# Proving |N(P)| for Prime P in S(p) - Dummit & Foote

• hermanni
In summary, the conversation discusses proving that if P is a subgroup of S(p) of order p, then the normalizer of P in S(p), denoted as N(P), has an order of p(p-1). This can be argued by showing that every conjugate of P contains exactly p-1 p-cycles and using the formula for the number of p-cycles to compute the index of N(P) in S(p). The conversation also considers the possible elements of P and how to determine the elements of its conjugacy class. It is determined that P must be cyclic and generated by an element of order p, which is a p-cycle. The conversation concludes that all conjugates of P must also be p-cycles.
hermanni
I'm reading my textbook (Dummit & Foote) and having trouble at conjugacy section , here's the question:

Prove that if p is a prime and P is a subgroup of S(p) of order p, then
| N (P) | = p(p-1). (Argue that every conjugate of P contains exactly p-1 p-cycles and use the formula for the number of p-cycles to compute the index of
N(P) in S(p) .
N(P) : normalizer of P in S(p).

I reaaly have no clue , can someone help??

Suppose, as you say, that $$P$$ is a subgroup of $$\mathfrak{S}_p$$ of order $$p$$. What can the elements of $$P$$ be, in concrete terms? There aren't that many choices.

Also, given a permutation $$\sigma \in \mathfrak{S}_p$$, there is a general way to write down the elements of its conjugacy class $$\{ \pi\sigma\pi^{-1} \mid \pi \in \mathfrak{S}_p \}$$. (It helps to write out $$\sigma$$ in cycle decomposition first.)

I think that P must be cyclic (since it's of prime order ) so it's generated by an element of order p , which must be an p-cycle.Am I right?

Right. So if $$\kappa$$ is a $$p$$-cycle, and $$P = \langle\kappa\rangle$$ is the subgroup generated by $$\kappa$$, what can the conjugates $$\pi P \pi^{-1}$$ look like?

Well, conjugates have the same cycle structure.Conjugate of P has also p elements , but I still can't see why they must be all p-cycles.

## 1. How do you prove |N(P)| for prime P in S(p)?

To prove |N(P)| for prime P in S(p), we can use the Sylow theorems, which state that for any finite group G and any prime factor p of its order, there exists a subgroup of G whose order is a power of p. By applying these theorems to the Sylow p-subgroups of S(p), we can prove that |N(P)| = p^(n-1), where n is the number of prime factors of p in the prime factorization of |S(p)|.

## 2. Why is it important to prove |N(P)| for prime P in S(p)?

Proving |N(P)| for prime P in S(p) is important because it helps us understand the structure and properties of S(p), a fundamental group in abstract algebra. It also has important applications in various fields, such as number theory, cryptography, and physics.

## 3. What are the Sylow p-subgroups of S(p)?

The Sylow p-subgroups of S(p) are the subgroups of S(p) whose orders are powers of p, where p is a prime factor of |S(p)|. These subgroups are important in proving |N(P)| for prime P in S(p) and can be found using the Sylow theorems.

## 4. How does proving |N(P)| for prime P in S(p) relate to other concepts in abstract algebra?

Proving |N(P)| for prime P in S(p) is related to other important concepts in abstract algebra, such as group actions, conjugacy classes, and normal subgroups. By understanding the structure and properties of S(p), we can also gain insights into other groups and algebraic structures.

## 5. Are there any real-world applications of proving |N(P)| for prime P in S(p)?

Yes, there are many real-world applications of proving |N(P)| for prime P in S(p). One example is in cryptography, where the structure and properties of S(p) are used to generate and analyze secure encryption algorithms. It also has applications in physics, specifically in the study of symmetry and group theory.

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