So I'm looking through solutions for a practice problem in my modern algebra textbook that has me completely stumped.
Let $S_{4}$ be the group of permutations of the set $\{ 1 , 2 , 3 , 4 \}$, and let $A_{4}$ the subgroup of even permutations.
(a) Write down all the left cosets of $A_{4}$ in $S_{4}$.
(b) Write down all the right cosets of $A_{4}$ in $S_{4}$.
The answer I've found has $A_{4}$, $(12)A_{4}$ for part (a) and $A_{4}$,$A_{4}(12)$, what has me confused because wouldnt say $(34)A_{4}$ be another left coset? in fact wouldnt I have $(23)A_{4}$ or $(13)(23)A_{4}$ also be another coset ?
I understand that the the index of $A_{4}$ with $S_{4}$ is $2$ so that means their should be $2$ cosets in total no? I 'm very confused..
$\endgroup$ 21 Answer
$\begingroup$If $G$ is a group and let $H$ be a subgroup of $G$ then the two cosets $aH$ and $bH$ are same iff $ab^{-1}\in H.$ So in your case cosets of $(34)$, $(23)$, $(12)$ are same and the coset of $(13)(23)$ is $A_4.$
$\endgroup$ 2
