normal p-subgroups of a finite group

Let $\displaystyle G$ be a finite group. Let $\displaystyle H$ and $\displaystyle K$ be normal $\displaystyle p$-subgroups of $\displaystyle G$ for some prime $\displaystyle p$ with $\displaystyle H$ being a proper subgroup of $\displaystyle K$ and $\displaystyle H/K$ is elementary abelian p-group. Let $\displaystyle S$ be a normal subgroup of $\displaystyle K$ with $\displaystyle H<S$. If $\displaystyle H$ contain every element of order $\displaystyle p$ of $\displaystyle S$ and$\displaystyle K=<S^{g},g \in G>$, then $\displaystyle H$ contain every element of order $\displaystyle p$ of $\displaystyle K$.

I need to prove the above statement. Thanks in advance.

Re: normal p-subgroups of a finite group

your question doesn't make sense, as written. if H < K, then what does H/K mean?

Re: normal p-subgroups of a finite group

I fix the mistake in the question with some additions.

Let $\displaystyle G$ be a finite solvable group. Let $\displaystyle K/H$ be a chief factor of $\displaystyle G$ that is not of prime order, where $\displaystyle K$ is a $\displaystyle p$-subgroup of $\displaystyle G$ for some prime $\displaystyle p$ divides the order of $\displaystyle G$. Let $\displaystyle S$ be a proper normal subgroup of $\displaystyle K$ with $\displaystyle H <S$ and $\displaystyle |S/H|=p$. If $\displaystyle H$ contain every element of order $\displaystyle p$ of $\displaystyle S$ and $\displaystyle K=<S^{g},g \in G>$, then $\displaystyle H $contains every element of order $\displaystyle p$ of $\displaystyle K$.

I need to prove the above statement.

Here is what I know.

Since $\displaystyle G$ is solvable then $\displaystyle K/H$ is abelian $\displaystyle p$-group of exponent $\displaystyle p$. $\displaystyle \bigcap_{g \in G}S^{g}$ is a normal subgroup of $\displaystyle G $that contains $\displaystyle H$. So H=$\displaystyle \bigcap_{g \in G}S^{g}$.

Thanks in advance.