In Dummit and Foote, Section 6.2 (Applications in Groups of Medium Order), it states that the Feit-Thompson Theorem asserts that there are no simple groups of odd composite order.
The Feit-Thompson was mentioned in Section 6.1: If $|G|$ is odd, then $G$ is solvable.
The definition of solvable group is that each of its composition factors is abelian.
I feel like I am not able to put these together to arrive at the conclusion that there are no simple groups of odd composite order. Here is my attempt: If $|G|$ is odd and a composite number, then because it is odd, then $G$ must be solvable. Then each of its composition factors are abelian. What next?
Best Answer
From its composition series, such a group $G$ would have a normal subgroup of prime index.