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Let $ G $ be a soluble group. If $ P_{G}(M) = \langle y\in G | \langle y \rangle M = M\langle y \rangle \rangle > M $for any subgroup $ M $ of prime power index in $ G $, then every chief factor of $ G $ has order $ 4 $ or a prime.

For proof let $ G $ be a counterexample of minimal order and choose a minimal normal subgroup $ A $ of $ G $. Hence $ A $ is elementary abelian group of prime power order. since $ G/A $ is a soluble group and $ \vert G/A \vert < \vert G \vert $, so each chief factor of $ G/A $ has order $ 4 $ or a prime. Now i need to prove $ A $ is the unique minimal normal subgroup of $ G $. Suppose $ A_{1} $ and $ A_{2} $ are minimal normal subgroups of $ G $ that $ A_{1} \neq A_{2} $. So $ A_{1} \cap A_{2} = 1 $ and each chief factor of $ G/A_{1} $ and $ G/A_{2} $ have order $ 4 $ or prime. Since $ G/A_{1} \times G/A_{2} = G/A_{1} \cap A_{2} \cong G $, now can say every chief factor of $ G $ has order $ 4 $ or a prime? If it is true, then it is contradiction.

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  • $\begingroup$ It is not true that $G/A_1 \times G/A_2 = G/A_1 \cap A_2$. It is true that $G \cong G/A_1 \cap A_2$ is isomorphic to a subgroup of $G/A_1 \times G/A_2$. $\endgroup$ – Derek Holt Aug 31 '15 at 16:04
  • $\begingroup$ @DerekHolt How prove $ A $ is unique minimal normal subgroup? $\endgroup$ – Soroush Aug 31 '15 at 16:08
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In general, if $H \le G$, and $K/L$ is a chief factor of $G$, then $(H \cap K)/(H \cap L) \cong (H \cap K)L/L$ is a subgroup of $K/L$.

So, if every chief factor of $G$ has order $4$ or prime, then the same is true of any subgroup $H$ of $G$.

So, in your situation, every chief factor of $G/A_1$ and of $G/A_2$ has order $4$ or prime, so the same is true of $G/A_1 \times G/A_2$, and hence the same is true of $G \cong G/A_1 \cap A_2$, which is isomorphic to a subgroup of $G/A_1 \times G/A_2$.

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