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

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 $$(0,1,2,3,4)^{(1,2,4,3)} = (0,2,4,1,3)= (0,1,2,3,4)^2\in P$$.
(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.)