\documentclass{article}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{ifthen}
\usepackage{graphicx}

\setlength{\unitlength}{0.1in}

\setlength{\oddsidemargin}{0.0in}
\setlength{\evensidemargin}{0.0in}
\setlength{\topmargin}{0.0in}
\setlength{\headheight}{0.0in}
\setlength{\headsep}{0.0in}
\setlength{\topskip}{0.0in}
\setlength{\textheight}{8.9in}
\setlength{\textwidth}{6.5in}

\newcommand{\comp}{ \,{\scriptstyle \stackrel{\circ}{}}\, } 
\newcommand{\nullset}{\mathrm{O}\!\!\!\!\big/\,}
\newcommand{\divides}{\,\Big|\,}
\newcommand{\myspace}{\mbox{\parbox[c][0.35in]{0.35in}{\hfill}}}
\newcommand{\varspace}[2]{\mbox{\parbox[c][#1]{#2}{\hfill}}}
\newcommand{\diagquotient}[2]{\displaystyle{{#1 \atop \;}\hspace*{-0.1in} \mbox{\put(0,0){\line(2,1){2}}} \hspace*{0.1in} {\; \atop #2}}}

\begin{document}

%\pagestyle{empty}

\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill April $20^{\mathrm{th}}$, 2010}

\begin{center} \sc \large Answer Key \end{center}

\vspace{0.1in}

\noindent
{\bf \large 1. (20 points)} For each of the following pairs of groups, if the groups are isomorphic, circle
   $G_1 \cong G_2$ and explain why they are isomorphic. If the groups aren't isomorphic, circle $G_1 \not\cong G_2$ and explain why not.
   
\begin{enumerate}[(a)]
   \item $\mathbb{C} \not\cong \mathrm{GL}_2(\mathbb{R})$

Both of these are infinite groups (in fact, their cardnality is ``continuum'' -- the size of the real numbers). On the other
hand, $\mathbb{C}$ is abelian whereas $\mathrm{GL}_2(\mathbb{R})$ is not:

\vspace{-0.2in}

\[ a+b = b+a \quad \forall a,b \in \mathbb{C}  \hspace{1in}
   \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 2 & 1 \end{bmatrix}
   \not= 
   \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 1 \end{bmatrix} \]

\vspace{-0.1in}
   
   \item $\mathbb{Z}_{6} \cong U(7)$

Remember that $U(n)$ is the set of all equivalence classes in $\mathbb{Z}_n$ whose representatives are relatively prime to $n$. Since $7$ is prime,
the only thing we have to throw out is $0$. $U(7) = \{ 1,2,3,4,5,6 \}$. So we have 2 abelian groups of order 6 (both $\mathbb{Z}_n$ and $U(m)$
are always abelian since both modular addition and multiplication are commutative). $\mathbb{Z}_6$ is cyclic. What about $U(7)$? Let's look for a generator in $U(7)$...
\begin{itemize}
\item $1^1 = 1$ (order 1).
\item $2^1 = 2$, $2^2=4$, $2^3=1$ (order 3).
\item $3^1 = 3$, $3^2=2$, $3^3=6$ so $3$ does have order 1, 2, or 3. This leaves only one last choice (order 6).
\end{itemize}
We can stop here. $U(7)$ has an element of order 6 (the order of $U(7)$). Therefore, $U(7)$ is cyclic. Any two cyclic groups
of the same order are isomorphic. Done.

\vspace{0.05in}
   
\noindent {\it Remark:} Although it's not exactly easy to show, it is true that $U(p)$ is a cylic group (of order $p-1$) if $p$ is prime.

\vspace{0.1in}

   \item $S_4 \not\cong D_{12}$

$|S_4| = 4! = 24 = 2 \cdot 12 = |D_{12}|$. So they are both non-abelian (hence non-cyclic) groups of the same order.
We should consider orders of elements. 

When we consider orders of elements, the fact that they're not isomorphic becomes clear. $D_{12}$ has rotations of order 1, 2, 3, 4, 6, and 12.
On the other hand $S_4$'s elements have orders 1, 2, 3, and 4. $S_4$ has no elements of order 6 or 12. Thus these groups are not isomorphic.

Of course, we could have arrived at this conclusion using other data. Like, for example, $D_4$ has only 2 elements of order 3 (rotations of $120^\circ$
and $240^\circ$), but $S_4$ has a whole bunch of 3 cycles: $(123), (132), (124), (142),$ ... So the number of elements of order 3 does not match.
OR, notice that $D_{12}$ has 12 reflections. Thus it has at least 12 elements of order 2 (in fact, it has exactly 13 elements of order -- don't forget the
$180^\circ$ rotation). On the other hand, $S_4$'s elements of order 2 are: $(12), (13), (14), (23), (24), (34), (12)(34), (13)(24),$ $(14)(23)$ (not enough). 

\end{enumerate}

\vspace{0.1in}

% \newpage
\noindent
{\bf \large 2. (20 points)} A very normal problem. [Assume $G$ is a group with identity $e$.]

\begin{enumerate}[(a)]

\item Let $H$ and $K$ be normal subgroups of $G$. Show that $H \cap K$ is a normal subgroup of $G$.

\vspace{0.1in}

First, we need to show that $H \cap K$ is a subgroup. Then we need to show it's normal in $G$.
\begin{itemize}
   \item $e \in H$ and $e \in K$ (since both $H$ and $K$ are subgroups and subgroups always contain the identity). Thus $e \in H \cap K$, so
            the intersection is non-empty.
   \item Let $a,b \in H \cap K$ this implies that $a,b \in H$ and $a,b \in K$. Now $ab \in H$ and $ab \in K$ (since $H$ and $K$ are subgroups
            and thus are closed under the operation). So we have $ab \in H \cap K$.
   \item Let $a \in H \cap K$ this implies that $a \in H$ and $a \in K$. Now $a^{-1} \in H$ and $a^{-1} \in K$ (since $H$ and $K$ are subgroups
            and thus contain inverses). So we have that $a^{-1} \in H \cap K$.
   \item Let $a \in H \cap K$ and $g \in G$. Then $a \in H$ and $a \in K$. But $H$ and $K$ are normal subgroups of $G$, so $gag^{-1} \in H$
            and $gag^{-1} \in K$. Therefore, $gag^{-1} \in H \cap K$.
\end{itemize}
Therefore, $H \cap K$ is a normal subgroup of $G$.

\vspace{0.1in}

\item Let $H$ be a subgroup of $G$ and let $N$ be a normal subgroup of $G$. 

          Let $HN = \{ hn \,|\, h \in H \mbox{ and } n \in N \}$
          The following facts are true (you do not have to prove them): $HN$ is a subgroup of $G$ and $N$ is a normal subgroup of $HN$.
          
          Let $\phi : H \rightarrow \diagquotient{HN}{N}$ be defined by $\phi(h)=hN$. 
          
          This map is onto since $hnN \in \diagquotient{HN}{N}$ implies that $\phi(h)=hN=hnN$ ($n \in N$ implies $nN=N$).
          
          \begin{enumerate}[i.]
          \item Show $\phi$ is a homomorphism.
          
\vspace{0.1in}

$\phi(xy) = (xy)N = (xN)(yN) = \phi(x)\phi(y)$

\vspace{0.1in}
          
          \item Show $\mathrm{Ker}(\phi) = H \cap N$.
          
\vspace{0.1in}

Let $x \in \mathrm{Ker}(\phi)$. Then $\phi(x)=N$ (remember that $N$ is the identity in the quotient group $HN/N$).
Thus $xN=N$ which implies that $x \in N$. Since $x \in H$ ($H$ is the domain of $\phi$), we have that $x \in H \cap N$.
Therefore, $\mathrm{Ker}(\phi) \subseteq H \cap N$.

Conversely, let $x \in H \cap N$. Then $x \in H$ and $x \in N$. $\phi(x)=xN=N$ since $x \in N$. Thus $x \in \mathrm{Ker}(\phi)$.
Therefore, $H \cap N \subseteq \mathrm{Ker}(\phi)$.

Thus $\mathrm{Ker}(\phi) = H \cap N$.

\vspace{0.1in}
          
          \item State the first isomorphism theorem. Then apply this theorem to $\phi$.
                  
\vspace{0.1in}

Let $\psi:G \rightarrow G'$ be a homomorphism. The first isomorphism theorem says that $G / \mathrm{Ker}(\psi) \cong \psi(G)$
(that is ``$G$ mod the kernel of $\psi$ is isomorphic to the image of $\psi$'').

We have shown that $\mathrm{Ker}(\phi) = H \cap N$ and we were told $\phi$ is onto so that the image of $\phi$ is $HN/N$. Therefore,

\vspace{-0.2in}

\[ \diagquotient{H}{H \cap N} \cong \diagquotient{HN}{N} \]

This result is called the second isomorphism theorem.

\vspace{0.1in}

          \end{enumerate}

\end{enumerate}

\noindent
{\bf \large 3. (20 points)} Computing with Quotients.

\begin{enumerate}[(a)]

\item Recall $A_4 = \{ (1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23) \}$. 
          Let $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$. $H$ is a subgroup of $A_4$ (you don't need to prove this).
          
          Find all of the left and right cosets of $H$. Is $H$ normal in $A_4$? If so, write a Cayley table for $\diagquotient{A_4}{H}$
          and determine if this quotient is abelian, cyclic, both or neither. If $H$ is not normal in $A_4$, explain why not.

\vspace{0.1in}

Remember that $H$ itself is a coset (just copy it). Then pick an element not in $H$, say $(123)$ and multiply everything in
$H$ on the left by $(123)$. At this point we know there are 12 elements in $A_4$ and 4 elements in $H$ so $H$ has $12/4=3$
cosets in $A_4$. Since we have found 2 cosets already, the final coset must be the ``leftovers'' (remember that cosets partition
the group).

\newpage

Left Cosets: \vspace{-0.1in}
\begin{itemize}
\item $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$
\item $(123)H = \{ (123)(1), (123)(12)(34), (123)(13)(24), (123)(14)(23) \} 
                          = \{ (123),      (134),                (243),                (142)  \}$
\item $(132)H = \{ (132),      (143),                (234),                (124)  \}$
\end{itemize}
Right Cosets: \vspace{-0.1in}
\begin{itemize}
\item $H = \{ (1), (12)(34), (13)(24), (14)(23) \}$
\item $H(123) = \{ (1)(123), (12)(34)(123), (13)(24)(123), (14)(23)(123) 
                          = \{ (123),      (243),                (142),               (134)  \}$
\item $H(132) = \{ (132),      (143),                (234),                (124)  \}$
\end{itemize}

Since the left and right cosets are equal, $H$ is a normal subgroup of $A_4$.

\begin{center}
\begin{tabular}{c||c|c|c|} 
$\diagquotient{A_4}{H}$ &          H & (123)H & (132)H \\ \hline \hline
                                         H &          H & (123)H & (132)H \\ \hline
                               (123)H & (123)H & (132)H &          H \\ \hline
                               (132)H & (132)H &           H & (123)H \\ \hline
\end{tabular}
\end{center}

Now $A_4/H$ is a group of order 3, so since 3 is prime, it must be cyclic and thus also abelian.

Alternatively, the Cayley table is obviously symmetric across the diagonal, so this quotient is abelian.
Also, $(123)H \not= H$, $((123)H)^2 = (123)^2H = (132)H \not= H$, and $((123)H)^3 = (123)^3H = (1)H = H$.
Therefore, $(123)H$ is an element of order 3, thus the quotient is cyclic.

\vspace{0.1in}

\item Compute the index of $\langle 4 \rangle$ in $\mathbb{Z}_{12}$. Then write down all of the cosets of $\langle 4 \rangle$ in
          $\mathbb{Z}_{12}$. Finally, write a Cayley table for $\diagquotient{\mathbb{Z}_{12}}{\langle 4 \rangle}$
          Is this quotient is abelian, cyclic, both or neither?

\vspace{0.1in}

Remember the operation is addition mod 12, so we will have additive cosets.

First, the index of $\langle 4 \rangle$ in $\mathbb{Z}_{12}$ is the number of cosets. By Lagrange's theorem, the number of cosets 
is the order of the group divided by the order of the subgroup. $\langle 4 \rangle = \{ 0, 4, 4+4, 4+4+4, \dots \} = \{ 0, 4, 8 \}$.
Thus the index is...

\vspace{-0.2in}

\[ [ \mathbb{Z}_{12} : \langle 4 \rangle ]  = \left|\diagquotient{\mathbb{Z}_{12}}{\langle 4 \rangle}\right| 
  = \frac{|\mathbb{Z}_{12}|}{|\langle 4 \rangle|} = \frac{12}{3} = 4 \]

Now let's find the 4 distinct cosets:
\[ \langle 4 \rangle = \{ 0, 4, 8 \}, \qquad 1 + \langle 4 \rangle = \{ 1, 5, 9 \}, \qquad 2 + \langle 4 \rangle = \{ 2, 6, 10 \}, \qquad \mbox{and} \qquad 3 + \langle 4 \rangle = \{ 3, 7, 11 \} \]

Next, the Cayley table. A sample calculation: $(2+\langle 4 \rangle) + (3+\langle 4 \rangle) = (2+3)+\langle 4 \rangle = 5+\langle 4 \rangle = 1 + \langle 4 \rangle$ since 1 and 5 are in the same coset.

\begin{center}
\begin{tabular}{r||r|r|r|r|} 
$\diagquotient{\mathbb{Z}_{12}}{\langle 4 \rangle}$ \varspace{0.4in}{0.01in}
                                       &      $\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$ \varspace{0.25in}{0.01in} \\ \hline \hline
    $\langle 4 \rangle$ &      $\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$ \varspace{0.25in}{0.01in} \\ \hline 
$1+\langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$ &  $    \langle 4 \rangle$ \varspace{0.25in}{0.01in} \\ \hline 
$2+\langle 4 \rangle$ & $2+\langle 4 \rangle$ & $3+\langle 4 \rangle$ &  $    \langle 4 \rangle$ & $1+\langle 4 \rangle$ \varspace{0.25in}{0.01in} \\ \hline 
$3+\langle 4 \rangle$ & $3+\langle 4 \rangle$ & $     \langle 4 \rangle$ & $1+\langle 4 \rangle$ & $2+\langle 4 \rangle$ \varspace{0.25in}{0.01in} \\ \hline 
\end{tabular}
\end{center}

Since $\mathbb{Z}_{12}$ is abelian, so is the quotient group $\mathbb{Z}_{12}/\langle 4 \rangle$. Also, notice that
$1 + \langle 4 \rangle$, $(1 + \langle 4 \rangle) + (1 + \langle 4 \rangle) = 2 + \langle 4 \rangle$, 
$(2 + \langle 4 \rangle) + (1 + \langle 4 \rangle) = 3 + \langle 4 \rangle$, and 
$(3 + \langle 4 \rangle) + (1 + \langle 4 \rangle) = \langle 4 \rangle$. So $1 + \langle 4 \rangle$ is a generator, thus the quotient is cyclic.
Alternatively, ANY quotient of a cyclic group is cyclic. So $\mathbb{Z}_{12}$ is cyclic implies $\mathbb{Z}_{12}/\langle 4 \rangle$ is cyclic.

\vspace{0.1in}

\end{enumerate}

\noindent
{\bf \large 4. (20 points)} Consider the ring $\mathbb{Z}_{10}$. For each of the following elements circle ``unit'' if it's a unit and/or ``zero divisor'' if it's a zero divisor. If an element has a (multiplicative) inverse, find it. If it's a zero divisor, show why it's a zero divisor.

\vspace{0.1in}

\hspace*{-0.4in}
\begin{tabular}{lll}
1 is a unit since $1 \cdot 1 = 1$ so $1^{-1}=1$ &
2 is a zero divisor since $2 \cdot 5 = 0$ &
3 is a unit since $3 \cdot 7 = 1$ so $3^{-1}=7$ \\
4 is a zero divisor since $4 \cdot 5 = 0$ &
5 is a zero divisor since $5 \cdot 2 = 0$ &
6 is a zero divisor since $6 \cdot 5 = 0$ \\
7 is a unit since $7 \cdot 3 = 1$ so $7^{-1}=3$ &
8 is a zero divisor since $8 \cdot 5 = 0$ &
9 is a unit since $9 \cdot 9 = 1$ so $9^{-1}=9$
\end{tabular}

\vspace{0.1in}

A few quick questions...

%\vspace{-0.1in}

\begin{enumerate}[(a)]

\item Is $\mathbb{Z}_{10}$  a commutative ring with unity? (Just ``Yes'' or ``No'' will suffice.)

%\vspace{0.1in}

\noindent  {\bf Answer:} ``Yes'' (This is true for all $\mathbb{Z}_n$.)

%\vspace{0.1in}

\item Is $\mathbb{Z}_{10}$ an integral domain? Why? or Why not?

%\vspace{0.1in}

\noindent {\bf Answer:} ``No'' $\mathbb{Z}_{10}$ has zero divisors. ($\mathbb{Z}_n$ is an integral domain if and only if $n$ is prime.)

%\vspace{0.1in}

\item Is $\mathbb{Z}_{10}$ a field? Why or Why not?

%\vspace{0.1in}

\noindent {\bf Answer:} ``No'' Several non-zero elements are not units. ($\mathbb{Z}_n$ is a field if and only if $n$ is prime.)

\vspace{0.1in}

\end{enumerate}

\noindent
{\bf \large 5. (20 points)} Some quick ring questions.

\begin{enumerate}[(a)]

\item Briefly explain (in a sentence or two) why all fields are integral domains.

\vspace{0.1in}

\noindent {\it Quick Answer:}
All non-zero elements in a field are units. Units are never zero divisors, so fields are integral domains.

\vspace{0.1in}

\noindent {\it More detail:} 
A field is a commutative ring with $1 \not= 0$ such that all non-zero elements are units. An integral domain is a
commutative ring with $1 \not= 0$ which has no zero divisors. So we need to explain why fields have no zero
divisors. Suppose that $a \not= 0$ is some element of the field. Then $a$ is a unit, so $a^{-1}$ exists in the field. 
Thus if $ab=0$ then $a^{-1}ab = a^{-1}0$ so that $b=0$. So $a$ is not a zero divisor. This shows
that there are no zero divisors in a field. Therefore, fields are integral domains.

\vspace{0.1in}


\item Let $\mathbb{F}$ be a field, let $r \in \mathbb{F}$, and suppose $r^2=r$. Prove that $r=0$ or $r=1$.\\
          Is this still true if $\mathbb{F}$ is an integral domain?
          
\vspace{0.1in}

Either $r=0$ (in which case $0^2=0$) or $r \not=0$. If $r \not= 0$, then $r^{-1} \in \mathbb{F}$ since $\mathbb{F}$ is a field.
Thus $r^{-1}r^2=r^{-1}r$ so that $r=1$ (notice $1^2=1$). Therefore, in a field, the only solutions to $r^2=r$ are $r=0$ and $r=1$.

This is still true for integral domains. In fact, in this context we are lead to a cleaner proof. If $r^2=r$, then $r^2-r=0$ so that
$r(r-1)=0$. Now integral domains don't have zero divisors so either $r=0$ or $r-1=0$. Thus either $r=0$ or $r=1$.

An alternate proof: Cancellation laws hold in integral domains, so either $r=0$ or $r \not= 0$ so we can cancel off $r$ in
the equation $r \cdot r = r \cdot 1$ and get $r=1$.

\vspace{0.1in}

\item Show $\displaystyle{ S = \left\{ \begin{bmatrix} 0 & x \\ 0 & 0 \end{bmatrix} \,\Big|\, x \in \mathbb{R} \right\} }$ is a subring
          of $\mathbb{R}^{2 \times 2}$ (the ring of $2 \times 2$ real matrices). 

\vspace{0.1in}

$S$ is obviously a non-empty subset of $\mathbb{R}^{2 \times 2}$. We need to check closure under subtraction and closue under
multiplication. Let $\displaystyle{ A = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix}, B = \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix} \in S}$.

\[ A - B = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}
              = \begin{bmatrix} 0 & a-b \\ 0 & 0 \end{bmatrix} \hspace{1in}
    AB = \begin{bmatrix} 0 & a \\ 0 & 0 \end{bmatrix} \cdot \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}
              = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

Both $A-B$ and $AB$ belong to $S$. Thus $S$ is a subring of $\mathbb{R}^{2 \times 2}$.


\vspace{0.1in}

\end{enumerate}

\end{document}


