By Application Of Cayley's Theorem to show Every non-abelian group of order 6 is isomorphic to $S_3$ By Application Of Cayley's Theorem

Question

I know this question already answered But I wanted to solve this question using Application Of Cayley's Theorem .This is mentained in Topics in Algebra Herstein's Book Page 74.I had Doubts regarding That .
Let G is non-abelian Group of order 6 .So there must be element of order 2 in it by argument that if not then there is no element in that which are self inverse in it .Which is contradict with fact that by pairing we left with identity and one element which must required element . So we can form subgroup H={e,a} where $a\neq e ,a^2=e$ .It is apparent that this H does not has nontrivial normal subgroup.From that G is isomorphic to T which is subset of A(S) ,Where S is right cosets of H in G .|A(S)|=|G:H|!=6 Therefore T=S . But Form this How to conclude $A(S)\approx S_3$ ?

Any Help will be appreciated.

Answer 1

$A(S)$ is presumably the set of permutations of $S$? In which case there are $3$ cosets, so $A(S)\cong S_3$. Since $G\cong T$ and $T$ is a subgroup of $A(S)$ of order $6$, we must have $T=A(S)$ so $G\cong T=A(S)\cong S_3$.