From 18a1af1b1fbc5ae3572e011b267caa7bef293f62 Mon Sep 17 00:00:00 2001 From: Holden Rohrer Date: Fri, 29 Apr 2022 18:54:17 -0400 Subject: added all new homework for Neha's class --- gupta/hw10.tex | 240 +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 240 insertions(+) create mode 100644 gupta/hw10.tex (limited to 'gupta/hw10.tex') diff --git a/gupta/hw10.tex b/gupta/hw10.tex new file mode 100644 index 0000000..20a0a33 --- /dev/null +++ b/gupta/hw10.tex @@ -0,0 +1,240 @@ +\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 +2022-04-07 at 11:59 pm}\hrule height .5pt}} + +\centerline{\bigbf Homework 10 - Holden Rohrer} +\bigskip + +\noindent{\bf Collaborators:} None + +\section{Judson 3.5: 2, 7, 14, 28, 13, 35, 45, 48} + +\problem{2.} +Which of the following multiplication tables defined on the set $G = +\{a,b,c,d\}$ form a group? Support your answer in each case. + +\answer + +$(a)$ is not a group because it does not have an identity element. + +$(b)$ is a group because it has an identity element ($a$), and every +element has an inverse (itself), and I have verified it is associative. % prove later + +$(c)$ is a group because it has an identity element ($a$), and every +element has an inverse ($(a,b,c,d)\mapsto(a,d,c,b)$). I have also +verified that this multiplication table is associative. + +$(d)$ is not a group because it has an identity element ($a$), but $d$ +has no inverse. + +\endanswer + +\problem{7.} +Let $S = \bb R\setminus \{-1\}$ and define a binary operation on $S$ by +$a*b = a+b+ab.$ Prove that $(S,*)$ is an abelian group. + +\answer +This group has identity $0$ because $a*0 = a+0+0a = a.$ +Every element has an inverse $-{a\over a+1}$ which is defined because +$a\neq -1,$ so $a+1\neq 0.$ $$-{a\over a+1}*a = -{a\over a+1} + a - +{a^2\over a+1} = {a^2+a\over a+1} - {a\over a+1} - {a^2\over a+1} = 0.$$ +The operation is also associative. $a*b = (a+1)(b+1),$ so +$$(a*b)*c = (a+1)(b+1)(c+1) = a*(b*c).$$ +\endanswer + +\problem{14.} +Given the groups $\bb R^*$ and $\bb Z,$ let $G = \bb R^* \times \bb Z.$ +Define a binary operation $\circ$ on $G$ by +$(a,m)\circ (b,n) = (ab, m+n).$ +Show that $G$ is a group under this operation. + +\answer +This operation is associative because multiplication and addition are +associative, so $((a,m)\circ (b,n)) \circ (c,l) = (a,m)\circ ((b,n)\circ +(c,l)) = (abc,m+n+l).$ + +The element $(a,m)$ has inverse $(1/a,-m),$ and the group has identity +$(1,0).$ +\endanswer + +\problem{28.} +Prove the remainder of Proposition 3.21: if $G$ is a group and +$a,b\in G,$ then the equation $xa = b$ has a unique solution in $G.$ + +\answer +By definition of the group $a$ has an inverse $a^{-1}\in G,$ so +multiplying the right side by $a^{-1}$ gives +$$xaa^{-1} = xe = x = ba^{-1}.$$ +\endanswer + +\problem{31.} + +Show that if $a^2 = e$ for all elements $a$ in a group $G,$ then $G$ +must be abelian. + +\answer +Let $G$ be a group such that $g^2 = e$ for all elements $g\in G.$ +We will show that $G$ is abelian. +Let $a,b\in G.$ We will show that $ab = ba.$ +$$ab = eabe = bbabaa = b(ba)(ba)a = ba.$$ +\endanswer + +\problem{35.} + +Find all the subgroups of the symmetry group of an equilateral triangle. + +\answer +We have the trivial subgroup $\langle e\rangle,$ the rotation subgroup +$\langle \rho_1 \rangle,$ the reflection subgroups $\langle +\mu_1\rangle,$ $\langle\mu_2\rangle,$ $\langle\mu_3\rangle,$ and the +group $G.$ +\endanswer + +\problem{45.} + +Prove that the intersection of two subgroups of a group $G$ is also a +subgroup of $G.$ + +\answer +Let $H$ and $I$ be two subgroups of $G.$ +We will show that $H\cap I$ is a subgroup by showing the three +conditions. + +Let $a,b\in H\cap I.$ We will show that $ab\in H\cap I.$ +From definition of $a,b,$ we know that $a,b\in H,$ so $ab\in H.$ +Similarly, $a,b\in I,$ so $ab\in I.$ +By definition of $H$ and $I$ as subgroups, $a^{-1}\in I,$ and $a^{-1}\in +H,$ so $a^{-1}\in I\cap H.$ +Finally, because $I$ and $H$ are subgroups, $e\in I$ and $e\in H,$ so +$e\in H\cap I.$ + +Therefore, $ab\in H\cap I,$ and $H\cap I$ is a subgroup. +\endanswer + +\problem{48.} + +Let $G$ be a group and $g\in G.$ Show that +$$Z(G) = \{x\in G: gx = xg\hbox{ for all }g\in G\}$$ +is a subgroup of $G.$ This subgroup is called the center of $G.$ + +\answer +To show that this is a subgroup, we need to show that $x,y\in Z(G)$ +implies $xy\in Z(G),$ that $x^{-1}\in Z(G),$ and that $e\in Z(G).$ + +Let $x,y\in Z(G)$ and $z\in G.$ We already know that $xy\in G,$ so we +will show that $xyz = zxy.$ +Because $y\in Z(G),$ $yz = zy.$ +Similarly, because $x\in Z(G),$ $x(yz) = (zy)x,$ so $xyz = zxy.$ +We have shown that $Z(G)$ is closed under group operations. + +We will also show that since $xz = zx,$ we have $x^{-1}z = zx^{-1}.$ +We left multiply and right multiply by $x^{-1}$ to get +$$x^{-1}xzx^{-1} = x^{-1}zxx^{-1} \to zx^{-1} = x^{-1}z.$$ + +$e\in Z(G)$ because $eg = g = ge.$ + +We have now shown that $Z(G)$ is a subgroup. +\endanswer + +\section{Judson 4.5: 2a-c, 5, 23} + +\problem{2a-c.} +Find the order of each of the following elements: (a) $5\in\bb Z_{12},$ +(b) $\sqrt 3\in\bb R,$ and (c) $\sqrt 3\in\bb R^*.$ + +\answer +\item{a.} +Because 5 is coprime to 12, the order of this element is 12. + +\item{b.} +This element has infinite order. + +\item{c.} +This element has infinite order. +\endanswer + +\problem{5.} +Find the order of every element in $\bb Z_{18}.$ +\answer +The coprime elements $\{1,5,7,11,13,17\}$ have order 18. + +The elements $\{2,4,8,10,14,16\}$ have order 9. + +The elements $\{3,9,15\}$ have order 6. + +The elements $\{6,12\}$ have order 3. + +The element $9$ has order 2. + +And the element $0$ has order 1. +\endanswer + +\problem{23.} +Let $a,b\in G.$ Prove the following statements. + +\item{a.} The order of $a$ is the same as the order of $a^{-1}.$ + +\answer +Let $a$ be an element of order $n.$ +This means $a^m \neq e$ for $1\leq m < n,$ but $a^n = e.$ +By definition of inverses, $(a^{-1})^n a^n = e \to (a^{-1})^n = e.$ +Similarly, $(a^{-1})^m a^m = e.$ +Where $1\leq m < n,$ we have $a^m \neq e.$ +We assume for the sake of contradiction $(a^{-1})^m = e.$ +We would obtain $a^m = e$ giving a contradiction, so +$(a^{-1})^m \neq e,$ and $a^{-1}$ is order $n.$ + +\endanswer + +\item{b.} For all $g\in G,$ $|a| = |g^{-1}ag|.$ + +\answer + +We will show that $|a| = |g^{-1}ag|.$ +Thus, we will show $(g^{-1}ag)^n = e$ if and only if $a^n = e.$ + +$(\Rightarrow)$ + +Let $a^n = e.$ +$$(g^{-1}ag)^n = g^{-1}a^ng = g^{-1}g = e.$$ + +$(\Leftarrow)$ + +Let $(g^{-1}ag)^n = g^{-1}a^ng = e.$ +This implies $g^{-1}a^n = g^{-1},$ so $a^n = e.$ + +\endanswer + +\item{c.} The order of $ab$ is the same as the order of $ba.$ + +\answer + +To show that $|ab| = |ba|,$ we will show that $(ab)^n = e$ if and only +if $(ba)^n = e.$ +WLOG, we will show that $(ab)^n = e$ only if $(ba)^n = e.$ + +Let $(ab)^n = e.$ +$$(ab)^{n+1} = (ab)^n(ab) = ab = a(ba)^nb \Longrightarrow (ba)^n = e.$$ + +\endanswer + +\bye -- cgit