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I'm dealing with the following problem in Isaacs Finite Group Theory [6A.5], I would appreciate if you could help:

Let $G$ be a nonabelian solvable group in which the centralizer of every nonidentity element is abelian. Show that $G$ is a Frobenius group where $F(G)$ is the Frobenius kernel. [Here $F(G)$ is the Fitting subgroup of $G$]

My attempt: I think if we can show that centralizer $C_G(n)$ of any nonidentity element $n$ of $F(G)$ is contained in $F(G)$, the result will immediately follow. But I could not show that.

Martin Brandenburg
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Yılmaz
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1 Answers1

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Since $F(G)$ is nilpotent, we have $Z(F(G)) \ne 1$, and $F(G) \le C_G(g)$ for all $g \in Z(F(G))$, so $F(G)$ is abelian.

Let $1 \ne g \in F(G)$. Since $C_G(g)$ is abelian and contains $F(G)$, we have $C_G(g) \le C_G(F(G))$.

But $G$ solvable implies $C_G(F(G)) \le F(G)$, so $C_G(g) =F(G)$, and you can do it from there.

Nicky Hekster
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Derek Holt
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    Here, I think we should add where we use the assumption that G is not abelian. Since G is nonabelian and solvable we can directly say that its fitting subgroup is nontrivial. – Yılmaz Jun 30 '22 at 21:29
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    I did not use the assumption that $G$ is nonabelian in my answer. I just showed that $C_G(g) \le F(G)$ for all $g \in F(G)$, because you said that you could complete the proof provided that you could prove that. You need the assumption that $G$ is nonabelian in order to complete the proof. – Derek Holt Jun 30 '22 at 21:34