# Statement(s) > [!NOTE] Statement 1 ($A_{n}$ has order $n!/2$) > Let $n\in N^{+}.$ The [[nth Alternating Group|nth alternating group]] has [[Finite Group|order]] $\frac{n!}{2}.$ # Proof(s) ###### Proof of statement 1: We construct an injection from odd to even permutations and another from even to odd permutations. Suppose $\alpha\in S_{n}$ is an even permutation. Let $\beta=(1, 2),$ an odd permutation. Then $\beta\alpha$ is an odd permutation. Now if $\alpha_{1}$ and $\alpha_{2}$ are distinct then so are $\beta a_{1}$ and $\beta a_{2}$ (since if not we could left multiply by $\beta$ and get that $\alpha_{1}=\alpha_{2}$). Similarly, suppose $\alpha\in S_{n}$ is odd. Then $\beta \alpha$ is even. Moreover, if $\alpha_{1}$ and $\alpha_{2}$ are distinct then so are $\beta \alpha_{1}$ and $\beta \alpha_{2}.$ Thus by [[Schroeder-Bernstein Theorem|Schroder-Bernstein]], the number of even and odd permutations are the same and so $|A_{n}|=\frac{1}{2}|S_{n}|=\frac{n!}{2}.$ $\square$ ###### Proof of statement 1: By [[Alternating Group has Index Two in Symmetric Group]], $[S_{n}:A_{n}]=2.$ By [[Lagrange's theorem (on Finite Groups)]], $\# A_{n}=\#S_{n}/2=n!/2.$ $\square$