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

Tags: elements, equation, satisfy
 HW Helper P: 6,189 Yep!
 HW Helper P: 6,189 So do you think you can deduce the number of 3-cycles now?
 Sci Advisor P: 906 this is how i count n-cycles in Sn (this works for any n, but i will use n = 4): since any 4-cycle in S4 involves 1, we may as well start with it: (1..... i have 3 choices for my next element (the image of 1): (1 2..... (1 3..... (1 4..... after i make my next choice, i'll have them all: (1 2 3 4) (1 2 4 3) (1 3 2 4) (1 2 4 2) (1 4 3 2) (1 4 2 3) that makes 6. in general, in Sn, we'll get (n-1)(n-2)....(2) = (n-1)! possible distinct n-cycles.
P: 320
 Quote by I like Serena So do you think you can deduce the number of 3-cycles now?
Yes. Actually I was thinking about generalizing this idea to other cases as well.
For 3-cycles, each permutation will come in 3 different variants, and we'll have 24 permutations that we can consider, so the answer should be 8 and the desired 3-cycles are as follows:
(1 2 3),(2 3 1),(3 1 2)
(1 3 2),(3 2 1),(2 1 3)
(1 2 4),(2 4 1),(4 1 2)
(1 4 2),(4 2 1),(2 1 4)
(2 3 4),(3 4 2),(4 2 3)
(2 4 3),(4 3 2),(3 2 4)
(3 4 1),(4 1 3),(1 3 4)
(3 1 4),(1 4 3),(4 3 1)

Right?
P: 906
 Quote by AdrianZ Yes. Actually I was thinking about generalizing this idea to other cases as well. For 3-cycles, each permutation will come in 3 different variants, and we'll have 24 permutations that we can consider, so the answer should be 8 and the desired 3-cycles are as follows: (1 2 3),(2 3 1),(3 1 2) (1 3 2),(3 2 1),(2 1 3) (1 2 4),(2 4 1),(4 1 2) (1 4 2),(4 2 1),(2 1 4) (2 3 4),(3 4 2),(4 2 3) (2 4 3),(4 3 2),(3 2 4) (3 4 1),(4 1 3),(1 3 4) (3 1 4),(1 4 3),(4 3 1) Right?
i prefer to think of how many ways we can choose 3 objects out of 4:

4!/(3!1!) = 4, and for each set {a,b,c}, we can form two different 3-cycles:

(a b c) and (a c b)
 HW Helper P: 6,189 Right!
P: 320
 Quote by Deveno i prefer to think of how many ways we can choose 3 objects out of 4: 4!/(3!1!) = 4, and for each set {a,b,c}, we can form two different 3-cycles: (a b c) and (a c b)
Yea, That's exactly what I did.
 Sci Advisor P: 906 so, to find the number of k-cycles in Sn, you take n choose k times (k-1)!, which is how many k-cycles you can form in Sk.
 P: 320 Hey man, I thought group theory sucks, but now I think group theory rocks. everything looks so beautifully consistent. I believe I should solve more problems in group theory rather than just dealing with the concepts abstractly. I'll try to find how many solutions the equation x5=e can have in S5. then I'll try to guess how many solutions the equation xn=e can have in Sn and will write down my thoughts here. it looks to be a good food for thought. Thank you guys for your helps, especially I like Serena. One more question, Is it always possible to solve an equation like axn=b in Sn? When it's possible?
HW Helper
P: 6,189
 Quote by AdrianZ Hey man, I thought group theory sucks, but now I think group theory rocks. everything looks so beautifully consistent. I believe I should solve more problems in group theory rather than just dealing with the concepts abstractly. I'll try to find how many solutions the equation x5=e can have in S5. then I'll try to guess how many solutions the equation xn=e can have in Sn and will write down my thoughts here. it looks to be a good food for thought. Thank you guys for your helps, especially I like Serena.
Looks like a good plan!

 Quote by AdrianZ One more question, Is it always possible to solve an equation like axn=b in Sn? When it's possible?
No, it's not always possible.
It depends on the order of the permutations involved, and it also depends on whether the permutations are even or odd.
P: 320
 Quote by I like Serena No, it's not always possible. It depends on the order of the permutations involved, and it also depends on whether the permutations are even or odd.

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?
Mentor
P: 18,346
 Quote by AdrianZ Would you explain more please? 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?
Because there are three elements we missed: (1 2)(3 4) is one of them. Can you find the others?
HW Helper
P: 6,189
 Quote by AdrianZ One more question, Is it always possible to solve an equation like axn=b in Sn? When it's possible?
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.

 Quote by AdrianZ 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.

As a challenge (when you find them), how should you count how many there are?
P: 320
 Quote by micromass 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).

 Quote by I like Serena 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.
HW Helper
P: 6,189
 Quote by AdrianZ 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.

 Quote by AdrianZ 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.
P: 320
 Quote by I like Serena Didn't you just proof that? You found 8 3-cycles, and the other 16 permutations obey x^4=id.

 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.
 HW Helper P: 6,189 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.
P: 320
 Quote by 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.
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 Sn 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 xk=e in Sn 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.

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