If $G$ is nonabelian & solvable s.t. the centralizer of each nontrivial element is abelian, then $G$ is Frobenius with kernel its Fitting subgroup

Question

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.

Answer 1

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.