Let $G$ be a finite solvable group. Let $K/H$ be a chief factor of $G$ that is not of prime order, where $K$ is a $p$-subgroup of $G$ for some prime $p$ divides the order of $G$. Let $S$ be a proper normal subgroup of $K$ with $H and $|S/H|=p$. If $H$ contain every element of order $p$ of $S$ and $K= \langle S^{g},g \in G \rangle$, then $H$ contains every element of order $p$ of $K$.
I need to prove the above statement.
Here is what I know.
Since $G$ is solvable then $K/H$ is abelian $p$-group of exponent $p$. $\bigcap_{g \in G}S^{g}$ is a normal subgroup of $G$ that contains $H$. So $H= \bigcap_{g \in G}S^{g}$.
Thanks in advance.
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