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 March 12th, 2011, 08:08 AM #1 Newbie   Joined: Mar 2011 Posts: 26 Thanks: 0 Minimal Normal subgroup of a solvable group Let $G$ be a finite solvable group and $H$ a minimal normal subgroup of $G$. 1. Show $H$ is abelian 2. Show $H$ is a p-subgroup for some prime number p. (without using the characteristic group, hint - you can look at a p-sylow group of $H$). 3. Let $K=\{ x\in H: x^p=1\}$ show that $K=H$. 4. Show that $H$ is a direct sum of cyclic grouops of order p. I was able to show 1. Currently stuck on #2, need help there. Thanks
 March 12th, 2011, 02:10 PM #2 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Re: Minimal Normal subgroup of a solvable group Okay, here's a way to do it for #2: Assume that H is not a p-group, and consider a k-Sylow subgroup K of H. Since all Sylow subgroups of H are conjugated and H is abelian, K is the unique k-Sylow subgroup of H, and K is different from H (since K is not a p-group). If we manage to prove that K is normal in G then we will be done since then H will not be a minimal normal subgroup of G (absurd). Proof: considering an element a in G, $aKa^{-1}$ is a p-Sylow of H, and is therefore equal to K.
March 12th, 2011, 04:30 PM   #3
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Re: Minimal Normal subgroup of a solvable group

Quote:
 Originally Posted by julien Proof: considering an element a in G, $aKa^{-1}$ is a p-Sylow of H, and is therefore equal to K.
did you mean $a^{-1}Ka$ is a k-sylow subgroup of H? I think I understand everything else

Thanks.

 March 13th, 2011, 01:31 AM #4 Site Founder     Joined: Nov 2006 From: France Posts: 824 Thanks: 7 Re: Minimal Normal subgroup of a solvable group They are the same group, since H has a unique p-Sylow ...

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