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 $F_{p(p-1)}$?

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.


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.)


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