# Solvable Frobenius Group

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.

• 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? – Levent Mar 9 '18 at 14:20
• Yes. In general this is false. But I was wondering what conditions one would impose to make it abelian. – J. R. Mar 9 '18 at 14:25
• If order of the complement is even then the kernel is abelian. – Levent Mar 9 '18 at 14:27
• Interesting. So if we had an odd order kernal what structure of the group would be needed. – J. R. Mar 9 '18 at 14:31