Few But Ripe Tex

\documentclass{article} \usepackage{amsmath,amsthm,amssymb} \renewcommand{\thefootnote}{\fnsymbol{footnote}} \newcommand{\level}[1]{#1} \newcommand{\levell}[1]{#1} \newcommand{\levelll}[1]{#1} %%%\newcommand{\nlevel}[2]{#2} \newtheorem{exc}{Exercise} \newtheorem{lem}[exc]{Lemma} \newtheorem{thm}[exc]{Theorem} \newtheorem{cor}[exc]{Corollary} \theoremstyle{remark} \newtheorem{rem}[exc]{Remark} \newtheorem{note}[exc]{Note} \theoremstyle{definition} \newtheorem{defn}[exc]{Definition} \newtheorem*{eg}{Example} \newenvironment{soln}{\begin{proof}[Solution]}{\end{proof}} \title{Few but ripe} %%\author{Shane D'Mello} \begin{document} \maketitle \begin{thm} An automorphism $T$ on a finite group $G$ that takes more than three fourth of the elements to their inverses, must take all the elements to their inverses. The group will be Abelian. \end{thm} \begin{proof} Let $S=\{x \in G : Tx=x^{-1}\}$ \textit{If $G$ is abelian}, $T$ takes the product of elements in $S$ to their inverses. Since $S$ has more than half the elements of $G$, it must generate the group, and so $T$ takes all the elements of $G$ to their inverses. \textit{If $G$ is non-abelian} at least one element most have a proper normalizer. Observe that if $T(gx)=(gx)^{-1}$, then $g$ and $x$ must commute. In other words, if $x \in S$ does not commute with $g \in S$, $T(gx) \in G \setminus S$. So for a $g$ whose normalizer $N(g)$ is proper, the map $\theta_g: S \setminus N(g) \to G \setminus S$ defined as $\theta_g(x)=gx$ is well defined. This map is easily seen to be injective. But since $S$ has more than three fourth the elements of the groups, the order of the domain is greater than the order of the range, contradicting the injectivity of the map. So G must be abelian. \end{proof} \end{document}