SomethingRelatedToAlternatingGroupIsANormalSubgroupOfTheSymmetricGroup 2007-03-09 06:18:57 2007-06-23 12:00:18 Proof alternating group is a normal subgroup of the symmetric group 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here We prove that the alternating group $A_n$ has index 2 in the symmetric group $S_n$, i.e., $A_n$ has the same cardinality as its complement $S_n\setminus A_n$. The proof is function-theoretic. Its idea is similar to the proof in the parent topic, but the focus is less on algebraic aspect. Let $\pi\in S_n\setminus A_n$. Define $\pi:S_n \setminus A_n\rightarrow A_n$ by $\pi(\sigma)=\pi\sigma$, where $\pi\sigma$ is the product of $\pi$ and $\sigma$. One-to-one: \begin{equation*} \pi(\sigma)=\pi(\delta) \Longrightarrow \sigma=\delta \end{equation*} since $\pi^{-1}$ exists and $\pi^{-1}\pi\sigma=\pi^{-1}\pi\delta$. Onto: Given $\alpha\in A_n$, there exists an element in $S_n\setminus A_n$, namely $\lambda=\pi^{-1}\alpha$, such that \begin{equation*} \pi(\alpha)=\lambda. \end{equation*} (The element $\lambda$ is in $S_n\setminus A_n$ because $\pi^{-1}$ is and the product of an odd permutation and an even permutation is odd.) The function $\pi:S_n \setminus A_n\rightarrow A_n$ is, therefore, a one-to-one correspondence, so both sets $S_n \setminus A_n$ and $A_n$ have the same cardinality.