Hall's Theorem of Solvable Groups (6.1.33)

Dummit and Foote Abstract Algebra, section 6.1, exercise 33:

Let $\pi$ be any set of primes. A subgroup $H$ of $G$ is called a Hall $\pi$-subgroup if the only primes dividing $|H|$ are in $\pi$ and $(|H|,|G:H|)=1$. When $\pi=\{p\}$, these groups are known as Sylow $p$-subgroups. Prove the following generalization of Sylow's Theorem for solvable groups: If $G$ is a finite solvable group, then for any set of primes $\pi$, the set of Hall $\pi$-subgroups is nonempty and all of its elements are conjugate to each other in $G$. [Let $M$ be a minimal normal subgroup of $G$, so $M$ is a $p$-group. If $p∈\pi$, then proceed by induction on $G/M$. If $p \not ∈\pi$, then reduce to the case where $|G|=p^an$, where $n$ is the order of a Hall $\pi$-subgroup. In this case let $N/M$ be a minimal normal $q$-subgroup of $G/M$, and let $Q∈Syl_q(N)$. If $Q \trianglelefteq G$, then proceed with $Q$ in place of $M$. If not, then use Frattini's argument to show $N_G(N)$ is a Hall $\pi$-subgroup and establish conjugacy as well.]

Proof: Let the overbar denote passage into $G/M$. If $p∈\pi$, then by induction there is a Hall $\pi$-subgroup of $\overline{G}$ of order $n/p^a$, so that its preimage in $G$ is of order $n$, and is thus a Hall $\pi$-subgroup. Furthermore, if any Hall $\pi$-subgroup $H$ does not contain $M$, then $H∩M≠M$, so that $p$ divides $| M : H∩M |$, so that by the Second Isomorphism Theorem $p$ divides $| HM : H |$, which implies $p$ divides $|G:HM| \cdot |HM : H|=|G:H|$, a contradiction. Now, for any two Hall $\pi$-subgroups $H_1$ and $H_2$, we have $\overline{H_1}$ and $\overline{H_2}$ are Hall $\pi$-subgroups of $\overline{G}$, and by induction they are conjugate, so that $\overline{g}\overline{H_1}\overline{g}^{-1}=\overline{gH_1g^{-1}}=\overline{H_2}$. Since $|gH_1g^{-1}|=n$ we have $gH_1g^{-1}$ is a Hall $\pi$-subgroup, and now both these groups contain $M$ so that $gH_1g^{-1}=H_2$.

Therefore assume $p \not ∈ \pi$. Taking the Hall $\pi$-subgroup $\overline{H}$ of $\overline{G}$ and taking its preimage $HM$ of order $p^an$, we can be assured that if $HM < G$, then by induction we can recognize Hall $\pi$-subgroups of $HM$ for the existence condition. For any two Hall $\pi$-subgroups $H_1$ and $H_2$ of $G$, since $H_1∩M=H_2∩M=1$, we have Hall $\pi$-subgroups $\overline{H_1}$ and $\overline{H_2}$ of $\overline{G}$, so that $\overline{g}\overline{H_1}\overline{g}^{-1}=\overline{H_2}$ by induction, and now $gH_1g^{-1}M=gH_1Mg^{-1}=H_2M$, so that $gH_1g^{-1}≤H_2M$. This reveals $H_1$ is conjugate to some Hall $\pi$-subgroup of $H_2M$, and by induction this one is conjugate to $H_2$, so that finally $H_1$ is conjugate to $H_2$.

So assume $HM=G$ and $|G|=p^an$. If $G$ is a $p$-group the proposition is evident, so we can take a minimal normal subgroup $\overline{N}$ (of order $q^b$ with $q≠p$) of $\overline{G}$, and observe $Q∈Syl_q(N)$. If $Q \trianglelefteq G$, then we can argue with $Q$ in place of $M$ as above. Therefore assume $N_G(Q) < G$. Since $\overline{N} \trianglelefteq \overline{G} ⇒ N \trianglelefteq G$, we can apply Frattini's Argument to $N$. Notice: $$|G|=p^an=\dfrac{|N| \cdot |N_G(Q)|}{|N∩N_G(Q)|}=\dfrac{p^aq^b \cdot |N_G(Q)|}{|N∩N_G(Q)|}$$ Since $Q≤N$ and $Q≤N_G(Q)$, we have $q^b$ divides $|N∩N_G(Q)|$. Therefore, letting $n=q^{\alpha_0}p_1^{\alpha_1}...p_r^{\alpha_r}$, we observe that $q^{\alpha_0}$ and $p_i^{\alpha_i}$ divides $|N_G(Q)|$ for all $i$, ensuring that $n$ divides $|N_G(Q)|$ and now by induction there exist Hall $\pi$-subgroups of $G$. Establish conjugacy: Take an arbitrary Hall $\pi$-subgroup $T$ of $G$. Since $MT=G$ and $M≤N$, we have:$$\dfrac{|N| \cdot |T|}{|N∩T|}=|G|⇒\dfrac{p^aq^b \cdot n}{|N∩T|}=p^an⇒|N∩T|=q^b=|Q|$$So that $N∩T$ is a Sylow $q$-subgroup of $N$ and is thus conjugate to $Q$; let $g(N∩T)g^{-1}=Q$. Since clearly $T≤N_G(N∩T)$, we have $gTg^{-1}≤gN_G(N∩T)g^{-1}=N_G(g(N∩T)g^{-1})=N_G(Q)$, so that every Hall $\pi$-subgroup of $G$ is conjugate to some Hall $\pi$-subgroup of $N_G(Q)$, whose Hall $\pi$-subgroups are inductively assumed to be conjugate to each other.$~\square$