Show that a set is a right transversal for $S_3$ in $S_4$

Question

I am given the set A ={e,(14),(24),(34)}

I'm supposed to show that this set is a right transversal for S3 in S4, meaning that every right coset of S3 contains exactly one element of A.

I'm getting stuck because (14) is not an element in S3...any help would be appreciated.

Answer 1

Let $G=S_4,\ H=S_3$ where $H$ is regarded as pemutations of $\{1,2,3\}$ and so is a subset (and a subgroup) of $G$ regarded as permutations of $\{1,2,3,4\}.$

Then none of $(14),(24),(34)$ are in $H,$ so we can form the three possibly distinct right cosets $H(14),\ H(24),\ H(34)$ and none of these are $H.$

If we can show any two of these are in fact distinct, we can finish. Suppose for example that $H(14)=H(24).$ Then multiplying each on the right by $(14)$ and using that $(14)(14)=e,$ we would have $H=H(24)(14)=H(124).$ But that would imply that $(124)$ is in $H$ which it is not.

Other cases are similar. We now have four pairwise disjoint cosets $He,\ H(14),\ H(24),\ H(34)$ each of size 6 so these are the distinct right cosets of $H$ in $G.$ We have already shown that no two of these right cosets can contain two elements of $A=\{e,\ (14),\ (24),\ (34)\}.$