From 18a1af1b1fbc5ae3572e011b267caa7bef293f62 Mon Sep 17 00:00:00 2001
From: Holden Rohrer <hr@hrhr.dev>
Date: Fri, 29 Apr 2022 18:54:17 -0400
Subject: added all new homework for Neha's class

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 gupta/hw11.tex | 134 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++
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+\newfam\bbold
+\def\bb#1{{\fam\bbold #1}}
+\font\bbten=msbm10
+\font\bbsev=msbm7
+\font\bbfiv=msbm5
+\textfont\bbold=\bbten
+\scriptfont\bbold=\bbsev
+\scriptscriptfont\bbold=\bbfiv
+\font\bigbf=cmbx12 at 24pt
+
+\def\answer{\smallskip{\bf Answer.}\par}
+\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
+depth 0pt{\parfillskip0pt\medskip}}
+\let\endanswer\endproof
+\def\section#1{\vskip18pt plus 6pt minus 6pt\vskip0pt plus 1in\goodbreak\vskip 0pt plus -1in%
+\noindent{\bf #1}}
+\let\impl\to
+\def\nmid{\hskip-3pt\not\hskip2.5pt\mid}
+\def\problem#1{\bigskip\par\penalty-100\item{#1}}
+
+\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
+2022-04-14 at 11:59 pm}\hrule height .5pt}}
+
+\centerline{\bigbf Homework 11 - Holden Rohrer}
+\bigskip
+
+\noindent{\bf Collaborators:} None
+
+\section{Judson 5.4: 1b, 2c, 2p, 3a, 3c, 24}
+
+\problem{1b.}
+
+Write the following permutation in cycle notation:
+$$\pmatrix{1&2&3&4&5\cr 2&4&1&5&3}.$$
+
+\answer
+This is $(1 2 4 5 3).$
+\endanswer
+
+\problem{2c.}
+
+Compute $(1 4 3)(2 3)(2 4).$
+
+\answer
+$$\pmatrix{1&2&3&4\cr 2&3&1&4}$$
+\endanswer
+
+\problem{2p.}
+
+Compute $[(1 2 3 5)(4 6 7)]^{-1}.$
+
+\answer
+$$\pmatrix{1&2&3&4&5&6&7\cr 5&1&2&7&3&4&6}$$
+\endanswer
+
+\problem{3a.}
+Express the following permutation as a product of transpositions and
+identify it as even or odd: $(1 4 3 5 6).$
+
+\answer
+$$(1 4 3 5 6) = (1 4)(4 3)(3 5)(5 6),$$
+and since there are 4 transpositions, this is an even permutation.
+\endanswer
+
+\problem{3c.}
+
+Express the following permutation as a product of transpositions and
+identify it as even or odd: $(1 4 2 6)(1 4 2).$
+
+\answer
+$$(1 4 2 6)(1 4 2) = (1 4)(4 2)(2 6)(1 4)(4 2),$$
+and since there are 5 transpositions, this is an odd permutation.
+
+\problem{24.}
+
+Show that a 3-cycle is an even permutation.
+
+\answer
+Let us have a 3-cycle $(a_1 a_2 a_3).$
+This can be written as $(a_1 a_2)(a_2 a_3)$ because $a_2$ ends in the
+position of $a_3,$ $a_1$ ends in the position of $a_2,$ and $a_3$ ends
+in the position of $a_1.$
+This is two transpositions, so this is an even permutation.
+\endanswer
+
+\section{Judson 6.5: 5d, 5b}
+
+\problem{5d.}
+
+List the left and right cosets of the subgroups of $A_4$ in $S_4.$
+
+\answer
+The left and right cosets are $\{A_4, (1\,2)A_4\}.$
+\endanswer
+
+\problem{5b.}
+
+List the left and right cosets of the subgroups of $\langle 3\rangle$ in
+$U(8).$
+
+\answer
+$\langle 3\rangle = \{{\rm id}, 3\},$ and since this group is abelian,
+it has the same left and right cosets.
+$5\langle 3\rangle = \{5, 7\},$ and this completes the partition of the
+group.
+\endanswer
+
+\section{Problem not from the textbook}
+
+\problem{1.}
+
+Let $H$ be a subgroup of $G$ and suppose that $g_1,g_2\in G.$
+Prove that $g_1H = g_2H$ if and only if $g_2\in g_1H.$
+
+\answer
+Let $H$ be a subgroup of $G$ and $g_1,g_2\in G.$
+We will show that $g_1H = g_2H$ if and only if $g_2\in g_1H.$
+
+$(\Rightarrow)$
+
+Let $g_1H = g_2H.$
+We will show that $g_2\in g_1H.$
+
+$e\in H,$ so $g_2e = g_2\in g_2H,$ and by equality, $g_2\in g_1H.$
+
+$(\Leftarrow)$
+
+Let $g_2 \in g_1H.$
+This means there is $h\in H$ such that $g_2 = g_1h.$
+We then know that $g_2H = g_1hH = g_1H$ because $hH = H$ by group
+closure.
+
+\endanswer
+\bye
-- 
cgit 

