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 the diagram in the proof of Theorem 2.20 on page 63 of the book. This is in continuation of my previous question. My apologies for another question.
Here is the set-up:
- $G$ is a non-trivial finite group and let $\mathbb F(G)$ be the Fitting subgroup of $G$.
- $A$ is a cyclic proper subgroup (with trivial core) of $G$ such that $|A|\ge |G:A|$ and $A\cap \mathbb F(G)>1$.
- $\mathbb F(G)>1$ and we can choose a minimal normal subgroup $E$ of $G$ with $E \subseteq Z(\mathbb F(G))$.
- $E$ is an elementary abelian $p$-group for some prime $p$.
- Further $A\cap \mathbb F(G)\trianglelefteq AE$ and $AE\subset G$ but $AE\ne G$.
The next part of the proof relies on this diagram given on page 63:
It is not explicitly mentioned in the book what is $M$. Later it is used to define $\bar M=\text{core}_{G/E}(\bar A)$ with $E\subseteq M$, $M\trianglelefteq G$ and $AM=AE$.
My question is:
What actually the subgroup $M$ is? In particular what is $\bar A$?
I was thinking maybe Theorem 2.18 (Zenkov) might be used which says that if $K$ is a finite group and $X$, $Y$ are abelian subgroups of $K$ then a minimal member $M$ of the set $\{ X\cap Y^g \mid g \in K\}$ satisfies $M\subseteq \mathbb F(K)$; here $M$ is a minimal member of this set in the sense that no member of the set is properly contained in $M$. But I am not sure if this minimal $M$ is somehow related to my original question.
Any help will be really appreciated.
With $\overline{G} = G/E$, the notation used in the book for a subgroup $H \leq G$ is that $\overline{H}$ is the image of $H$ in $G/E$. See page 22 and 23 there.
So to answer your question: $M$ is the unique subgroup $E \leq M \leq G$ such that $M/E = \overline{M}$.