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

1. ## 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 $X_o$ be the set of all odd cardinality subsets of the set $X$, and let $X_e$ be the set of all even cardinality subsets of the set $X$. Do the following:
- Find a bijection $F:X_o -> X_e$,
- Express $|X_o|$ and $|X_e|$ in terms of the number $|X|$.

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 $X_o$ and $X_e$ because $|X_o|=|X_e|$.

Also, we know that if $|X|=2^n$, we have $|Xe|=2^(n-1)$ and $|X_o|=2^(n-1)$. But we also know that in every set $X$, we only have odd and even subsets, so if $|X_o|=|X_e|$ then we can say

$|X|=|X_o|+|X_e|$, so $|X_o|=(1/2)*|X|$ and $|X_e|=(1/2)*|X|$"

2. Basically, you just need to show that $|X_o| = |X_e|$.

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 $Y=X\setminus\{x\}$ where $x \in X$ 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 $S \subset Y$ the $S\cup \{x\} \subset X$)).