Let G be any non-Abelian group of order 6. By Cauchy's theorem, G has an element, a, of order 2. Let H = a, and let S be the set of left cosets of H.
(a) Show H is not normal in G. (Hint: If H is normal, then H is a subset of Z(G), and then it can be shown that G is Abelian)
(b) Use the below result and part (a) to show that G must be isomorphic to Sym(S). Thus any non-Abelian group of order 6 is isomorphic to S_3.
"Let H be a subgroup of G, and let S denote the set of left cosets of H. For a,x an element of G define a(xH)=axH. The multiplication defined yields a group action of G on S. Let phi: G --> Sym(S) be a homomorphism that corresponds to the group action defined above then the ker(phi) is the largest normal subgroup of G that is contained in H. Assuming that G is finite and let [G:H]=n then if n! is not divisible by |G| then H must contain a nontrivial subgroup of G."