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

[AdrianZ]

well, I'm thinking to answer the question in a generalized way. I want to find all the elements of Sn that satisfy the equation xn=e.
well, if m|n and xm=e then xn=e, hence, if we raise x (which is of order m and m divides n) to the exponent n we'll have xn=e. so the answer will be this:
S={all cycles of the length m s.t. m|n}

Is that a true conclusion?

As an example, in S4, since divisors of 4 are 1,2 and 4 the solutions will be all cycles with lengths 1,2 and 4. namely, {e,(1 2),(1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3 4), (4 3 2 1), (1 3)(2 4), (1 2)(3 4), (1 4)(2 3)}
Is there anything I'm missing?

[micromass]

well, I'm thinking to answer the question in a generalized way. I want to find all the elements of Sn that satisfy the equation xn=e.
well, if m|n and xm=e then xn=e, hence, if we raise x (which is of order m and m divides n) to the exponent n we'll have xn=e. so the answer will be this:
S={all cycles of the length m s.t. m|n}

Is that a true conclusion?

As an example, in S4, since divisors of 4 are 1,2 and 4 the solutions will be all cycles with lengths 1,2 and 4. namely, {e,(1 2),(1 3), (1 4), (2 3), (2 4), (3 4), (1 2 3 4), (4 3 2 1), (1 3)(2 4), (1 2)(3 4), (1 4)(2 3)}
Is there anything I'm missing?

First of all something pedantic: I wouldn't call (1 2)(3 4) a cycle. It is rather the product of cycles. So I would say that your set would rather be the cycles generated by the cycles of 1, 2 and 4.

I think the conclusion holds true. To prove it, you should use that if x_i are disjoint cycles of length k_i, then the order of x_1...x_l is lcm(k_1,...,x_l).

[Deveno]

First of all something pedantic: I wouldn't call (1 2)(3 4) a cycle. It is rather the product of cycles. So I would say that your set would rather be the cycles generated by the cycles of 1, 2 and 4.



even this statement is not true. for example, (1 3)(1 2) = (1 2 3), so we don't want the set generated by cycles of length 1,2 and 4, but rather the set of products of disjoint cycles generated by cycles of length 1,2 and 4.

we can disregard cycles of length 1, as these are all the identity map.

[AdrianZ]

Yea. so, the correct statement would be: S={all cycles of Sn generated by disjoint cycles of length m such that m divides n} would be the solutions of the equation xn=e.
if we stop being pedantic, have I found the solutions of that equation correctly or there are more?

[I like Serena]

There are more.
Did you enumerate and check all of them?

[AdrianZ]

There are more.
Did you enumerate and check all of them?

well, maybe I should include (3 4 1 2) as well?
I didn't enumerate them, but tried to find them mentally. I considered all transpositions, because they are of order 2. They can be found easily, then I considered all cycles of the length 4 with their inverses (there was only one such cycle), and then I added all cycles that were products of disjoint cycles of order 2. I think I missed (3 4 1 2), it's of order 2 and it's its own inverse too.
S={e,(1 2),(1 3),(1 4), (2 3), (2 4), (3 4), (1 2 3 4), (4 3 2 1), (3 4 1 2), (1 2)(3 4),(1 3)(2 4), (1 4)(2 3)}
well, Is there any other solutions to the equation or I'm done?

[I like Serena]

Did you know that (1 2 3 4)=(3 4 1 2)?
How many cycles of length 4 did you find?

Do you know how many elements S4 has?
How many could you eliminate?
Do you have all the elements that you could not eliminate?

[AdrianZ]

Did you know that (1 2 3 4)=(3 4 1 2)?
Oops, Yea, I'm sorry. I thought (3 4 1 2) meant (1 3)(2 4). my bad.

How many cycles of length 4 did you find?
2.

Do you know how many elements S4 has? 4! = 24.

How many could you eliminate?
Do you have all the elements that you could not eliminate?
all elements of order 3 can be eliminated. right?
well, I have 12 elements. but I guess there must be more. I'm not sure. Could you tell me one element that is not included in my list?

[I like Serena]

You have 12 elements of order 3?
How did you count them?

And 2 elements of order 4?
There should be more.

[AdrianZ]

You have 12 elements of order 3?
How did you count them?
No, I have so far counted 12 solutions. I didn't say I have 12 elements of order 3. If I had 12 elements of order 3 then I was done. lol

And 2 elements of order 4?
There should be more.
That's exactly where I'm stuck.
Is there any theorem that tells us how many elements of a specific order we have in a group?

[I like Serena]

Suppose you were to create a 4-cycle from scratch.
How many choices do you have for the 1st number?
And the following numbers?

[AdrianZ]

Suppose you were to create a 4-cycle from scratch.
How many choices do you have for the 1st number?
And the following numbers?

well, I'm not so sure but I will just tell you my guess based on intuition.
I'll have 4 choices for the first letter, but just after I chose the first number, the arrange of the 3 next numbers would be known, I can either shift all the remaining numbers to right or I can shift them to left depending on the first number I've chosen. Is that right? so if that's right, we might have 8 cycles of order 4 at maximum.

[I like Serena]

Let's not make assumptions about the other 3 numbers yet.
Let's start with counting how many sequences of 4 different numbers you can make.

On another angle, consider (1 2 3 4).
Which other 4-cycles are identical to this one?

[AdrianZ]

Let's not make assumptions about the other 3 numbers yet.
Let's start with counting how many sequences of 4 different numbers you can make.
What kind of sequences you mean? just to understand the terminology.

On another angle, consider (1 2 3 4).
Which other 4-cycles are identical to this one?
(1 2 3 4),(2 3 4 1), (3 4 1 2), (4 1 2 3) are identical to this one.

[I like Serena]

What kind of sequences you mean? just to understand the terminology.

I mean sequences like:
1 2 3 4
1 2 4 3
...

How many are there?


(1 2 3 4),(2 3 4 1), (3 4 1 2), (4 1 2 3) are identical to this one.

Good!
So a given 4-cycle comes in 4 variants.

[AdrianZ]

I mean sequences like:
1 2 3 4
1 2 4 3
...

How many are there?
If you're putting no additional conditions, there are 4!=24 permutations on 4 letters.

Good!
So a given 4-cycle comes in 4 variants.
True. well?

[I like Serena]

If you're putting no additional conditions, there are 4!=24 permutations on 4 letters.


True. well?

Yes. Both true!

Since there are 24 sequences of 4 letters and since each 4-cycle comes in 4 flavors.
How many different 4-cycles do you think there are?

Could you enumerate them now?

[AdrianZ]

Yes. Both true!

Since there are 24 sequences of 4 letters and since each 4-cycle comes in 4 flavors.
How many different 4-cycles do you think there are?

Could you enumerate them now?

(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! :smile:

[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]

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?

[Deveno]

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)

[I like Serena]

Right!

[AdrianZ]

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.

[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 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?

[I like Serena]

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!


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.

[AdrianZ]

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.

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]

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?

[I like Serena]

One more question, Is it always possible to solve an equation like axn=b in Sn? 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 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.

[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 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.

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.