I was revisiting group theory in detail and reading Isaacs's Finite Group Theory book in my own time. I need a little clarification on one part in the proof of Theorem 2.20. This is on page 63 of the book.
Here is the set-up:
$G$ is a non-trivial finite group and let $\mathbb F(G)$ be the Fitting subgroup of $G$. Suppose $A$ is a cyclic proper subgroup (with trivial core) of $G$ such that $|A|\ge |G:A|$ and $A\cap \mathbb F(G)>1$. For any subgroup $H$ of $G$, we denote the center of $H$ as $Z(H)$.
Now we do the following:
Since $A\cap \mathbb F(G)>1$, it is clear that $\mathbb F(G)>1$. So we can choose a minimal normal subgroup $E$ of $G$ with $E \subseteq \mathbb F(G)$.
My question is about the following claims made in the book; any help why these are true will be greatly appreciated.
(1) Why $E\cap Z(\mathbb F(G)) >1$?
(2) Why it follows from (1) and minimality of $E$ that $E\subseteq Z(\mathbb F(G))$?
(3) How (using minimality of $E$ and $E$ is abelian) one can show that $E$ is an elementary abelian $p$-group for some prime $p$?
Thanks in advance!
(1) The Fitting subgroup is nilpotent; finite nilpotent groups have the property that every normal subgroup intersects the center nontrivially. Since $E$ is normal in $G$, it is also normal in $\mathbb{F}(G)$ and so must intersect the center of $\mathbb{F}(G)$ nontrivially.
(2) But the center of a group $H$ is characteristic in the subgroup $H$. Thus, since $Z(\mathbb{F}(G))$ is characteristic in the normal (in fact, characteristic) subgroup $\mathbb{F}(G)$, it follows that $Z(\mathbb{F}(G))$ is normal in $G$, and so $E\cap Z(\mathbb{F}(G))$ is normal in $G$. Since it is nontrivial, the minimality of $E$ guarantees that it cannot be a proper (nontrivial) subgroup of $E$, hence it must equal $E$. Thus, $E\subseteq Z(\mathbb{F}(G))$.
(3) If $E$ is not a $p$-group, then one of its $p$-parts is proper, nontrivial, and characteristic in $E$, so it would be normal in $G$, contradicting minimality of $E$. So $E$ must be a $p$-group. Also, $E^p$ is characteristic in $E$, hence normal in $G$, and since $E$ is a $p$-group, it is a proper subgroup of $E$. Again, minimality of $E$ tells us that $E^p$ is trivial, so $E$ is of exponent $p$. Since $E$ is abelian of exponent $p$, it is elementary abelian.