# Structure of a finite group $G$, given layer $E(G)=1$

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 $$C_G(F(G)) \le F(G)$$.