Why the normalizer of the Sylow $p$-subgroups of the symmetric group of degree $p$ has order $p(p-1)$ and is known as Frobenius group $F_{p(p-1)}$?

Question

Why the normalizer of the Sylow $p$-subgroups of the symmetric group of degree $p$ has order $p(p-1)$ and is known as Frobenius group $F_{p(p-1)}$?

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I am trying to understand the statements on Wikipedia about Sylow subgroups of the symmetric group, where the above statement has been made.

Of course, the Sylow $p$-subgroup $C_p$ of $S_p$ is a normal subgroup of a group of order $p(p-1)$ by Sylow theorems. But how to show that it is the maximal subgroup normalizing $C_p$ in $S_p$?

Moreover, what is the relation between this normalizer and the Frobenius group $F_{p(p-1)}$?

Thank you.

Answer 1

The main question seems to be about the order of the normalizer $N:=N_{S_p}(P)$, where $P$ is a Sylow $p$-subgroup of $S_p$, for example $P=\langle(123\ldots p)\rangle\simeq C_p$.

I view this as a simple counting exercise.

1. There are $(p-1)!$ elements of order $p$ in $S_p$ (count the $p$-cycles "starting" from $1$).
2. Each Sylow $p$-subgroup of $S_p$ contains exactly $p-1$ such elements, so there are a total of $n_p:=(p-1)!/(p-1)=(p-2)!$ Sylow $p$-subgroups.
3. Let $X$ be the set of Sylow $p$-subgroups of $G=S_p$. $G$ acts on $X$ by conjugation with the stabilizer of $P$ being equal to $N$. By basic Sylow theory the action is transitive, so the orbit-stabilizer theorem tells us that $$ (p-2)!=n_p=|X|=[G:N]=\frac{p!}{|N|}. $$
4. Therefore $|N|=p(p-1)$.

The other answers seem to have addressed your other questions.

Answer 2

Sylow subgroups are conjugate. So pick your favorite example Sylow $p$-subgroup, say $P=\langle (0,\ldots,p-1)\rangle$. Then consider its normalizer. Well $P\cong \mathbb{Z}/p$ which is also a ring (this is why I choose to start counting at $0$. Notice acting by the generator is the same as adding 1 in the ring $\mathbb{Z}/p$. The additive automorphisms are multiplication by units in the ring, i.e the $(p-1)$ non-zero elements. These permute the points $0,\ldots,p-1$ in $\mathbb{Z}/p$ but those I have identified with the domain of the permutation. So each one of those can be recorded as a permutation. For example with $p=5$ then multiply by 2 would $(1,2,4,3)$ and if we conjugate $(0,1,2,3,4)^{(1,2,4,3)} = (0,2,4,1,3)= (0,1,2,3,4)^2\in P$.

Having exhibited a normalizer of the maximal order (and all such being conjugate) you get your result.

(What I'm really doing is observing $P$ is regular hence it is fair to identify the domain with the group elements, but the lables $0,\ldots,p-1$ seem intuitive without as much vocabulary.)

Answer 3

My approach is more combinatorial than algebraic. Any group isomorphic to $(\mathbb Z_p)^n$ can be written as a permutation group on $p^n$ points, with any Frobenius complement a group that fixes exactly one of these points, therefore, maximally, a group of order $p^n - 1$. It can be shown that this maximal group always exists. Of course, this is not the maximal normalizer of the $(\mathbb Z_p)^n$ group in the symmetric group on $n$ letters (though it is for $n = 1$); the other normalizers, however, fix more than one point and therefore do not serve as Frobenius complements.