Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 2 / 1

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

Write $F=F(G)$, $C=C_G(F)$ and $D=C\cap F=Z(F)$. Both $C$ and $D$ are normal in $G$. Suppose $C>D$. Choose a minimal normal subgroup $K/D$ of $G/D$ contained in $C/D$. In a finite soluble group, such a minimal normal subgroup is elementary abelian: the last nontrivial term of its derived series is an abelian characteristic subgroup, so minimality makes the whole subgroup abelian; its primary components and subgroup of elements killed by $p$ then force one prime and exponent $p$.

It follows that $K^{\prime }\le D$. Moreover $K\le C$ centralizes $D\le F$, so $K^{\prime }\le Z(K)$. Therefore $K$ is a nilpotent group of class at most two. Since $K$ is normal in $G$, the definition of the Fitting subgroup gives $K\le F$. But then $K\le C\cap F=D$, a contradiction. This proves the Fitting subgroup is self-centralizing in soluble groups:

It suffices to centralize each minimal normal subgroup $M$. The subgroup $M\cap F(G)$ is normal in $G$, so it is either $1$ or $M$. In the first caseIn the second case $M$ is a nilpotent group. Its nontrivial group center is a characteristic subgroup of $M$, hence normal in $G$, so minimality makes $M$ abelian. The primary components, which are characteristic subgroups, and the subgroup killed by a prime show that $M$ is elementary abelian of exponent $p$. Thus $M\le O_p(G)$.

The finite p-group $O_p(G)$ acts by conjugation on $M$. Its orbits have prime-power sizes; hence the number of fixed elements is congruent to $|M|=0$ modulo $p$. The identity is fixed, so there is a nonidentity fixed element. Therefore $M\cap Z(O_p(G))\ne 1$. This intersection is normal in $G$, and minimality makes it $M$. The other p-cores commute with $O_p(G)$, and hence with $M$. Thus every factor of $F(G)$ centralizes $M$. Since the group socle is generated by these minimal normal subgroups,

Put $\Phi =\Phi (G)$ and let $H$ be the full preimage of $F(G/\Phi )$ in $G$. The preceding part gives $\Phi \le F(G)$, so the normal nilpotent image $F(G)/\Phi$ lies in $F(G/\Phi )$.

For the reverse inclusion, choose a Sylow subgroup $P$ of $H$. Its image $P\Phi /\Phi$ is the unique Sylow subgroup of the nilpotent group $H/\Phi$, hence a characteristic subgroup there and normal in $G/\Phi$. Thus $P\Phi ⊴G$. The subgroup $P$ is also a Sylow subgroup of $P\Phi$: the p-part of $\Phi$ lies in $P$, because $P\cap \Phi$ is a Sylow subgroup of the normal subgroup $\Phi$ of $H$. Apply the Frattini argument to $P\Phi$:The Frattini subgroup property forces $N_G(P)=G$. Hence every Sylow subgroup of $H$ is normal, so $H$ is a nilpotent group and $H\le F(G)$. Together the inclusions proveThis is Frattini lifting of nilpotence; simply calling an extension of nilpotent groups nilpotent would not establish it.