Difference between revisions of "Chapter 5 43" - Rhea

Revision as of 10:22, 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