A problem in Finite Group Theory

[the_fox]

This is a problem I encountered in Martin Isaacs' 'Finite Group Theory'. It's located at the end of Chapter II which deals with subnormality, and the particular paragraph is concerned with a couple of not so well-known results which I quote for reference:

(In what follows F is the Fitting subgroup)

Theorem (Zenkov)
Let A and B be abelian subgroups of the finite group G, and let M be a minimal (in the sense of containment) member of the set {A \cap B^g | g in G}. Then M is a subgroup of F(G).

An easy corollary follows which establishes the existence of a subnormal subgroup:

Corollary
If A is an abelian subgroup of the finite group G and |A|>=|G:A|, then A intersects F(G) non-trivially.

In fact, if A is cyclic then a normal subgroup is guaranteed:

Theorem (Lucchini)
Let A be a cyclic proper subgroup of a finite group G, and let K=core_G(A). Then |A:K|<|G:A|, and in particular, if |A|>=|G:A|, then K>1.

Problem
Let G be a finite group such that G=AN, where A is abelian, N is normal in G, the centralizer of N in A is trivial as is the Fitting subgroup of N. Show that |A|<|N|.

Note that, since N is normal in G, the Fitting subgroup of N is the intersection of N with the Fitting subgroup of G. So, if |A|>=|N| in the problem, then |A|>=|N:N \cap A|=|NA:A|=|G:A| and the corollary applies to give a non-trivial intersection of A with F(G). How does one proceed from here to obtain a contradiction? In particular, how can the condition on the centralizer be utilized effectively?

 

[fresh_42]

Let $1\neq g \in A\cap F(G)$. Then there is a nilpotent normal subgroup $U\triangleleft G$, which contains $g$ and is thus not trivial. Then $U\cap N \subseteq U$ is also nilpotent and normal, i.e. included in $F(N)=\{\,1\,\}$. Hence $U\cap N = \{\,1\,\}$ and $U\times N$ is a direct product which contains $g$. But as $1\neq g \in A \cap U$ we have $g \in C_A(N) = \{\,1\,\}$, the required contradiction.