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\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