(New page: 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) <pre> If we do this for A5 we get: --(5)...) |
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--20 ord=3, | --20 ord=3, | ||
--15 ord=2 | --15 ord=2 | ||
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</pre> | </pre> | ||
| + | --[[User:Bcaulkin|Bcaulkin]] 04:43, 29 January 2009 (UTC) | ||
Revision as of 23:43, 28 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)
