$A_5$ is just the alternating group, i.e. the even permutation subgroup of $S_5$.
$H=\langle(12345)\rangle\cong\mathbb Z_5$.
I know there are exactly 12 different cosets of $H$ in $A_5$, but I have no idea how to write out them with a simple way.
Is there a simple method to determine these 12 different cosets without writing out all the elements $\alpha H$, for every $\alpha\in A_5?$
Thanks a lot :)
Update:
@Derek has given a method to find all the different representatives of cosets of such $H$(i.e. $H=\langle(12345)\rangle\cong\mathbb Z_5$) in $A_5$.
But his method fails when $H=\langle(123),(12)(45)\rangle \cong S_3$. I have a try to find similar method, but I fails. So this problem is still open, for the case $H=\langle(123),(12)(45)\rangle \cong S_3$.
Any help will be appreciated.
In general there is no magic formula for writing down coset representatives. There are of course computer algorithms used by software such as GAP and Magma, and they use a variety of technical ideas to reduce the search space. I won't go into that in any detail here.
But by applying some of the ideas used in the algorithms, I managed to write down left coset reps of the subgroup $\langle (1,2,3),(1,2)(4,5) \rangle$ of $A_5$ without too much difficulty or calculation.
The stabilizer of the point $1$ in $H$ is the subgroup $\langle (2,3)(4,5) \rangle$, so I started by writing down $6$ coset reps of that in $A_4$. That was not too hard, and I took $1$, $(2,4)(3,5)$, $(2,3,4)$, $(2,4,3)$, $(2,3,5)$, $(2,5,3)$.
We can take the other reps from permutations mapping $1$ to $4$ or $5$. I noticed that the three elements $(1,4)(2,3)$, $(1,4)(2,5)$ and $(1,4)(3,5)$ all lie in different cosets from the $6$ already chosen and from each other, so now we have $9$.
I was afraid the final one might be hard to find, but then I noticed that $(1,5)(2,3)$ is not in any of the other cosets, so we are done.