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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 $M\subseteq F(G)$.

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

If $A$ is an abelian subgroup of the finite group $G$ and $|A|\geq|G:A|$, then $A \cap F(G)>1$.

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 $|A|\geq|G:A|$, then $K>1$.

Let G be a finite group such that $G=AN$, where $A$ is abelian, $N \unlhd G$, $C_A(N)=1$ and $F(N)=1$. Show that $|A|<|N|$.

Note that, since $N \unlhd G$, it follows that $F(N)= N \cap F(G)$. So, if $|A|\geq|N|$ in the problem, then $|A|\geq|N:N \cap A|=|NA:A|=|G:A|$ and the corollary applies to give $A \cap F(G)>1$.

How does one proceed from here to obtain a contradiction? In particular, how can the condition on the centralizer be utilized effectively?

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I'm not sure this is the right site to visit for this kind of question, which is somewhat advanced but based on a textbook problem. I hope it's not a homework problem. (The issue would be different if you thought you found a serious gap or error in a published proof. Anyway, Isaacs himself is still active in mathematics and could be consulted in that case.) – Jim Humphreys Feb 17 '11 at 18:17
up vote 1 down vote accepted

You were almost there: Since $N$ and $F(G)$ are two normal subgroups intersecting trivially, they commute. But now take a non-trivial element $a \in A \cap F(G)$; then by the previous observation, $a$ commutes with every element of $N$. But this contradicts the fact that $C_A(N) = 1$.

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Indeed. What if we replace the condition on the Fitting subgroup with the requirement that A and N have relatively prime orders. Is it still true that |A|<|N|? – user13040 Feb 20 '11 at 23:15
Yes: if $A$ is a $\pi$-group (where $\pi$ is a set of primes), then the fact that $A \cap F(G) > 1$ gives $A \cap O_\pi(G) > 1$. Now proceed as in the previous problem with $F(G)$ replaced by $O_\pi(G)$. – Tom De Medts Feb 21 '11 at 13:19

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