If alpha is an r-cycle, show that alpha^r = (1)
Can someone show this? Thanks in advance
it sometimes helps to see an example. let's consider a 4-cycle in S7: α = (1 2 5 7).
this is the 1-1 function (actually, bijection) which maps:
1-->2
2-->5
3-->3
4-->4
5-->7
6-->6
7-->1
if we compose this function with itself, we can see that 3,4 and 6 are never affected, so we only need to worry about 1,2,5 and 7.
by direct calculation α^2 takes:
1-->2-->5
2-->5-->7
5-->7-->1
7-->1-->2, that is α^2 is the disjoint pair of 2-cycles: (1 5)(2 7). note that α^2 just takes whatever number is in the n-th place to the (n+2)-th place (mod 4).
similarly, α^3 takes the number in the n-th position to the number in the (n+3)-th place (mod 4), that is α^3 is the 4-cycle (1 7 5 2).
now n+3 = n - 1 (mod 4) so α^3 = α^(-1). multiplying both sides by α, or performing the direct calculation again shows α^4 = 1.
with a k-cycle, you can think of the numbers (some subset of {1,2,...,n}) as being isomorphic to the set {1,2...,k}. which k numbers we are permuting isn't really relelvant, what matters is how many cards the deck we are shuffling has (the cards being the individual numbers, the deck being the set of them-in my example this is the set {1,2,5,7}). so think of a k-cycle as turning a dial with k numbers 1 click. after k clicks, you've come full circle. in my example, we had 4 numbers we were shuffling, so 4 is the relevant piece of information about that cycle. "which" 4 numbers we used doesn't mean that much, all 4-cycles "act alike".
A 1-1 function is an injection, but not necessarily a surjection. If two sets are of the same cardinality there is a one-to-one correspondence between them. So, although technically you are correct, you might want to edit your post accordingly...
(Certainly in the UK we use 1-1 to mean an injection all the time, and wikipedia warns readers of this possible misunderstanding. Personally, I try to avoid using 1-1 for either thing, so as to avoid all this.)
yes, technically it would be better to read "1-1 correspondence" or "bijection". however, all bijections are 1-1 functions (although not necessarily vice versa). since i am explicitly defining the function for all elements in the domain, there shouldn't be any confusion. in fact i could have omitted the "1-1" entirely and just described it as a function.