How to find all elements of S4 that satisfy the equation x^4=e?

[I like Serena]

One more question, Is it always possible to solve an equation like axⁿ=b in S_n? When it's possible?

Would you explain more please?

Explained with orders:

In S4, x^4 is either identity or a 3-cycle (with order 3).
If a and b differ in order, but not by 3, there is no solution.Explained with even and odd permutions:
(Do you know what even and odd permutations are?)

In S4, x^4 is always an even permutation.
If a is odd and b is even, then there is no solution.

We found out that there are 1 one-cycle, 6 different 2-cycles, 8 different 3-cycles and 6 different 4-cycles in S₄. but if we add 1+6+8+6 it'd be equal to 21, not 24. How so?

Yes, you're missing 3 of them.
Did you already have them in your original solution?

As a challenge (when you find them), how should you count how many there are?

 

[AdrianZ]

Because there are three elements we missed: (1 2)(3 4) is one of them. Can you find the others?

Yup. (1 2)(3 4), (1 3)(2 4), (1 4)(2 3).

Explained with orders:

In S4, x^4 is either identity or a 3-cycle (with order 3).
If a and b differ in order, but not by 3, there is no solution.


Explained with even and odd permutions:

In S4, x^n is always an even permutation.
If a is odd and b is even, then there is no solution.

Why in S4, x^4 is either identity or a 3-cycle?

Yes, you're missing 3 of them.
Did you already have them in your original solution?

As a challenge (when you find them), how should you count how many there are?

well, in this case it's easy. I want to have 2 disjoint cycles, each cycle is of order 2, once I choose the first 2-cycle, the second 2-cycle will be automatically determined. I can choose the first cycle in 3 different ways, so I'll miss 3 solutions of the equation.

 

[I like Serena]

Why in S4, x^4 is either identity or a 3-cycle?

Didn't you just proof that?
You found 8 3-cycles, and the other 16 permutations obey x^4=id.

well, in this case it's easy. I want to have 2 disjoint cycles, each cycle is of order 2, once I choose the first 2-cycle, the second 2-cycle will be automatically determined. I can choose the first cycle in 3 different ways, so I'll miss 3 solutions of the equation.

Not quite. You can choose the first cycle in 6 different ways.

 

[AdrianZ]

Didn't you just proof that?
You found 8 3-cycles, and the other 16 permutations obey x^4=id.

Then let me re-read your post, maybe I misunderstood it.

Not quite. You can choose the first cycle in 6 different ways.

In the general case you're right, it'll be 4 choose 2. but here it won't differ.
(1 3)(2 4), (1 4)(2 3), (1 2)(3 4)
(2 3)(1 4), (2 4)(1 3)
(3 4)(1 2)

the last 2 rows are not new permutations. that's why I counted it that way for this particular case.

 

[I like Serena]

Okay, you counted it right, but then, you already knew it should be 3.

Anyway, I believe you were going to set up a generalized formula for k-cycles in Sn.
When you have that, you may want to revisit this problem.

 

[AdrianZ]

Okay, you counted it right, but then, you already knew it should be 3.

Anyway, I believe you were going to set up a generalized formula for k-cycles in Sn.
When you have that, you may want to revisit this problem.

Actually I wanted to set up a generalized formula for the number of solutions of the equation xⁿ=e, but now I see that it can be a little bit more tricky when n is not a prime number, because then I'll have to count the number of the generated cycles as products of disjoint cycles. that would make it harder.

The number of k-cycles in S_n is (n,k)*(k-1)! as someone else mentioned. the reason is that first we have to choose k letters out of n letters for forming k-cycles, then we fix the first element and permute the others and that can be done in (k-1)! ways. so the answer will be (n,k)*(k-1)! where (n,k) is n choose k.

this gives us the ability to predict the solutions of xⁿ=e when n is prime. since the only divisors of n are 1 and itself, we'll have (1 + (n-1)!) solutions.
so, x⁵=e we'll have 25 solutions in S₅.

we can also predict the number of solutions of x^k=e in S_n when n is prime. the answer will be (1 + (n,k)*(k-1)!).
the number of solutions of x⁴=e in S₅ is 1 + 5*24 = 121.
the number of solutions of x²=e in S₃ is 1 + 3*1 = 4. those solutions namely are: {e,(1 2),(1 3),(2 3)}.

the case where n is not prime is a bit tricky, but I'll think about it. first I'll need to prove some theorems, for example if p and q are two disjoint cycles, then o(pq)=o(p)o(q). it needs more considerations, I'll think about it later.

Thanks guys for the help, and have a nice thanksgiving holiday tomorrow.

 

[I like Serena]

this gives us the ability to predict the solutions of xⁿ=e when n is prime. since the only divisors of n are 1 and itself, we'll have (1 + (n-1)!) solutions.
so, x⁵=e we'll have 25 solutions in S₅.

Right.

we can also predict the number of solutions of x^k=e in S_n when n is prime. the answer will be (1 + (n,k)*(k-1)!).
the number of solutions of x⁴=e in S₅ is 1 + 5*24 = 121.

These are only the 4-cycles.
What about solutions to x²=e in S₅?

the number of solutions of x²=e in S₃ is 1 + 3*1 = 4. those solutions namely are: {e,(1 2),(1 3),(2 3)}.

Yes.