Thread: Group Theory - Find a bijection F: Xo -> Xe (Xo-odd, Xe-even)

Hi, I have the following problem. I have solved the second part, but I don't know how to solve the first one.

Let be the set of all odd cardinality subsets of the set , and let be the set of all even cardinality subsets of the set . Do the following:
- Find a bijection ,
- Express and in terms of the number .

So the part I don't know is the first one, but here is what I got for the second one. Is it OK? And do you know a bijection for the first part? Thanks!


"There is a bijection between and because .

Also, we know that if , we have and . But we also know that in every set , we only have odd and even subsets, so if then we can say

, so and "

Basically, you just need to show that .

You can, if you wish, note that if |X| is odd then the complement of every even set us a unique odd set, and your done (taking the complement of X with X to be the empty set, and even set). This is just an observation, it isn't needed to prove the result per se.

Now, if |X|=n is then induct on n. n=1 can be checked manually (note that the empty set is take to be even).

Let where is some arbitrary element. The result holds for Y. Can you conclude that the result thus holds for X? (The subsets of X are the subsets of Y together with all the subsets of Y with x (as in, if the )).