Feit-Thompson Theorem's Logically Equivalent Representation (3.4.12)

Dummit and Foote Abstract Algebra, section 3.4, exercise 12:

Prove (without invoking the validity of the Feit-Thompson Theorem) that the following are logically equivalent:
(i) every group of odd order is solvable
(ii) the only simple groups of odd order are those of prime order

Proof: (i) ⇒ (ii) Let T be a simple group with odd order pn, and 1 < n∈Z⁺. Since T is of odd order, T is solvable (i), but since T is simple, T has no proper normal subgroups. By the solvability of T, T/1 ≅ T is abelian. By Cauchy's theorem, ∃x∈T, | x | = p. Now, < x > < T, but since T is abelian, < x > is normal in T, which violates the simplicity of T. Modus tollens, no simple groups with composite odd orders exist.

(ii) ⇒ (i) Let T be a group of odd order. Take the composition series of T (existent by part 1 of the Jordan-Hölder Theorem). By Lagrange's Theorem, all of its component link groups must be of odd order, therefore all of its factors are of odd order. By the definition of a composition series, these factors are simple, and by (ii) they are of prime order. Thus they are cyclic, thus abelian, thus the composition series fulfills the requirement to render T solvable.

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