Alternating Group
The alternating group AnA_n is the subset AnSnA_n \subseteq S_n of the symmetric group consisting of only even elements.
The alternating group AnA_n accounts for exactly half of the elements of SnS_n (and thus the other half are odd). That is, An=12Sn\lvert A_n \rvert = \frac 1 2 \lvert S_n \rvert pf
Fix nn. Let AnA_n and BnB_n respectively denote the even and odd elements of SnS_n. We will show that AnBnA_n \cong B_n; since together they form a partition of SnS_n, this implies that each contains exactly half of the elements. That AnBnA_n \cong B_n is witnessed by the following two injections: f:AnBn=(σ(1 2)σ)f~:BnAn=(σ(1 2)σ)\begin{align*} f : A_n \to B_n &= (\sigma \mapsto (1\ 2)\sigma) \\ \tilde f : B_n \to A_n &= (\sigma \mapsto (1\ 2)\sigma) \end{align*} Both of these are injective because (1 2)σ=(1 2)ρ(1 2)(1 2)σ=(1 2)(1 2)ρσ=ρ\begin{align*} (1\ 2)\sigma &= (1\ 2)\rho \\ (1\ 2)(1\ 2)\sigma &= (1\ 2)(1\ 2)\rho \\ \sigma &= \rho \end{align*} Since there is an injection in both directions AnBnA_n \leftrightarrow B_n, we have that AnBnA_n \cong B_n.

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