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How to find all elements of S4 that satisfy the equation x^4=e? 
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#20
Nov2311, 04:46 PM

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P: 6,187

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



#21
Nov2311, 04:49 PM

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P: 906

this is how i count ncycles in Sn (this works for any n, but i will use n = 4):
since any 4cycle 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 (n1)(n2)....(2) = (n1)! possible distinct ncycles. 


#22
Nov2311, 04:51 PM

P: 320

For 3cycles, 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 3cycles 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
Nov2311, 04:53 PM

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P: 906

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


#25
Nov2311, 04:54 PM

P: 320




#26
Nov2311, 05:00 PM

Sci Advisor
P: 906

so, to find the number of kcycles in Sn, you take n choose k times (k1)!, which is how many kcycles you can form in Sk.



#27
Nov2311, 05:01 PM

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


#28
Nov2311, 05:15 PM

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P: 6,187

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


#29
Nov2311, 05:22 PM

P: 320

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


#30
Nov2311, 05:28 PM

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#31
Nov2311, 05:29 PM

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P: 6,187

In S4, x^4 is either identity or a 3cycle (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. 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
Nov2311, 05:34 PM

P: 320




#33
Nov2311, 05:38 PM

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P: 6,187

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


#34
Nov2311, 05:43 PM

P: 320

(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
Nov2311, 05:57 PM

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P: 6,187

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 kcycles in Sn. When you have that, you may want to revisit this problem. 


#36
Nov2311, 06:30 PM

P: 320

The number of kcycles in S_{n} is (n,k)*(k1)! as someone else mentioned. the reason is that first we have to choose k letters out of n letters for forming kcycles, then we fix the first element and permute the others and that can be done in (k1)! ways. so the answer will be (n,k)*(k1)! where (n,k) is n choose k. 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 + (n1)!) solutions. so, x^{5}=e we'll have 25 solutions in S_{5}. 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)*(k1)!). the number of solutions of x^{4}=e in S_{5} is 1 + 5*24 = 121. 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)}. 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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