n| = n!$. We have that $|H_1| = 4$ by counting the number of elements in $H_1$. Moreover, we know that $|H_1|$ divides $|S_4|$ by Lagrange's Theorem. Then the index of $H_1$ in $S_4$ is given by $\frac{|S_4|}{|H_1|} = \text{number of right cosets in $H_1$}$
In this case we then have $\frac{4!}{4} = 6$ cosets.
So we just plug in an element $x\in S_4\setminus H_1$, which generates a right coset. Then we repeat with an element $y\in S_4\setminus(H_1 \cup xH_1)$. We repeat this until we have $6$ cosets and then we know we are done.