I have an exercise where I am supposed to find the left and right cosets of $H = \{(1), (12), (34), (12) \circ(34)\}$ in $S_4$.
But how do I generate the cosets?
As I have understood it you are supposed to pick a number that is not in the set $H$ and multiply it with every number in $H$. But this does not exactly give the right answer.
I would really appreciate it if someone gave an easy to understand explanation of how to generate the left and right cosets.
Best Answer
Assume your group is $G$ and the subgroup is $H$. By definition $gH$={$gh$|$h\in H$} is a left coset of $H$ respect to $g$, in $G$ and $Hg$={$hg$|$h\in H$} is a right coset of $H$ respect to $g$, in $G$. Here your group is $S_4$={$(),(3,4),(2,3),(2,3,4),(2,4,3),(2,4), (1,2),(1,2)(3,4),(1,2,3),(1,2,3,4),(1,2,4,3),(1,2,4),(1,3,2),(1,3,4,2),(1,3), (1,3,4),(1,3)(2,4), (1,3,2,4), (1,4,3,2), (1,4,2), (1,4,3), (1,4), (1,4,2,3), (1,4)(2,3)$ }, and the $H$ is as you pointed. According to Group theory, the number of right cosets of a subgroup in its group called index is $\frac{|G|}{|H|}$. $|S_4|=4!$ and $|H|=|\langle(1,2),(3,4)\rangle|=4$ so you have atlast $\frac{4!}{4}=6$ cosets right or left for the subgroup. Here there is no matter what $g$ is taken in group $G$. For example, if you take $(1,2,4,3)$ in group, then $(1,2,4,3)H$={$(1,2,4,3)(),(1,2,4,3)(1,2),(1,2,4,3)(3,4),(1,2,4,3)(1,2)(3,4)$}. Hope to help.