Question 1: In $A_4$, does $(01)(23) = (01)(23)$?
I think you'll have to try again on that. I'm not sure what you meant to ask. However, note that in this class, we would use the numbers 1, 2, 3, 4 to make permutations in $S_4$ and $A_4$, rather than 0, 1, 2, 3. I'm guessing that you got this notation from Group Explorer. It doesn't really matter what we use, but we should try to be consistent and be sure that we can translate back and forth if we are going to start changing formats.
Question 2: Can we copy the cosets for a group from Group Explorer, as I did for 5.5.d for $A_4$?
Well, I guess you could, but the point is to understand how to compute them by hand. If you are confident that you know how to compute them, then using Group Explorer to speed up the process is fine. Of course, you can use Group Explorer to check your work. However, in the case of $A_4$ in $S_4$, I'm way too lazy to write them all out by hand. Instead, it pays to be clever. Remember, the cosets partition the big group. Since $A_4$ is exactly half of the elements in $S_4$ and all of the cosets turn out to be the same size (we haven't officially proved this yet, but in each specific case, it is easy to convince yourself that it has to be true), there can only be one other coset, namely the set of odd permutations. We can represent this coset using any odd permutation. For example, since $(12)$ is odd, $(12)A_4$ (this is of the form $gH$) is one of the many ways to name this second coset. In summary, there are precisely two left cosets (and hence right cosets since there isn't any "wiggle" room): $A_4$ and $(12)A_4$.
