Fix n. Let
An and
Bn respectively denote the
even and
odd elements of
Sn.
We will show that
An≅Bn; since together they form a
partition of
Sn, this implies that each contains exactly half of the elements.
That
An≅Bn is witnessed by the following two
injections:
f:An→Bnf~:Bn→An=(σ↦(1 2)σ)=(σ↦(1 2)σ)
Both of these are
injective because
(1 2)σ(1 2)(1 2)σσ=(1 2)ρ=(1 2)(1 2)ρ=ρ
Since there is an
injection in both directions
An↔Bn, we have that
An≅Bn.