Number of Elements of order p in S_{p} An exercise from Herstein asks to prove that the nu

Kymani Shepherd 2022-04-24 Answered
Number of Elements of order p in ${S}_{p}$
An exercise from Herstein asks to prove that the number of elements of order p, p a ' in ${S}_{p}$, is $\left(p-1\right)!+1$. I would like somebody to help me out on this, and also I would like to know whether we can prove Wilson's theorem which says $\left(p-1\right)!\equiv -1$ (mod p) using this result
You can still ask an expert for help

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it

Answers (2)

Olive Guzman
Answered 2022-04-25 Author has 16 answers
Step 1
Maybe you mean the number of elements of order dividing p (so that you are including the identity)? (Think about the case $p=3$ - there are two three cycles, not three of them.) For the general question, think about the possible cycle structure of an element of order p in ${S}_{p}$.
You can go from the formula in your question to Wilson's theorem by counting the number of p-Sylow subgroups (each contains $p-1$ elements of order p), and then appealing to Sylow's theorem. (You will find that there are $\left(p-2\right)!$ p-Sylow subgroups, and by Sylow's theorem this number is congruent to $1\text{mod}p$. Multiplying by $p-1$, we find that $\left(p-1\right)!$ is congruent to $-1\text{mod}p$.)
Not exactly what you’re looking for?
Aliana Porter
Answered 2022-04-26 Author has 6 answers
Every element of order p in ${S}_{p}$ is a p-cycle. The symmetric group ${S}_{p-1}$ acts transitively on these p cycles.
Not exactly what you’re looking for?

Expert Community at Your Service

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Available 24/7
• Math expert for every subject
• Pay only if we can solve it