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This question is from Dummit and Foote's Abstract Algebra, page 638, question 20. It gives a nice paragraph of hints that basically guides one through the problem, but I'm very stuck at a crucial junction. Any useful hint is much appreciated. I have detailed what I know and what I do not know, but if you just want the tl;dr, just read the question, which is the following sentence.

"Let $p$ be a prime. Show that any solvable subgroup of $S_p$ of order divisible by $p$ is contained in the normalizer of a Sylow $p$-subgroup of $S_p$. [...]

Hint: Let $G \leq S_p$ be a solvable subgroup of order divisible by $p$. Then $G$ contains a $p$-cycle, hence is transitive on $\{1, \ldots, p\}$. Let $H < G$ be the stabilizer in $G$ of the element $1$, so $H$ has index $p$ in $G$. Show that $H$ contains no nontrivial normal subgroups of $G$ (note the conjugates of $H$ are the stabilizers of the other points). Let $G^{(n-1)}$ be the last nontrivial subgroup in the derived series for $G$. Show that $H \cap G^{(n-1)} = 1$ and conclude that $\lvert G^{(n-1)}\rvert = p$, so that the Sylow $p$-subgroup of $G$ (which is also a Sylow $p$-subgroup of $S_p$) is normal in $G$."

Here are the things I do know:

  1. $H$ has an order that divides $(p-1)!$ since it has index $p$ in $G$, and $G$ has order $pu$ for some $u$ not divisible by $p$.
  2. Everything up to and excluding the part where I am asked to prove that $H \cap G^{(n-1)} = 1$.
  3. I know how to prove the next part where I'm asked to prove that $|G^{(n-1)}| = p$ provided I know how to do that previous part!
  4. I know that $\lvert S_p \rvert = p!$, so any Sylow $p$-subgroup of $S_p$ has size $p^1 = p$, since no other factors of $p!$ can contain $p$ as a prime factor.

Now here are the things I do not know:

  1. I am terribly stuck at the step where I have to show $H \cap G^{(n-1)} = 1$. I tried showing that this is normal, so I can use the result immediately preceding to conclude that it is trivial. But I'm having major problems. I may just be missing something extremely obvious.
  2. Even if I can do that part, the next part asks us to conclude that this Sylow $p$-subgroup is normal in $G$, which I can't immediately see how to derive. I'm assuming ``this Sylow $p$-subgroup'' is referring to the size $p$ subgroup $G^{(n-1)}$---it has the right size to be a Sylow $p$-subgroup.
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    Just to define some terms in case the definition you are used to is different from mine. A finite group $G$ is solvable if the derived series $G^{(0)} := G, G^{(k)} := [G^{(k-1)}, G^{(k-1)}]$ for $k \geq 1$ eventually becomes the trivial subgroup $\{e\}$. Here, $[G^{(k)}, G^{(k)}]$ is the subgroup generated by all the "commutators" of the form $g^{-1}h^{-1}gh$, where $g, h \in G^{(k)}$. This condition is equivalent to saying that the finite group $G$ has a composition series whose factors are Abelian. The derived series is not to be confused with the lower central series.2011-03-25
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    Note that $G^{(n-1)}$ is an abelian group, and $H$ is maximal; Thus $H\cap G^{(n-1)}$ is a normal subgroup of $G$. From what came before, this is the trivial subgroup; It then follows that $G=HG^{(n-1)}$ and so $|G| = |H|\cdot |G^{(n-1)}|$.2011-03-25
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    Thanks for your reply! I didn't realize that $G^{(n-1)}$ was Abelian until you mentioned it. But I can't see why an intersection of a maximal subgroup and an Abelian one means it is normal.2011-03-25
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    Yeah, I don't see why Abelian intersect maximal implies normal. Any suggestions?2011-03-25
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    Thanks for everyone's help! In case this is of any use to anyone else in the future, I will summarize various hints on how to tackle (1) and (2) in "things I do not know": (1) Try looking ahead. In order to prove $|G^{(n-1)}| = p$, what new subgroup are you going to construct? This construction, along with the maximality of $H$, may give you a hint as to how to prove $H \cap G^{(n-1)}$ is normal in $G$. (2) Is $G^{(n-1)}$ normal in $G$? What is its order (size)? What _should_ the size of a Sylow $p$-subgroup be?2011-03-25

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