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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?

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    $\begingroup$ If $G$ is simple then $G > 1$ is a composition series of $G$, so $G$ is a composition factor of $G$ and hence, by the Feit-Thompson Theorem, $G$ is abelian. Now it just remains for you to prove that the only abelian simple groups have prime order. $\endgroup$ – Derek Holt Dec 3 at 8:24
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From its composition series, such a group $G$ would have a normal subgroup of prime index.

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