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The largest nilpotent normal subgroup of a finite group $G$. If $O_p(G)$ denotes its p-core, distinct p-cores commute, and $F(G)={\prod }_p{O}_p(G)$. Every normal nilpotent group has Sylow subgroups which are characteristic subgroups, so it is contained in this product.

In a finite soluble group, $C_G(F(G))\le F(G)$. Put $C=C_G(F)$ and $D=C\cap F=Z(F)$. If $C>D$, choose an elementary abelian minimal normal subgroup of $G/D$ inside $C/D$, and let $K$ be its preimage. Then $K^{\prime }\le D\le Z(K)$, so $K$ is a nilpotent normal subgroup, contradicting $K\nleqq F$.

A normal subgroup which is a nilpotent group. In a finite group its unique Sylow subgroups are characteristic subgroups and hence normal in the ambient group. Products of nilpotent normal subgroups are nilpotent groups: their normal Sylow subgroups of different prime characteristics commute.

The Frattini subgroup is the intersection of the maximal subgroups of $G$, with $\Phi (1)=1$. A useful equivalent property isIndeed, a proper subgroup $H$ lies in a maximal subgroup, which also contains $\Phi (G)$. The Frattini subgroup is a characteristic subgroup, since automorphisms permute the maximal subgroups.

The Fitting subgroup is the largest nilpotent normal subgroup. To prove its existence, first form the p-core $O_p(G)$: the product of all normal p-subgroups. For two such subgroups, their product is normal and has order $|AB|=|A||B|/|A\cap B|$, a power of $p$. There are only finitely many subgroups, so $O_p(G)$ exists. For distinct primes $p,r$,Thus the product of the p-cores is a nilpotent normal subgroup, a direct product of groups of its Sylow subgroups. Conversely, a finite nilpotent normal subgroup $K$ has unique Sylow subgroups, each a characteristic subgroup of $K$. Each is normal in $G$ and lies in the corresponding p-core. Consequently the required largest subgroup exists, and

Put $\Phi =\Phi (G)$ and choose a Sylow subgroup $P$ of $\Phi$. Since $\Phi$ is normal, the Frattini argument gives $G=\Phi N_G(P)$: for $g\in G$, the Sylow subgroups $P^g$ and $P$ of $\Phi$ are conjugate by an element of $\Phi$, leaving an element of $N_G(P)$. The non-generator property of the Frattini subgroup proved above forces $N_G(P)=G$. Thus every Sylow subgroup of $\Phi$ is normal. Distinct such Sylow subgroups commute because their group commutators lie in their trivial intersection. Hence $\Phi$ is a nilpotent normal subgroup, and