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Thread: normal p-subgroups of a finite group

  1. #1
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    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.
    Last edited by moont14263; Aug 17th 2012 at 10:41 AM.
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  2. #2
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    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?
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  3. #3
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    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.
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