normal p-subgroups of a finite group

**moont14263** · August 17th 2012, 10:15 AM

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

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

**Deveno** · August 17th 2012, 12:30 PM

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

**moont14263** · August 17th 2012, 03:21 PM

I fix the mistake in the question with some additions.
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 $H <S$ and $|S/H|=p$ . If $H$ contain every element of order $p$ of $S$ and $K=<S^{g},g \in G>$ , 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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