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

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 $S_p$.

  • $\begingroup$ 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. $\endgroup$ – 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 $G$.

  • $\begingroup$ 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. :) $\endgroup$ – vwxf Mar 25 '11 at 23:12
  • $\begingroup$ @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 $gS_1g^{-1}=S_1$. $\endgroup$ – Arturo Magidin Mar 26 '11 at 21:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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