An exercise from Herstein asks to prove that the number of elements of order p, p a ' in

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 $(p-1)!+1$ . 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 $(p-1)!\equiv -1$ (mod p) using this result

An exercise from Herstein asks to prove that the number of elements of order p, p a ' in

You can still ask an expert for help

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 $(p-2)!$ p-Sylow subgroups, and by Sylow's theorem this number is congruent to $1\text{mod}p$ . Multiplying by $p-1$ , we find that $(p-1)!$ is congruent to $-1\text{mod}p$ .)

Maybe you mean the number of elements of order dividing p (so that you are including the identity)? (Think about the case

You can go from the formula in your question to Wilson's theorem by counting the number of p-Sylow subgroups (each contains

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.

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Can I just automatically say that

by just dividng the first polynomial by 2?

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Let the roots be $\zeta}^{1},{\zeta}^{2},\dots ,{\zeta}^{10$, 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].

$\begin{array}{rl}{A}_{0}& ={x}_{1}+{x}_{2}+{x}_{3}+{x}_{4}+{x}_{5}\\ {A}_{1}& ={x}_{1}+\zeta {x}_{2}+{\zeta}^{2}{x}_{3}+{\zeta}^{3}{x}_{4}+{\zeta}^{4}{x}_{5}\\ {A}_{2}& ={x}_{1}+{\zeta}^{2}{x}_{2}+{\zeta}^{4}{x}_{3}+\zeta {x}_{4}+{\zeta}^{3}{x}_{5}\\ {A}_{3}& ={x}_{1}+{\zeta}^{3}{x}_{2}+\zeta {x}_{3}+{\zeta}^{4}{x}_{4}+{\zeta}^{2}{x}_{5}\\ {A}_{4}& ={x}_{1}+{\zeta}^{4}{x}_{2}+{\zeta}^{3}{x}_{3}+{\zeta}^{2}{x}_{4}+\zeta {x}_{5}\end{array}$

Once one has $A}_{0},\dots ,{A}_{4$ one easily gets $x}_{1},\dots ,{x}_{5$. It's easy to find $A}_{0$. The point is that $\tau$ takes $A}_{j$ to $\zeta}^{-j}{A}_{j$ and so takes $A}_{j}^{5$ to $A}_{j}^{5$. Thus $A}_{j}^{5$ can be written down in terms of rationals (if that's your starting field) and powers of $\zeta$. Alas, here is where the algebra becomes difficult. The coefficients of powers of $\zeta$ in $A}_{1}^{5$ 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 $A}_{1$ as a fifth root of a certain explicit complex number. Then one can express the other $A}_{j$ in terms of $A}_{1$. 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.

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