\documentclass[12pt]{amsart} \usepackage{amssymb} \usepackage{amsmath} \voffset=-1in \hoffset=-1in \textheight=10in \textwidth=6in \begin{document} \author{} \title{Math 720\\Fall 2010\\Homework 5} \date{} \maketitle \thispagestyle{empty} \pagestyle{empty} \centerline{\it Due Friday, October 29, 2010.} \centerline{} {\it This assignment will be devoted to showing that the only nonabelian simple group of order less than or equal to 100 is $A_5$.} \centerline{} \noindent 1. Suppose that $\vert G\vert=60$ and that $G$ is simple. \begin{itemize} \item[a)] (5 pt) Find the possibilities for the number of Sylow 2-subgroups ($n$) of $G$ and show that we only need concern ourselves with $n=5$ or $n=15$. \item[b)] (5 pt) Show that if $n=5$ then $G$ is isomorphic to a subgroup of $S_5$ and conclude that $G\cong A_5$. \item[c)] (5 pt) Show that if $n=15$ then there are two Sylow 2-subgroups (say $P$ and $Q$) that must intersect in a subgroup of $G$ of order 2. \item[d)] (5 pt) If $H=N_G(P\bigcap Q)$ show that $4\vert\ \vert H\vert$ and conclude that $\vert H\vert\geq 12$. (Hint: any group of order 4 is abelian so $P\subseteq N_G(P\bigcap Q)$.) \item[e)] (5 pt) Show that since the index of $H$ in $G$ less than or equal to 5, $G\cong A_5$. \end{itemize} \centerline{} \noindent 2. Assume that $\vert G\vert\leq 100$ and that $G$ is simple and nonabelian. \begin{itemize} \item[a)] (5 pt) Use the results from this (and previous assignments) to make a list of the possible orders of $G$. \item[b)] (5 pt) Eliminate all possibilities except for $\vert G\vert=60$ or $90$ (most of these should be almost immediate). \end{itemize} \centerline{} \noindent 3. Now assume that $\vert G\vert=90$ and is simple. \begin{itemize} \item[a)] (5 pt) Show that $G$ must have $6$ Sylow 5-subgroups. \item[b)] (5 pt) Show that $G$ is necessarily isomorphic to a subgroup of $A_6$. (Hint: $G$ can be considered a simple subgroup of $S_6$; consider $G\bigcap A_6$). \item[c)] (5 pt) Derive a contradiction by showing that $A_6$ has no subgroup of order $90$. (Hint: if $A_6$ has a subgroup of order $90$, look at the orbit of this group under conjugation action...what is the order of its normalizer?) \end{itemize} \centerline{} \noindent 4. (5 pt) Conclude that if $\vert G\vert\leq 100$ and $G$ is nonabelian and simple, then $G\cong A_5$. \end{document}