Thank you for taking the time to read this and help me out. This was a question I missed on my past test:
"Let $G = S_3$ and $H = A_3$ ($H \subseteq G$ is a subgroup). Compute the left and right cosets -
1.) $(12)$$A_3$
2.)$A_3$$(12)$
I've been having trouble trying to understand, the alternating group, but all my book says about it is, "The group of even permutations of n symbols is denoted by $A_n$" Is it addressing permutations such as (12) or (23)? So would the left cosets look something like (12)(23), (13)(12)?
The alternating group is the group of all permutations that are products of an even number of transpositions. Thus the cosets in question consist only of products of odd numbers of transpositions, and they have three elements. In this case there are only three elements that are products of odd numbers of transpositions, and these are just the transpositions $(12),(23),(13)$, so both cosets consist of those three elements.
You can compute this explicitly by noting that $A_3=\{e,(12)(23),(23)(12)\}$. Thus $(12)A_3=\{(12),(23),(12)(23)(12)=(13)\}$ and $A_3(12)=\{(12),(12)(23)(12)=(13),(23)\}$.