Making sure a list of cosets is exhaustive I have a basic group theory question but I can't seem to find a satisfying answer.
For example, say I'm interested in the subgroup $H$ of $S_5$ generated by the cycle $(1\, 2\,3)$, and I want to list all its left cosets (ie all the sets of the form $xH$ with $x \in S_5$).
And I can use a few properties such that $xH = yH \iff xy^{-1} \in H$ to simplify my research (since it states that if two permutations differ by a $(1\,2\,3)^k$ cycle the coset is going to be the same).
But it only makes the listing barely any quicker. And when working by hand with relatively big groups like $S_5$, I can't really find a listing method that allows me to say « that's it, the list is complete I don't have to try with all the remaining elements ».
Hope I made my question clear enough, thank you guys for reading.

Since $(1\ \ 2\ \ 3)$ generates a group $H$ with $3$ elements, there will be $40\left(=\frac{120}3\right)$ cosets. One of them will be $H$, of course.
Now, take $\sigma\in S_5\setminus H$. Say that you take $\sigma=(1\ \ 4)$. Now, compute $\sigma H$. That will give you a second coset.
Then, you take $\eta\in S_5\setminus(H\cup\sigma H)$ (note that $H$ and $\sigma H$ are disjoint) and you start all over again.
After having obtained $40$ cosets, you're done.
