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Thus, in A8, there is a permutation such that ord(p)=15. | Thus, in A8, there is a permutation such that ord(p)=15. | ||
--[[User:Bcaulkin|Bcaulkin]] 14:08, 28 January 2009 (UTC) | --[[User:Bcaulkin|Bcaulkin]] 14:08, 28 January 2009 (UTC) | ||
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| + | ----- | ||
| + | Can someone help me out with [[Cycle Notation]]? | ||
Latest revision as of 16:00, 28 January 2009
I found what an element of order 15 in A_8 looks like using Theorem 5.3 on page 101 of the textbook (sixth edition). Is there a way to prove such an element's existence without just guessing until you find an example?
On Jan 22 we talked about permutations (example was a pentagram). We saw that the order of a permutation is the smallest non-zero number of iterations such that the result is the identity permutation. (order = number of times the permutation is repeated before producing the identity)
If we take a set of 3 {1, 2, 3} and permute them such that 1->2, 2->3, and 3->1, then it would take 3 iterations before the 'output' is the original ordering. Similarly with a set of 5, rotating by one each time. We also know that any integer multiple of the order will also produce the identity (for a 3-rotation, 6, 9, 12 etc. all produce the identity)
So, if A8 is a set of 8 numbers, {1, 2, 3..., 8} and the permutation of these 8 numbers has a 3-rotation and a 5-rotation, the order of the set cannot be 3 (the 5-rotation has not gone enough) nor can it be 5 (the 3-needs one more iteration). Thus, for this 'composition' of a 3-rotation and a 5-rotation, the order must be the LCM of the individual orders. LCM(3, 5) is 15.
Thus, in A8, there is a permutation such that ord(p)=15. --Bcaulkin 14:08, 28 January 2009 (UTC)
Can someone help me out with Cycle Notation?
