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Deveno

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- Thread starter AdrianZ
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- #26

Deveno

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I'll try to find how many solutions the equation x

One more question, Is it always possible to solve an equation like ax

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I like Serena

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Looks like a good plan!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 x^{5}=e can have in S_{5}. then I'll try to guess how many solutions the equation x^{n}=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.

No, it's not always possible.One more question, Is it always possible to solve an equation like ax^{n}=b in S_{n}? When it's possible?

It depends on the order of the permutations involved, and it also depends on whether the permutations are even or odd.

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Would you explain more please?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 S

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Because there are three elements we missed: (1 2)(3 4) is one of them. Can you find the others?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 S_{4}. but if we add 1+6+8+6 it'd be equal to 21, not 24. How so?

- #31

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One more question, Is it always possible to solve an equation like ax^{n}=b in S_{n}? When it's possible?

Explained with orders:Would you explain more please?

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

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

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

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

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.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?

- #33

I like Serena

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Didn't you just proof that?Why in S4, x^4 is either identity or a 3-cycle?

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.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.

- #34

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Then let me re-read your post, maybe I misunderstood it.Didn't you just proof that?

You found 8 3-cycles, and the other 16 permutations obey x^4=id.

In the general case you're right, it'll be 4 choose 2. but here it won't differ.Not quite. You can choose the first cycle in 6 different ways.

(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

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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.

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Actually I wanted to set up a generalized formula for the number of solutions of the equation x

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.

The number of k-cycles in S

this gives us the ability to predict the solutions of x

so, x

we can also predict the number of solutions of x

the number of solutions of x

the number of solutions of x

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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- #37

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Right.this gives us the ability to predict the solutions of x^{n}=e when n is prime. since the only divisors of n are 1 and itself, we'll have (1 + (n-1)!) solutions.

so, x^{5}=e we'll have 25 solutions in S_{5}.

These are only the 4-cycles.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 x^{4}=e in S_{5}is 1 + 5*24 = 121.

What about solutions to x

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

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