Abstract Algebra- Cosets I'm stuck on the following questions "In the alternating group $A_4$, let $H$ be the cyclic subgroup generated by $(123)$. Find all the right cosets of $H$ and all the left cosets of $H$. Is every right coset also a left coset."
The answer is as follows "$H = [idx_4 , (123), (132)]$ so each coset will have order 3, and since
$|A_4| = 12$ there will be 4 distinct cosets. We find the following right
cosets:
$H = [idx_4 , (123), (132)]$
$H(234) =[ (234), (12)(34), (134)]$
$H(124) = [(124), (13)(24), (243)]$
$H(143) = [(143), (14)(23), (142)]$
and the following left cosets:
$H = [idx_4 , (123), (132)]$
$(234)H = [(234), (13)(24), (142)]$
$(124)H = [(124), (14)(23), (134)]$
$(143)H = [(143), (12)(34), (243)]$
and we see that $H$ is the only coset that is both a left and a right coset."
Now I understand how you find the left and right cosets, the thing I'm stuck on is why did they choose $(234), (124), (143)$ out of the possible options from the alternating group $A_4$ as the permutations that $H$ is used on?
 A: This is probably because $3$-cycles generate the alternating group. In the case of $A_4$ there are only $8$ $3$-cycles, $(123),(124), (134), (234)$ and their inverses. So taking $3$ of them (different from $(123)$ and its inverse) is enough.
A: The choice was to an extent arbitrary. The fact is any single element from each coset can be chosen to represent it. 
If your question is as to how to find these elements, you can work recursively. The easiest coset to find is $H$ itself. To find the next coset, pick any element $a_1$ not in $H$. This will give you a new coset. To find all of the elements in this coset, multiply $a_1$ on the left or right (depending on whether you're doing right or left cosets) by every element of $H$.
Suppose you've found $n$ cosets $Ha_0,\ldots,Ha_n$. Following this procedure, you'll also know all of the elements in the union of these cosets. If you pick one element $a_{n+1}$ not in the union, this will give you a new coset. If the union is the entire group, then you've found all of the cosets.
A: I do not know how much group theory you have covered until now, but if you consider the subgroup of $A_{4}$ 
$$
V = \{ 1, (12)(34), (13)(24), (14)(23) \},
$$
you will have $$\{ v h : v \in V, h \in H\} 
=
V H = A_{4} = H V = \{ h v : v \in V, h \in H\},$$ with $V \cap H = \{ 1 \}$.
So, although as noted in other answers there are many choices for sets of representatives of the left and right cosets (actually $3^{4}$ choices for each set!), the set $V$ is somewhat a natural choice.
