# Structure of Groups : Cauchy's Theorem, Order, Abelian Groups, Non-Abelian Groups, Isomorphisms and Subgroups

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."

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#### Solution Preview

Proof:

(a) If H is normal, then for any g in G, we have gHg^(-1)=H. Since |H|=2 and H=<a>={e,a}, then gag^(-1)=e or gag^(-1)=a. But if gag^(-1)=e, then a=e. This is a contradiction. So we must have gag^(-1)=a or ga=ag. So H is a subgroup of Z(G). Now I claim Z(G)=G. If not, we consider h in G but not in ...

#### Solution Summary

Structure of groups, Cauchy's Theorem, Order, Abelian Groups, Non-Abelian Groups, Isomorphisms and Subgroups are investigated. The solution is detailed and well presented.