HW3.tex

\documentclass[11pt]{article} \usepackage{verbatim} \usepackage[pdftex]{hyperref} \usepackage{amssymb} \usepackage{amsmath} \usepackage{graphicx} \voffset= -0.5in \textheight=8.8in \textwidth=6.5in \oddsidemargin=0in \parindent=0pt \begin{document} \pagestyle{empty} \begin{center} {\large \bf \sc MATH 421/521 Section B Intro to Abstract Algebra HW3 --- Fall 2020} \\ \vspace{0.5em} \end{center} All homework are required to be typed in LaTeX. You can use the free online editor\\ \url{http://www.overleaf.com}. See \url{https://www.overleaf.com/learn/} for a brief introduction. \\ HW3 is due Tuesday September 29, by 11:59pm. Please upload your solutions on canvas under Assignments by the due time. \medskip 1. List the left cosets of the subgroups in each of the following. Here $\langle a\rangle$ denotes the subgroup generated by the element $a$. \medskip (a) $\langle 8 \rangle$ in $(\mathbb{Z}_{24},+)$ \medskip (b) $\langle 3 \rangle$ in $U(8)$ \medskip (c) $3\mathbb{Z}$ in $\mathbb{Z}$\quad (where $3\mathbb{Z}=\{3k: k\in \mathbb{Z}\}$.) \medskip (d) $A_n$ in $S_n$ \quad (where $A_n$ is the set of all even permutations on $\{1,\dots, n\}$.) \bigskip 2. Find all the left cosets of $H=\{1,19\}$ in $U(30)$. \bigskip 3. Given a finite group $G$ and $H$ a subgroup. Using the same argument as in the Lemma proved in class, one can show all the statements of the lemma hold analogously for right cosets of $H$ in $G$. In particular, $Ha=H$ if and only if $a\in H$ and the distinct right cosets of $H$ form a partition of $G$. Now, suppose that $|H|=|G|/2$. \medskip (a) Show that for every $a\in G$, $aH=Ha$. (comment: in general $aH$ and $Ha$ are not necessarily equal. But with our condition $|H|=|G|/2$ here, this indeed holds.) \medskip (b) Suppose $a,b\in G$ are two elements of $G$ that are not in $H$. Prove that $ab\in H$. \bigskip 4. Let $G$ be a group of order $63$. Prove that $G$ must have an element of order $3$. \medskip 5. Let $G$ be a group of order $155$. Suppose $a,b$ are two nonidentity elements of $G$ that have different orders. Prove that the only subgroup of $G$ that contains both $a$ and $b$ must be $G$ itself. (hint: By Larange's theorem, the order of any nonidentity element must be one of $5,31, 155$. If one of $a,b$ has order $155$ then the statement is quite easy to prove. So one may assume $|a|=5$ and $|b|=31$. Consider how Theorem 7.2 might be relevant.) \bigskip 6. (Graduates only) Prove that every subgroup of $D_n$ that has an odd order must be cyclic. \end{document}