Let $G$ be a finite group. $H \leq G$ is a component of $G$ if $H$ is quasisimple and subnormal in $G$. $E(G)$ is the group generated by all the components of $G$.
What can be said about the structure of a group $G$ if $E(G)=1$? (i.e $G$ has no components) I know that finite solvable groups and simple groups have no components, but are those the only cases?
Thank you in advance for the help.
No, there are non-solvable groups with $E(G)=1$.
There is an example of order $8 \times 168$, which is easily described. This is a group denoted by ${\rm AGL}(3,2)$, which is a semidirect product $G=N \rtimes H$ with $N$ elementary abelian of order $8$, $H = GL(3,2)$, with the action of $H$ on $N$ is just the action on its natural module. Then $F(G) = F^*(G) = 1$.
In fact, there are smaller examples of order $960$ with the same basic structure $N \rtimes H$, but with $N$ elementary abelian of order $15$ and $H \cong A_5$. There are two irreducible modules for $H$ of dimension 4 over ${\mathbb F}_2$, and you can use either of those to define the action of $H$ on $N$.
One thing you can say about groups with $E(G)=1$ is that $F(G) = F^*(G)$, and hence $C_G(F(G)) \le F(G)$.