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<br>I think I can contribute to this brilliant explanation of the homework solution. | <br>I think I can contribute to this brilliant explanation of the homework solution. | ||
In the first part of explanation, about 5 legnth cycles. For each cycle there are 4 other repetitions. For example, (1 2 3 4 5) is the same as (2 3 4 5 1).. and there are other 3 of this kind. So if we have 5! permutations in all, for each we have 5 equal cycles, leaving it as 5!/5 = 4!. | In the first part of explanation, about 5 legnth cycles. For each cycle there are 4 other repetitions. For example, (1 2 3 4 5) is the same as (2 3 4 5 1).. and there are other 3 of this kind. So if we have 5! permutations in all, for each we have 5 equal cycles, leaving it as 5!/5 = 4!. | ||
| − | The same goes for the second part. For each of 3 length cycle there are other 2 repetitions. So if there are 5*4*3 permutations of length 3, then we should divide by 3 since there are 3 equal cycles, so the result will be 20. | + | <br>The same goes for the second part. For each of 3 length cycle there are other 2 repetitions. So if there are 5*4*3 permutations of length 3, then we should divide by 3 since there are 3 equal cycles, so the result will be 20. |
| − | --[[User:Guteulin|Guteulin]] | + | <br>--[[User:Guteulin|Guteulin]] |
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| + | [[Category:MA453Spring2009Walther]] | ||
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| + | Thanks a lot for the explanation on this problem. I was confused for a long time on this problem, but this makes much more sense now. | ||
| + | --[[User:Nswitzer|Nswitzer]] 16:36, 29 January 2009 (UTC) | ||
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| + | [[Category:MA453Spring2009Walther]] | ||
Latest revision as of 11:36, 29 January 2009
Looking at Example 4 by Theorem 5.4 (on page 103 in 7th Ed) we see that we can show all possible n-cycle combinations of Sn. (Example uses S7)
If we do this for A5 we get: --(5) --(3)(1)(1) --(2)(2)(1) with orders 5, 3, and 2 respectively.
For the (5) set, recall combinatorics. The number of possible permutations of 5 objects is 5!. However, for this to be a 5-cycle, no item may map to itself. Thus, each of the 5 positions in the permutation has only 4 possible places to map. Thus the number of ways to map a 5-cycle is 4! which is 24!
For (3)(1)(1), the first obstacle is to choose which 3 to permute. You have 10 choices. Similar to (5), each position has only 2 options. We know that the number of possible k-cycle mappings is (k-1)! So, we have 10 * (3-1)! = 20!!
Lastly, for (2)(2)(1), we choose from 5, 2 to permute first. From the remaining 3, we choose 2 again to permute. Since these two sets of choices are independent, they are multiplicative. So, we have (5 nCr 2)*(3 nCr 2) = 30. However, each combination of 2-cycles is counted twice (since (1,2)(3,4)=(3,4)(1,2)). Thus, we divide by 2 to eliminate the duplicates and we get 15!
Thus we have shown that there exists: --24 ord=5, --20 ord=3, --15 ord=2
--Bcaulkin 04:43, 29 January 2009 (UTC)
I think I can contribute to this brilliant explanation of the homework solution.
In the first part of explanation, about 5 legnth cycles. For each cycle there are 4 other repetitions. For example, (1 2 3 4 5) is the same as (2 3 4 5 1).. and there are other 3 of this kind. So if we have 5! permutations in all, for each we have 5 equal cycles, leaving it as 5!/5 = 4!.
The same goes for the second part. For each of 3 length cycle there are other 2 repetitions. So if there are 5*4*3 permutations of length 3, then we should divide by 3 since there are 3 equal cycles, so the result will be 20.
--Guteulin
Thanks a lot for the explanation on this problem. I was confused for a long time on this problem, but this makes much more sense now. --Nswitzer 16:36, 29 January 2009 (UTC)
