MissMathsxo
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Could someone please tell me how you calculate how many of the elements of the symmetric group (1,2,3,4,5,6) have order 2 and order 3. I understand why we use 6C2 to find the number of order 2 permutations of the form (m,n) but after that I'm not sure of the method. Thank you
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Cryptokyo
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(Original post by MissMathsxo)
Could someone please tell me how you calculate how many of the elements of the symmetric group (1,2,3,4,5,6) have order 2 and order 3. I understand why we use 6C2 to find the number of order 2 permutations of the form (m,n) but after that I'm not sure of the method. Thank you
The problem is a delicate one.

If an element of S6 has order 2 then it cannot have any cycles of size greater than 2 and it must contain at least one two cycle. An element of order 2 in S6 can thus have 1,2 or 3 disjoint 2-cycles in it. The number of elements with 1 disjoint 2-cycle is \binom{6}{2}\binom{4}{4}=15 - we choose the 2-cycle and let the rest be 1-cycles. The number of elements with 2 disjoint 2-cycles is \frac{1}{2}\binom{6}{2}\binom{4}  {2}\binom{2}{2}=45 - we choose the first 2-cycle then the second 2-cycle then leave the rest as 1-cycles, we divide by two as the order of the cycles doesn't matter. The number of elements with 3 disjoint 2-cycles is \frac{1}{3!}\binom{6}{2}\binom{4  }{2}\binom{2}{2}=15 - we choose the first 2-cycle, then the second, then the third but divide by 3! as the order of the cycles does not matter. Thus, there are 15+45+15=75 elements of order 2.

If an element of S6 has order 3 then it cannot have any cycles greater than 3 nor can it contain a 2-cycle (why?!) but it must have at least one 3-cycle. Thus, it must have either 1 or 2 3-cycles. Can you see how to do this using a similar argument as above?
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Vinny C
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****... try the impossible cycle. Is a wheel with pedals but people master it.
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Cryptokyo
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(Original post by MissMathsxo)
Could someone please tell me how you calculate how many of the elements of the symmetric group (1,2,3,4,5,6) have order 2 and order 3. I understand why we use 6C2 to find the number of order 2 permutations of the form (m,n) but after that I'm not sure of the method. Thank you
Scrap the last thing I said. Although the answer is correct the methodology was poor as one cannot freely apply combinations for the case of 3-cycles.

First of all let me clarify what I mean by a k-cycle. A k-cycle is a set of k-elements that permute in a given order and return to the starting value after k steps. E.g. 1\rightarrow 5 \rightarrow 4 \rightarrow 1\rightarrow 5 \rightarrow\cdots is a 3-cycle denoted (1 5 4).

Let us find a formula that given k distinct elements, how many k-cycles can we make out of them. For k=1, it is trivially 1. We can arrange the k objects in k! ways. However, shifting each element along 1 space in the arrangement and moving the end element to the front makes an equivalent k cycle i.e 1\rightarrow 5 \rightarrow 4 \rightarrow 1\rightarrow 5 \rightarrow\cdots is equivalent to 5\rightarrow 4 \rightarrow 1 \rightarrow 5\rightarrow 4\rightarrow\cdots as the mappings are equivalent. We could do this k times before getting back to the same arrangement. Thus, we have that there are \frac{k!}{k}=(k-1)! possible k-cycles from k-elements.

As I said previously a element of S_{6} that has order 3 must have at least one 3-cycle and no cycle of length greater than 3 and no 2-cycles. So it either has one 3-cycle and three 1-cycles or two 3-cycles.

For the case of the single 3-cycle, we can choose the elements in the 3-cycle in \binom{6}{3} ways. There are 2! possible 3-cycles that can be made from this. Thus, there are \binom{6}{3}\times 2!=40 such elements. (Note we ignore the 1-cycles as they can only be chosen in one way).

Using a similar logic can you do the case for two 3-cycles?
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