Odd-order implies solvable

Original proof
This result was proved by Feit and Thompson, and is called the Feit-Thompson Theorem or the Odd order theorem.

Computer verification of proof
In September 2012, it was announced that the Feit-Thompson theorem proof had been completely verified using the proof assistant Coq.

Statement
There are two versions of the statement:


 * 1) There is no finite simple non-abelian group that has odd order.
 * 2) Every odd-order group is a solvable group, i.e., all its composition factors are abelian groups (and hence, cyclic of prime order).

Applications

 * Coprime implies one is solvable: If two finite groups have relatively prime orders, then one of them is solvable.

There are several applications of the result that given two groups of coprime order, one of them is solvable. In fact, many theorems in group theory can be proved modulo the assumption that among two given groups of coprime order, one is solvable. For a list of such facts, refer:

Category:Facts about groups of coprime order whose proof requires the assumption that one of them is solvable

Facts used

 * 1) uses::Lagrange's theorem
 * 2) uses::Solvability is extension-closed

Proof that (1) implies (2)
Suppose (1) holds, i.e., there is no finite simple non-abelian group of odd order. Then, we want to show that (2) holds. Consider a minimal counterexample to (2), i.e., an odd-order group $$G$$ that is not solvable and has the minimum possible order among such groups.

Proof of (1)
The complete proof (which forms the subject of a 255-page paper) is beyond the scope of this page. However, we will attempt to describe the key idea.

The idea is to attempt to construct a "minimal counterexample" to (1), i.e., a simple non-abelian group of odd order that has the smallest possible order among such groups. Any such minimal counterexample must in particular be a minimal simple group: every proper subgroup is solvable (note therefore that the classification of finite minimal simple groups would settle this question; however, such a classification is itself conditional to first proving this fact, hence it does not help). In particular, it is a N-group: every local subgroup (the normalizer of a nontrivial solvable subgroup) is solvable. Or equivalently, for every prime $$p$$, every p-local subgroup is solvable (this follows from the fact that local subgroup of finite group is contained in p-local subgroup for some prime p).

The CN-group case
An example that is relatively easy to follow is the proof that odd-order and CN implies solvable.

A CN-group is a group where the centralizer of every non-identity element is nilpotent. The structure of CN-groups allows us to define an equivalence relation on the prime divisors of the group order based on commuting (see commuting of non-identity elements defines an equivalence relation between prime divisors of the order of a finite CN-group). For each equivalence class $$\omega$$ under the equivalence relation on the prime divisors of the order of a finite CN-group $$G$$, $$G$$ has a nilpotent $$\omega$$-Hall subgroup.

What we have said so far applies to all finite CN-groups. Restricting to the "minimal odd-order counterexample" that we want to show does not exist, we can obtain more structural restrictions on the nature of the Hall subgroups. The structural restrictions obtained from the proof mimic those used in the proof that finite non-abelian and every proper subgroup is abelian implies not simple. However, they are not quite as strong, and ultimately, we need to use ideas from character theory to complete the proof.