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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 $K= \langle S^{g},g \in G \rangle$, 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.

I got my question from the shaded area here

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The conclusion is not true. SmallGroup(160,99), 5 |x SmallGroup(32,50), is a counterexample for p=2. K is the 2-core, K/H is rank 4, K=Omega(K) but K/H is not the union of only 5 1-dimensional subspaces. Your hypotheses have some redundancies. What was the original question? – Jack Schmidt Aug 18 '12 at 14:59
The shaded line of seems like a gap in the proof to me. This is a proof by contradiction, so unless the theorem itself is false, it is hard to guarantee there is a mistake. However, the group I mention satisfies every hypothesis mentioned on this page (so other than being a minimal counterexample to the theorem), but in this case $L \neq \bigcup_{g \in G} T^g$. – Jack Schmidt Aug 19 '12 at 5:29
@JackSchmidt You are right about the research. I thought that L is generated by those conjugate but it seems that L is the union of those conjugate. I tried to prove that L equals that union but I failed. – user28083 Aug 19 '12 at 6:57

$G$ acts transitively on $K/H$, and so if $x\in K$ is of order $p$, there is some $s\in S^g$ with $sx^{-1}$ in $H$, implying $x\in S^g$, so that we must have had $x\in H$.

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G acts irreducibly, but need not be transitive. (Central factors are easiest examples, but K/H here is not central; a more real example might be an index 2 subgroup of AGL(1,9), smallgroup 36,9). – Jack Schmidt Aug 17 '12 at 21:30
Yes, you're right, this doesn't work. Maybe irreducible is enough though, I have to think about it. – user641 Aug 17 '12 at 22:47
@JackSchmidt I post where I got the question from. – user28083 Aug 18 '12 at 22:05

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