I know $S_4=\{(),(3,4),(2,3),(2,3,4),(2,4,3),(2,4),(1,2),(1,2)(3,4),(1,2,3),(1,2,3,4),(1,2,4,3),(1,2,4),(1,3,2),(1,3,4,2),(1,3),(1,3,4),(1,3)(2,4),(1,3,2,4),(1,4,3,2),(1,4,2),(1,4,3),(1,4),(1,4,2,3),(1,4)(2,3) \}$ and I know there are 8 cosets to be found (I could be wrong) since $24 \div 3=8$ but I have trouble finding out how solve this. Can someone please help? I don't understand how I should multiply each $H$ with the corresponding S₄.
[Math] Find the right and left cosets of H = {(1), (123), (132)} in S4
abstract-algebra
Best Answer
For example: $$ (34)H = \{(34)(1),(34)(123),(34)(132)\} = \{(34),(124),(1432)\} $$ is the left coset of $H$ associated with $(3,4)$. We "multiply $H$" by (in this case, left-) multiplying each element in $H$ by the relevant element of $S_4$.
Note that some cosets end up being the same. For example, $$ [(34)(123)]H = (124)H = \{(124),(1432),(34)\} = (34)H $$
You are indeed correct about the number of (distinct) cosets; this is a consequence of Lagrange's theorem.