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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
+2022-04-07 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