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Every finite soluble group has a Hall subgroup for each prime set. Induct on the group order using an elementary abelian minimal normal subgroup. In the coprime hard case, lift a minimal normal subgroup of the quotient, choose its Sylow subgroup, and apply the Frattini argument. A proper normalizer reduces the order; a normal Sylow subgroup allows induction in its quotient. This proves existence without assuming an independent complement theorem.

We prove Hall subgroup existence in soluble groups by induction on $|G|$. The trivial group is immediate. Choose a nontrivial minimal normal subgroup $N$, elementary abelian of order $p^a$ by part (c). Induction gives a Hall $\pi$-subgroup of $G/N$; let $K$ be its full preimage.

If $p\in \pi$, then $K$ itself is the required subgroup. If $p\notin \pi$ and $K<G$, apply induction inside the soluble subgroup $K$ to obtain a Hall $\pi$-subgroup $H$ of $K$. Since $[G:K]$ and $[K:H]$ are both ${\pi }^{\prime }$-numbers, $H$ is Hall in $G$ too.

It remains to treat $p\notin \pi$ and $K=G$. Then $G/N$ is a $\pi$-group. If $G=N$, the subgroup one works. Otherwise choose a minimal normal subgroup $M/N$ of $G/N$, an elementary abelian $q$-group with $q\in \pi$. Let $Q$ be a Sylow $q$-subgroup of $M$. As $|M|=|N|q^b$ and $p\ne q$, we have $M=NQ$. The permitted Frattini argument givesThe last equality uses $Q\le N_G(Q)$ and the normality of $N$.

If $T=N_G(Q)<G$, then $[G:T]=[N:N\cap T]$ is a power of $p$, hence a ${\pi }^{\prime }$-number. Apply induction to $T$ and multiply indices as before. If $T=G$, then $Q$ is a nontrivial normal $\pi$-subgroup. Induction in $G/Q$ gives a Hall $\pi$-subgroup whose full preimage in $G$ is Hall, since its additional factor $|Q|$ is a $\pi$-number. These cases exhaust the possibilities, proving existence for every prime set. No unproved complement theorem or conjugacy theorem for Hall subgroups was inserted into the proof.

If $|G|=p^{a}m$ with $p\nmid m$, a Sylow subgroup for $p$ is a subgroup of order $p^a$. The Sylow theorems assert: such subgroups exist; every $p$-subgroup is contained in a Sylow subgroup; all Sylow $p$-subgroups are conjugate; and their number $n_p$ divides $m$ and satisfies $n_p\equiv 1(\mathrm{mod}p)$.

For groups of order p squared q are not simple, if $p>q$, then $n_p\mid q$ and $n_p\equiv 1(\mathrm{mod}p)$ force $n_p=1$. Its unique Sylow subgroup is a nontrivial proper normal subgroup. Suppose instead $p<q$. If either Sylow count is one, the conclusion already follows. Otherwise $n_p=q$, while $n_q$ is either $p$ or $p^2$. The value $p$ is impossible because $0<p-1<q$. Thus $n_q=p^2$, and $q\mid (p^2-1)=(p-1)(p+1)$. Primality and $q>p$ imply $q\mid p+1$, hence $q=p+1$. The only consecutive primes are $p=2,q=3$.

In that exceptional order-$12$ case, four distinct Sylow $3$-subgroups contribute eight distinct nonidentity elements: their intersections are trivial. Only three nonidentity elements remain. Every subgroup of order four must contain exactly those remaining three, so there can be only one Sylow $2$-subgroup, contradicting the assumption that both counts were nontrivial. Therefore no group of order $p^{2}q$ with distinct primes is simple.

For the final factorization, normality of $H$ ensures $g^{-1}Pg$ is a Sylow $p$-subgroup of $H$. By Sylow conjugacy within $H$, choose $h\in H$ such that $g^{-1}Pg=h^{-1}Ph$. Then $k=g{h}^{-1}$ satisfiesThus $k\in N_G(P)$ and $g=kh$. We have proved the Frattini argumentThe normaliser factorization includes the case $P=\{1\}$, when its normalizer is all of $G$.