# prove unique minimal normal subgroup of soluble group by $P_{G}(M) > M$

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.

• 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 $G/A_1 \times G/A_2$. – Derek Holt Aug 31 '15 at 16:04
• @DerekHolt How prove $A$ is unique minimal normal subgroup? – Soroush Aug 31 '15 at 16:08

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 $G/A_1 \times G/A_2$.