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Let $G$ be a finite supersolvable group. Then a minimal normal subgroup of $G$ is has prime order. Is it true in general that $G$ has a minimal normal subgroup for each prime divisor of $G$?

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No. For example, the supersolvable group $S_3$ has only one minimal normal subgroup, of order 3. A similar situation occurs in the nonabelian group of order $pq$, where $p$ and $q$ are any primes with $p | q-1$.

All you can guarantee is that $G$ has a normal subgroup of order $p$, where $p$ is the largest prime dividing $|G|$. Reason: $G$ has a normal Sylow $p$-subgroup (the bottom of a Sylow tower), and this subgroup necessarily contains a minimal normal subgroup.

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