Fitting subgroup

Remarks

Fitting's theorem says that if $M$ and $N$ are nilpotent normal subgroups of $G$, then $MN$ is a nilpotent normal subgroup. Hence $F(G)$ is itself nilpotent and normal, and it contains every nilpotent normal subgroup of $G$: it is the unique largest one. $F(G)$ is a characteristic subgroup, and it is the product over the primes $p$ of the largest normal $p$-subgroups $O_p(G)$. If $G$ is solvable, then the centraliser of $F(G)$ in $G$ is contained in $F(G)$.

Example: $G = S_4$. Its normal subgroups are $\set{e}$, $V = \set{e, (1\,2)(3\,4), (1\,3)(2\,4), (1\,4)(2\,3)}$, $A_4$ and $S_4$. The group $V$ is abelian, hence nilpotent, and it is the Klein four-group. The groups $A_4$ and $S_4$ are not nilpotent, because their centres are trivial. So $F(S_4) = V$.

Non-example: $A_4$ is a normal subgroup of $S_4$ that is not nilpotent, so it is not contained in the generating family, and it is not $F(S_4)$.

If $G$ is a nonabelian simple group, then $F(G) = \set{e}$. If $G$ is nilpotent, then $F(G) = G$.