Recent content by thom

Ah, no worries. I'm pretty sure I can prove this without using the symmetry group after all. I have acquired a paper on derangements which seems to suffice!

Yes, your second sentence in the reply is what I meant. Thank you for your response. The cycle (x y) refers to a permutation which maps x to y and y to x. Here of course, f(x) = y and f(y) = x. This is the simplest bijection I can think of that holds the required conditions. Basically I...

Sorry I am not used to using the tex code, but will learn and explain in words for now! I am trying to prove that for all distinct x and y in a finite set X, there exists a function f in P(X) (the permutation group) such that f(x)=x' and f(y)=y'. Note: x' and y' are also distinct. I have...