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

AdrianZ

(1 2 3 4), (2 3 4 1), (3 4 1 2), (4 1 2 3)
(1 2 4 3), (2 4 3 1), (4 3 1 2), (3 1 2 4)
(1 3 2 4), (3 2 4 1), (2 4 1 3), (4 1 3 2)
(1 3 4 2), (3 4 2 1), (4 2 1 3), (2 1 3 4)
(1 4 3 2), (4 3 2 1), (3 2 1 4), (2 1 4 3)
(1 4 2 3), (4 2 3 1), (2 3 1 4), (3 1 4 2)

All the permutations in the same row are identical. so I think we'll have 6 different 4-cycles. Am I right?

I like Serena

Yep!

I like Serena

So do you think you can deduce the number of 3-cycles now?

Deveno

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.

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?

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)

I like Serena

Right!

AdrianZ

Yea, That's exactly what I did.

Deveno

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.

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 S_n 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 S_n? When it's possible?

I like Serena

Looks like a good plan!


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.

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?

micromass

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

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:
(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.



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

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

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


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

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.

AdrianZ

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

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.