Cycle Notation of Permutations - Abstract Algebra
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- Опубликовано: 14 окт 2018
- Cycle Notation gives you a way to compactly write down a permutation. Since the symmetric group is so important in the study of groups, learning cycle notation will speed up your work with the group Sn. In this lesson we show you how to convert a permutation into cycle notation, talk about the conventions, and discuss the key properties of cycles.
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We recommend the following textbooks:
Dummit & Foote, Abstract Algebra 3rd Edition
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Milne, Algebra Course Notes (available free online)
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The order of an n-cycle should be n. And consequently the order of a permutation is the LCM of the lengths of all its cycles.
What does order mean?
@@SameerKhan-nd5qb the number of elements in a group
@@SameerKhan-nd5qb In this case it's the smallest power of an element that yields the unit element of the group. So if you multiply the permutation (1 2 3) three times by itself you get the identity permutation e that does not permutate any elements, i.e. (1 2 3) (1 2 3) (1 2 3) = (1 2 3)^3 = (1) (2) (3) = e
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Conjectures
|( a b )| = 2 = 2 because there is 2 elements in the cycle, a maps to b and b maps to a
|( a b c)| = 3
|( a b c d)| = 4
…
|( 1 2 3 … n)| = n
came here to say this. thank you
Real thing is order of cycle is LCM of each cycle.
So LCM of 2, if there is 1 then it is 2.
And so on upto |n- cycle|
YES, if you multiply (a,b) with itself twice you will get (e) the identity cycle, the same thing if you multiply (a,b,c) with itself three times and the same thing for (a,b,c,d) multiplied with itself four times.
@@humamalsebai I tried this with a=(1,3,2) and after calculating (a cube) I got a^3=(1)(2)(3) . Please correct me if I am wrong or there is something else to consider.
@@rishidusad2985 your calculation seems right
It wasn't mentioned in the video, but if you think about it - what would be the identity element look like as a composition of cycles? Well, for Sn, it's (1)(2)(3)...(n), which means 1 maps to 1, 2 maps to 2 ... n maps to n (or that's the identity element) - you can clearly see it if you give yourself an example
The way you power up cycles with no repetitions of elements is that you basicly 'jump' as many times as the power
Example: (for Sn, where n = 3)
The notation (1 3 2) [2] = 1 means where to map the element 2 (as you can see, element 2 is mapped to 1 in the cycle (1 3 2))
if we apply this 3 times in a row, we get:
(1 3 2) [1] = 1 --> 1 'jumps forward' 3 times, meaning that 1 goes to 3, goes to 2, goes to 1, so 1 goes to 1
The same for (1 3 2) [2] = 2 and (1 3 2) [3] = 3
Result: 1 maps to 1, 2 maps to 2 and 3 maps to 3, so it's (1)(2)(3)
Another interesting thing which requres proof but is always true is:
For every cycle of length n, if you multiply it by itself k times, where k is such a number that: n=mk, where m is an integer (said othwerwide, k divides n), then the cycle is 'broken' into k independent cycles of length n/k (independent cycles meaning cycles with no common numbers)
Example: in Sn, n = 6, if we power up the following cycle (6 4 3 5 1 2) by 2, we get the cycles (5 3 1) (4 5 2)
if we power up the same cycle by 3, we get (6 3) (4 1) (3 2)
And therefore, if we power up it by 6, the same rule applies, becuase 6 divides 6 and we break our cycle of length 6 into 6 cycles of length 1, which is in fact our identity element.
EDIT for @Samuel Rho, when we map 1 element to itself, we get 1 cyce of length 1, so if a =(1,3,2), then a^3 is not (1,3,2), instead a^3 = (1)(2)(3) which is the identity element
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