= Solution
Let $N\ne1$ be a <minimal normal subgroup>. Its <commutator subgroup> $N'$ is characteristic in $N$, hence normal in $G$. Minimality makes $N'=1$ or $N'=N$. The latter would prevent the soluble group $N$ from having a terminating <derived series>, so $N'=1$ 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 $N$. Thus
$$
\boxed{N\cong C_p^d\text{ for some }d\geq1,}
$$
an <elementary abelian p-group>. Both abelianness and minimal normality are essential to the two <characteristic subgroup> arguments.
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