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 Sp
An exercise from Herstein asks to prove that the number of elements of order p, p a ' in Sp, is (p1)!+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 (p1)!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
Learn more

Solve your problem for the price of one coffee

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

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 Sp.
You can go from the formula in your question to Wilson's theorem by counting the number of p-Sylow subgroups (each contains p1 elements of order p), and then appealing to Sylow's theorem. (You will find that there are (p2)! p-Sylow subgroups, and by Sylow's theorem this number is congruent to 1modp. Multiplying by p1, we find that (p1)! is congruent to 1modp.)
Not exactly what you’re looking for?
Ask My Question
Aliana Porter
Answered 2022-04-26 Author has 6 answers
Every element of order p in Sp is a p-cycle. The symmetric group Sp1 acts transitively on these p cycles.
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

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

Relevant Questions

asked 2020-11-20
Prove that in any group, an element and its inverse have the same order.
asked 2022-01-13

When does the produt of two polynomials =xk?
Suppose f and g are are two polynomials with complex coefficents (i.e f, gC[x]). Let m be the order of f and let n be the order of g.
Are there some general conditions where
fg=αxn+m
for some non-zero αC

asked 2021-01-19
In group theory (abstract algebra), is there a special name given either to the group, or the elements themselves, if x2=e for all x?
asked 2022-01-14
Finding the Units in the Ring Z[t][t21]
asked 2022-04-22
How does cancellation work in polynomial quotient rings?
Z[x](2x6,6x15)
Can I just automatically say that
Z[x](2x6,6x15)=Z[x](x3,6x15)
by just dividng the first polynomial by 2?
asked 2022-04-01

How to solve a cyclic quintic in radicals?
Galois theory tells us that
z111z1=z10+z9+z8+z7+z6+z5+z4+z3+z2+z+1 can be solved in radicals because its group is solvable. Actually performing the calculation is beyond me, though - here what I have got so far:
Let the roots be ζ1,ζ2,,ζ10, following Gauss we can split the problem into solving quintics and quadratics by looking at subgroups of the roots. Since 2 is a generator of the group [2,4,8,5,10,9,7,3,6,1] we can partition into the five subgroups of conjugate pairs [2,9],[4,7],[8,3],[5,6],[10,1].
A0=x1+x2+x3+x4+x5A1=x1+ζx2+ζ2x3+ζ3x4+ζ4x5A2=x1+ζ2x2+ζ4x3+ζx4+ζ3x5A3=x1+ζ3x2+ζx3+ζ4x4+ζ2x5A4=x1+ζ4x2+ζ3x3+ζ2x4+ζx5
Once one has A0,,A4 one easily gets x1,,x5. It's easy to find A0. The point is that τ takes Aj to ζjAj and so takes Aj5 to Aj5. Thus Aj5 can be written down in terms of rationals (if that's your starting field) and powers of ζ. Alas, here is where the algebra becomes difficult. The coefficients of powers of ζ in A15 are complicated. They can be expressed in terms of a root of a "resolvent polynomial" which will have a rational root as the equation is cyclic. Once one has done this, you have A1 as a fifth root of a certain explicit complex number. Then one can express the other Aj in terms of A1. The details are not very pleasant, but Dummit skilfully navigates through the complexities, and produces formulas which are not as complicated as they might be. Alas, I don't have the time nor the energy to provide more details.

asked 2022-04-25
Show that gcd(a,b)=|a|ab?

Solve your problem for the price of one coffee

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