How to prove that every nonabelian group of odd order is not simple?

Question

How to prove or disprove that every nonabelian group of odd order is not simple? I have no ideas concerning that.

Answer 1

Hint. See Feit-Thompson Theorem and show that it is equivalent to the statement that finite nonabelian simple groups have even order.

Answer 2

The Feit-Thompson theorem states that every finite group of odd order is solvable. So any group of odd order will have a proper commutator subgroup, which is normal, essentially by the defintion of being solvable $( G \vartriangleright [G,G] \vartriangleright ... \vartriangleright 1).$ Since the group is not abelian, the commutator subgroup $[G,G]$ not the trivial group. That means that $[G,G]$ is a proper nontrivial normal subgroup, so $G$ is not simple.