Cardinality of Even and Odd Permutations on Finite Set
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Theorem
Let $n \in \N_{> 0}$ be a natural number greater than $0$.
Let $S$ be a set of cardinality $n$.
Let $S_n$ denote the symmetric group on $S$ of order $n$.
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 $R_o$ is $\dfrac {n!} 2$.
Proof
From Order of Symmetric Group:
- $\order {S_n} = n!$
where:
By definition:
- $\card {R_e} + \card {R_o} = \order {S_n}$
From Odd and Even Permutations of Set are Equivalent:
- $\card {R_e} = \card {R_o}$
The result follows.
$\blacksquare$
Sources
- 1968: Ian D. Macdonald: The Theory of Groups ... (previous) ... (next): Appendix: Elementary set and number theory