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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.

Let $N\ne 1$ be a minimal normal subgroup. Its commutator subgroup $N^{\prime }$ is characteristic in $N$, hence normal in $G$. Minimality makes $N^{\prime }=1$ or $N^{\prime }=N$. The latter would prevent the soluble group $N$ from having a terminating derived series, so $N^{\prime }=1$ and $N$ is abelian.

Choose a prime $p$ dividing $|N|$. In a finite abelian group its Sylow $p$-subgroup is characteristic, so minimal normality makes this subgroup all of $N$. The subgroup ${\Omega }_1(N)=\{n\in N:n^p=1\}$ is nontrivial, characteristic and hence normal in $G$. Minimality again makes it all of $N$. Thusan elementary abelian p-group. Both abelianness and minimal normality are essential to the two characteristic subgroup arguments.