[Math] Every simple group of odd order is isomorphic to $\mathbb{Z}_{p} $ iff every group of odd order is solvable

group-theorysimple-groups

I'm trying to prove the following claims are equivalent:

  1. Every simple group of odd order is of the type $\mathbb{Z}_{p}$
    for prime $p$
  2. Every group of odd order is solvable.

Getting from 2 to 1 was easy but I'm having problem with the other direction. Obviously I only need to show that given 1 every non-simple group of odd order is solvable. So if I assume $G$ is a non-simple group of odd order then it has a non-trivial normal subgroup and this is where I get stuck. I'd appreciate a hint that will lead me towards the solution without giving it up completely 🙂

Thanks in advance!

Best Answer

I will prove $1 \Rightarrow 2$.

Let $G$ be a finite group of odd order. Let $G = G_0 \supset G_1 \supset\cdots \supset G_{n-1} \supset G_n = 1$ be a composition series. Each $G_i/G_{i+1}$ is a simple group of odd order. Hence, by the assumption $1$, it is abelian. Hence $G$ is solvable.

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