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

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In summary: On another angle, consider (1 2 3 4).Which other 4-cycles are identical to this one?There are 3 more: (2 3 4 1), (3 4 1 2), and (4 1 2 3).
  • #1
AdrianZ
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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?
 
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  • #2
AdrianZ said:
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 [itex]x_i[/itex] are disjoint cycles of length [itex]k_i[/itex], then the order of [itex]x_1...x_l[/itex] is [itex]lcm(k_1,...,x_l)[/itex].
 
  • #3
micromass said:
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.
 
  • #4
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?
 
  • #5
There are more.
Did you enumerate and check all of them?
 
  • #6
I like Serena said:
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?
 
  • #7
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?
 
  • #8
I like Serena said:
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?
 
  • #9
You have 12 elements of order 3?
How did you count them?

And 2 elements of order 4?
There should be more.
 
  • #10
I like Serena said:
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?
 
  • #11
Suppose you were to create a 4-cycle from scratch.
How many choices do you have for the 1st number?
And the following numbers?
 
  • #12
I like Serena said:
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.
 
  • #13
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?
 
  • #14
I like Serena said:
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.
 
  • #15
AdrianZ said:
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?


AdrianZ said:
(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.
 
  • #16
I like Serena said:
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?
 
  • #17
AdrianZ said:
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?
 
  • #18
I like Serena said:
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?
 
  • #19
Yep! :smile:
 
  • #20
So do you think you can deduce the number of 3-cycles now?
 
  • #21
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.
 
  • #22
I like Serena said:
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?
 
  • #23
AdrianZ said:
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)
 
  • #24
Right!
 
  • #25
Deveno said:
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.
 
  • #26
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.
 
  • #27
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?
 
  • #28
AdrianZ said:
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!


AdrianZ said:
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.
 
  • #29
I like Serena said:
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?
 
  • #30
AdrianZ said:
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?
 
  • #31
AdrianZ said:
One more question, Is it always possible to solve an equation like axn=b in Sn? When it's possible?
AdrianZ said:
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.
AdrianZ said:
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?
 
  • #32
micromass said:
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).

I like Serena said:
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.
 
  • #33
AdrianZ said:
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.



AdrianZ said:
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.
 
  • #34
I like Serena said:
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.
 
  • #35
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.
 
<h2>1. How many elements are in the group S4?</h2><p>The group S4, also known as the symmetric group of degree 4, has 24 elements. This can be calculated by taking the factorial of 4, which is 4! = 4 x 3 x 2 x 1 = 24.</p><h2>2. What is the identity element in S4?</h2><p>The identity element in S4 is the element that has no effect on any other element when multiplied. In this case, the identity element is the permutation (1 2 3 4), which means that each element is mapped to itself.</p><h2>3. How do I find the elements of S4 that satisfy the equation x^4=e?</h2><p>To find the elements of S4 that satisfy the equation x^4=e, you can list out all the elements of the group and then raise each element to the 4th power. Any element that results in the identity element (e) is a solution to the equation.</p><h2>4. Are there any other ways to solve the equation x^4=e in S4?</h2><p>Yes, there are other ways to solve this equation. One method is to use the fact that the order of any element in S4 must divide the order of the group, which is 24. This means that the order of any element must be 1, 2, 3, 4, 6, 8, 12, or 24. By considering the possible orders of elements, you can narrow down the list of elements that satisfy the equation.</p><h2>5. Can I use a calculator to find the elements of S4 that satisfy the equation x^4=e?</h2><p>Yes, you can use a calculator to find the elements of S4 that satisfy the equation x^4=e. However, since the group S4 is relatively small, it may be easier to list out the elements and manually calculate their 4th powers. Additionally, using the calculator to check your answers can be helpful.</p>

1. How many elements are in the group S4?

The group S4, also known as the symmetric group of degree 4, has 24 elements. This can be calculated by taking the factorial of 4, which is 4! = 4 x 3 x 2 x 1 = 24.

2. What is the identity element in S4?

The identity element in S4 is the element that has no effect on any other element when multiplied. In this case, the identity element is the permutation (1 2 3 4), which means that each element is mapped to itself.

3. How do I find the elements of S4 that satisfy the equation x^4=e?

To find the elements of S4 that satisfy the equation x^4=e, you can list out all the elements of the group and then raise each element to the 4th power. Any element that results in the identity element (e) is a solution to the equation.

4. Are there any other ways to solve the equation x^4=e in S4?

Yes, there are other ways to solve this equation. One method is to use the fact that the order of any element in S4 must divide the order of the group, which is 24. This means that the order of any element must be 1, 2, 3, 4, 6, 8, 12, or 24. By considering the possible orders of elements, you can narrow down the list of elements that satisfy the equation.

5. Can I use a calculator to find the elements of S4 that satisfy the equation x^4=e?

Yes, you can use a calculator to find the elements of S4 that satisfy the equation x^4=e. However, since the group S4 is relatively small, it may be easier to list out the elements and manually calculate their 4th powers. Additionally, using the calculator to check your answers can be helpful.

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