Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
1  
The shaded line of ams.org/mathscinet-getitem?mr=2491730 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
add comment

1 Answer 1

$G$ 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 $x\in H$.

share|improve this answer
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.