Cardinality of Even and Odd Permutations on Finite Set

Theorem
Let $n \in \N_{> 0}$ be a natural number greater than $0$.

Let $S_n$ denote the symmetric group on $n$ letters.

Let $R_e$ and $R_o$ denote the subsets of $S_n$ consisting of even permutations and odd permutations respectively.

Then the cardinality of both $R_e$ and $R_o$ are $\dfrac {n!} 2$.

Proof
From Order of Symmetric Group:
 * $\left\vert{S_n}\right\vert = n!$

where:
 * $\left\vert{S_n}\right\vert$ denotes the cardinality of the underlying set of $S_n$
 * $n!$ denotes the factorial of $n$.

By definition:
 * $\left\vert{R_e}\right\vert + \left\vert{R_o}\right\vert = \left\vert{S_n}\right\vert$.

From Odd and Even Permutations of Set are Equivalent:
 * $\left\vert{R_e}\right\vert = \left\vert{R_o}\right\vert$

The result follows.