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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. –  vwxf Mar 25 '11 at 7:02
    
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 $|G| = |H|\cdot |G^{(n-1)}|$. –  user641 Mar 25 '11 at 8:04
    
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. –  vwxf Mar 25 '11 at 8:55
    
Yeah, I don't see why Abelian intersect maximal implies normal. Any suggestions? –  vwxf Mar 25 '11 at 12:01
    
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? –  vwxf Mar 25 '11 at 23:38

2 Answers 2

It is false in general that "abelian intersect maximal implies normal": $A_5$ is maximal in $S_5$, the subgroup generated by $(1,2,3,4)$ is abelian, but the intersection of the two is nontrivial (contains $(1,3)(2,4)$) and not normal in $S_5$.

However, it is true that the intersection of a maximal subgroup and an abelian normal subgroup is normal.

Proposition. Let $G$ be a group, and $H$ a maximal subgroup of $G$. If $N$ is an abelian normal subgroup of $G$, then $N\cap H$ is normal in $G$.

Proof. If $N\subseteq H$, then $N\cap H = N\triangleleft G$ and we are done.

If $N$ is not contained in $H$, then maximality of $H$ and normality of $N$ imply that $HN=G$ (since $HN$ is a subgroup). Let $x\in H\cap N$ and $g\in G$. Then we can write $g = hn$ with $h\in H$ and $n\in N$. Then $$gxg^{-1} = (hn)x(hn)^{-1} = h(nxn^{-1})h^{-1} = hxh^{-1}$$ with the last equality since $N$ is abelian. Now, $x,h\in H$, so $hxh^{-1}\in H$. And $x\in N$, so $hxh^{-1}\in N$. Thus, $hxh^{-1}\in H\cap N$, proving that $H\cap N$ is normal in $G$. QED

Added. More generally: note that $H\cap N$ is certainly normal in $H$. If $HN=G$, then you only need to show that $N$ normalizes $H\cap N$: then $(hn)x(hn)^{-1} = h(nxn^{-1})h^{-1}$, which will lie in $H\cap N$ if $nxn^{-1}\in H\cap N$. Therefore:

Proposition. Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. Then $H\cap N\triangleleft HN$ if and only if $N\subseteq N_{HN}(N\cap H)$, where $N_{NH}(N\cap H)$ is the normalizer of $N\cap H$ in $NH$.

Proof. If $N\subseteq N_{NH}(N\cap H)$, then the argument proceeds as above. Conversely, if $N\cap H\triangleleft NH$, then $N\subseteq NH=N_{NH}(N\cap H)$. $\Box$

In particular, if $N$ is abelian then $N\subseteq C_{NH}(N\cap H)\subseteq N_{NH}(N\cap H)$; and if $H$ is maximal, then this gives the proposition above in the nontrivial case.

Now apply this to $H$ and the abelian normal subgroup $G^{(n-1)}$ to conclude that $H\cap G^{(n-1)}\triangleleft G$.

Now, you know that $G^{(n-1)}$ is normal in $G$ (because the derived terms are always normal in $G$), and has order $p$. Since a $p$-Sylow subgroup of $G$ has order $p$, then $G^{(n-1)}$ is a $p$-Sylow subgroup of $G$. Since it is normal in $G$, the $p$-Sylow subgroup of $G$ is normal in $G$. So $G$ is contained in the normalizer of the $p$-Sylow subgroup $G^{(n-1)}$ of $S_p$.

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Thank you very much for your help! This step was in fact like a fibre bundle lol (locally trivial). I guess if I worked on it longer without thinking too hard about it, I guess I might be able to stumble across this solution. –  vwxf Mar 25 '11 at 23:10

Arturo's solution follows the hint and is correct. But I did not find the suggestion to show that $N \cap H = 1$ particularly helpful.

You could reason alternatively as follows. Use the same argument as Arturo to show that $NH = G$. Since $|NH| = |N||H|/|N \cap H|$, and $p$ does not divide $|H|$, it follows that $p$ divides $|N|$. Since $N$ is abelian, it has a unique Sylow $p$-subgroup of order $p$, which must be normal in $G$.

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Thanks for this additional speed-up. This would circumvent that step indeed. It does require the extra theorem about Abelian groups though, but I guess that is fair. Once again, I learn something new everyday. :) –  vwxf Mar 25 '11 at 23:12
    
@vwxf: It's not really a theorem, but an observation: any two Sylow subgroups are conjugate, but in an abelian group, all subgroups are normal; if you had $S_1$ and $S_2$ both Sylow $p$-subgroups, there is a $g$ with $S_2 = gS_1g^{-1}$, but $gS_1g^{-1}=S_1$. –  Arturo Magidin Mar 26 '11 at 21:46

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