I am asking this question because I am trying to solve this problem from a class note:
Assume that $R(G)$ is simple and not commutative, show that $G$ is a subgroup of $Aut(R(G)).$
The note's definition of $R(G)$ is not easy to digest, it consists of chains upon chains of sub-definitions like these:
(1) $R(G) = E(G)F(G)$;
(2) Where $F(G)$ is defined to be the (complex) product of all subgroups $O_p(G)$ with $p$ a prime number, this group $F(G)$ is called the Fitting subgroup of $G$;
(3) And $E(G)$ is defined to be the subgroup of $G$ generated by all components of $G$, this group $E(G)$ is called the Layer of $G$. And then another chain of definitions: A subnormal subgroup of $G$ is called a component of $G$ if it is quasisimple; the group $G$ is called quasisimple if $G’ = G$ and $G/Z(G)$ is simple.
Is it possible to take a break from formalism for just a little while and let me understand $R(G)$ in the most mundane and intuitive way? Thank you for your time and help.
PS. This chapter on Radical of Group is given before the next one on Solvable Group, so I assume the explanation does not involve solvability.