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I have a question regarding necessary conditions. I have been looking around in the literature and was wondering if there was any necessary conditions to ensure that the Frobenius kernel of a solvable group is abelian. I believe there may not be, but I just wanted to check with this community last. By the way, this is all self study from Issac's Finite Group Theory.

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  • $\begingroup$ Just to clarify : You have a group $G=NF$ where $N$ is the Frobenius kernel, $F$ is the Frobenius complement and $G$ is solvable and you wonder if $N$ is abelian or not? Am I correct? $\endgroup$ – Levent Mar 9 '18 at 14:20
  • $\begingroup$ Yes. In general this is false. But I was wondering what conditions one would impose to make it abelian. $\endgroup$ – J. R. Mar 9 '18 at 14:25
  • $\begingroup$ If order of the complement is even then the kernel is abelian. $\endgroup$ – Levent Mar 9 '18 at 14:27
  • $\begingroup$ Interesting. So if we had an odd order kernal what structure of the group would be needed. $\endgroup$ – J. R. Mar 9 '18 at 14:31

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