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:
- $H$ 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 $p$.
- Everything up to and excluding the part where I am asked to prove that $H \cap G^{(n-1)} = 1$.
- 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!
- 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:
- 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.
- 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.