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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!

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You can argue by induction in the order of the group starting with groups of (odd) prime order. – Diego Dec 26 '12 at 15:02
Yep that does solve it nicely when using the fact that if $H\vartriangleleft G$ and $H,\frac{G}{H}$ are solvable then $G$ is solvable – Serpahimz Dec 26 '12 at 15:49
up vote 1 down vote accepted

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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Oh, that is nice. – Serpahimz Dec 26 '12 at 17:30

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)

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If anyone finds an error with this I'd appreciate if you let me know :) – Serpahimz Dec 26 '12 at 16:14
The wording on your induction hypothesis should be fixed. – Alexander Gruber Dec 26 '12 at 17:12

$\,(1)\Longrightarrow (2)$:

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.

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