> [!NOTE] Lemma (Alternating Group has Index Two in Symmetric Group) > Let $n\geq2$. Then the [[nth Alternating Group|alternating group]] $A_{n}$ has [[Index of Subgroup|index]] $2$ as a subgroup of the [[Symmetric Groups of Finite Degree|symmetric group]] $S_{n}$. ###### Proof Consider the [[Sign of Permutation of n Letters|sign function]] $\text{sign}:S_{n}\to \{ 1,-1 \}.$ The [[Kernel of Homomorphism of Groups|kernel]] of $\phi$ is given by $\text{ker}(\phi)=\{ \sigma\in S_{n}: \text{sign}(\sigma)=1 \}=A_{n}.$ So it follows from [[Kernel of Homomorphisms of Groups is Normal Subgroup of Domain|kernel is normal subgroup of domain]] that $A_{n}\unlhd S_{n}.$ Applying [[First Isomorphism Theorem for Groups|the first isomorphism theorem]], gives an isomorphism $\begin{align}\widehat{\operatorname{sign}}: S_n / A_n &\rightarrow\{1,-1\} \\ \sigma A_n &\mapsto\operatorname{sign}(\sigma)\end{align}$thus $\#(S_{n}/A_{n})=\#\{1,-1\}=2.$ # Applications [[Order of nth Alternating Group]].