Every simple group of odd order is isomorphic to $\mathbb{Z}_{p} $ iff every group of odd order is solvable I'm trying to prove the following claims are equivalent:


*

*Every simple group of odd order is of the type $\mathbb{Z}_{p}$
for prime $p$

*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!
 A: 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.
A: $\,(1)\Longrightarrow (2)$:
Let $\,G\,$ be a group of minimal odd order that is not solvable. Thus $\,G\,$ cannot be abelian so $\,G'\neq 1\,$ . By (1), $\,G\,$ cannot be simple, so $\,\exists\,H\triangleleft G\,\,,\,1<H<G\,$ . Let us take $\,H\,$ maximal normal in $\,G\,$ .
By the minimality assumption  $\,G/H\,$ is solvable but also simple, by maximality of $\,H\,$ as normal subgroup and, of course, of odd order $\,\Longrightarrow\,G/H\cong\Bbb Z_p\,$,  for some prime$\,p\, $ . But this means $\,G'\leq H\,$ , and thus $\,1<G'<G\,$ and, again by minimality $\,|G|\,\,,\,\,G'\,$ is solvable. But this contradicts the assumption that $\,G\,$ isn't solvable (why?) and we're thus done.
A: Suppose every simple group of odd order is isomorphic to $\mathbb{Z}_{p}$
 for a prime $p$
 . Let $G$
  be a group of odd order. If $G$
  is simple then from the assumption $G$
  is solvable. Suppose $G$
  is not simple, we continue by complete induction on the odd numbers less than $\left|G\right|$.
Induction hypothesis: Every group $K$
  of odd order $\left|K\right|<\left|G\right|$
  is simple.
Since $G$
  is not simple there is proper normal subgroup $$\left\{ e\right\} \lneq H\lneq G$$
 From Lagrange's theorem we know that $\left|H\right|\vert\left|G\right|$
 and thus $\left|H\right|$
 is necessarily odd and since H
  is a proper subgroup $\left|H\right|<\left|G\right|$
 . Thus from the induction hypothesis $H$
  is solvable.
We also know that: $$\left|\frac{G}{H}\right|=\frac{\left|G\right|}{\left|H\right|}\overbrace{<}^{\left|H\right|>1}\left|G\right|$$
  So the quotient $\frac{G}{H}$
  is of odd order less than $\left|G\right|$
 and thus from the induction hypothesis it is also solvable. To conclude, H and $\frac{G}{H}$ 
  are solvable and thus so is $G$ 
  (according to a theorem not shown here)
