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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill November $6^{\mathrm{th}}$, 2015}

\vspace{0.1in}

\noindent {\large\bf Name:} \underline{\Large\sc\color{red} \quad Answer Key \quad} \hfill {\bf Be sure to show your work!}

\vspace{0.1in}

\noindent {\large 1. (15 points)} Getting things in order\dots
\begin{enumerate}[(a)]
   \item Let $G = A_4 \oplus D_3$ where $A_4 = \{ (1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23) \}$ and $D_3 = \langle x,y \;|\; x^3=1,y^2=1,xyxy=1 \rangle = \{ 1,x,x^2,y,xy,x^2y\}$. 

\vspace{0.05in}

The order of $G$ is $|G|=$ \underline{\quad $|A_4 \times D_3| = |A_4| \cdot |D_3| = 12 \cdot 6 = 72$ \quad}

\vspace{0.05in}

What is the largest element order in $A_4 \oplus D_3$? Give an example of such an element and explain why it has the largest possible order.

\vspace{0.05in}

The orders of the elements in $A_4$ are 1, 2 (eg. $(12)(34)$), and 3 (eg. $(123)$). In $D_3$ there are elements of orders 1, 2 (the reflections), and 3 (the non-trivial rotations) as well.
Recall that in a direct product group we have: $|(A,B)| = \mathrm{lcm}(|A|,|B|)$. Thus we can have orders coming the the least common multiple of 1, 2, and 3 with 1, 2, and 3. The largest possible order is thus $\mathrm{lcm}(2,3)=\mathrm{lcm}(3,2)=6$. 

For example: $|(\,(12)(34), x\,)| = \mathrm{lcm}(|(12)(34)|,|x|)=\mathrm{lcm}(2,3)=6$. Likewise, the order of $(\,y,(123)\,)$ also has order 6.


\vspace{0.05in}

\item Let $G$ be a group with subgroups $H$, $K$,  and $L$ such that $H \subseteq L$ and $K \subseteq L$. In addition, suppose that we know $|H|=2$, $|K|=3$, and $|G|=24$. What are the possible orders of $L$?

\vspace{0.05in}

We have $L$ is a subgroup of $G$, so by Lagrange's theorem $|L|$ divides $|G|=24$. Thus $|L|=1,2,3,4,6,8,12,$ or $24$. However, $K$ is a subgroup of $L$, so $|K|=3$ divides $|L|$. Thus $|L|$ is a multiple of $3$. This rules out $1,2,4,$ and $8$. So $|L|=3,6,12,$ or $24$. Finally, $H$ is a subgroup of $L$, so $|H|=2$ divides $|L$. Thus $|L|$ is even. Therefore, $|L|$  is \fbox{6, 12, or 24}.

\vspace{0.05in}

\end{enumerate}


\noindent {\large 2. (10 points)} Let $\displaystyle G = \left\{\begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix} \;\Big|\; a,b \in \mathbb{R} \mbox{ and } b \not=0 \right\}$. It turns out that $G$ is a group under matrix multiplication. Let $\displaystyle H = \left\{\begin{bmatrix} 1 & c \\ 0 & 1 \end{bmatrix} \;\Big|\; c \in \mathbb{R} \right\}$. Show that $H$ is a normal subgroup of $G$. As a help, notice that $\displaystyle \begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix}^{-1} = \begin{bmatrix} 1 & -a/b \\ 0 & 1/b \end{bmatrix}$.

\vspace{0.05in}

\force \hfill [You may {\bf not} assume that $H$ is a subgroup -- prove this as well.]

\vspace{0.05in}

Obviously $H$ is a non-empty subset of $G$. Let's check for: closure under matrix multiplication, closure under inverses, and invariance under conjugation.
\begin{itemize}
\item Let $A=\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}, B=\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} \in H$. Then $AB = \begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & b \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+b \\ 0 & 1 \end{bmatrix} \in H$.
\item Let $A=\begin{bmatrix} 1 & a \\ 0 & 1 \end{bmatrix} \in H$. Then $A^{-1} = \begin{bmatrix} 1 & -a/1 \\ 0 & 1/1 \end{bmatrix} = \begin{bmatrix} 1 & -a \\ 0 & 1 \end{bmatrix} \in H$.
\item Let $Y = \begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix} \in G$ and $X = \begin{bmatrix} 1 & c \\ 0 & 1 \end{bmatrix} \in H$. Then $Y^{-1}XY =  \begin{bmatrix} 1 & -a/b \\ 0 & 1/b \end{bmatrix} \begin{bmatrix} 1 & c \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & a \\ 0 & b \end{bmatrix} = \begin{bmatrix} 1 & -a/b \\ 0 & 1/b \end{bmatrix} \begin{bmatrix} 1 & a+bc \\ 0 & b \end{bmatrix}$

\force \hfill $= \begin{bmatrix} 1 & a+bc-(a/b)b \\ 0 & (1/b)b \end{bmatrix} = \begin{bmatrix} 1 & bc \\ 0 & 1 \end{bmatrix} \in H$.
\end{itemize}

\vspace{-0.15in}

Therefore, $H \triangleleft G$.


\vspace{0.1in}

\noindent {\large 3. (15 points)} A previous homework set showed that $H = \{ 1,x^3,x^6 \}$ is a normal subgroup of $D_9 = \langle x,y \;|\; x^9=1,y^2=1,xyxy=1 \rangle = \{ 1,x,\dots,x^8, y,xy,\dots, x^8y\}$. 

\begin{enumerate}[(a)]
   \item  Quick questions about $\diagquotient{D_9}{H}$.
\qquad   
  The order of $\diagquotient{D_9}{H}$ is \underline{\quad $\dfrac{|D_9|}{|H|} = \dfrac{18}{3} =6$ \quad}.

\vspace{0.05in}


   The identity of $\diagquotient{D_9}{H}$ is \underline{\quad $1H = H = \{1,x^3,x^6\}$ \quad}.
\qquad
   $(x^5H)^{-1}=$ \underline{\quad $(x^5)^{-1}H = x^4H = \{ x, x^4, x^7 \}$ \quad}.

\vspace{0.05in}

  The order of $xH$ in $\diagquotient{D_9}{H}$ is \underline{\quad 3 \quad}.
\qquad
 The size of the set $xH$ is \underline{\quad 3 \quad}.

\vspace{0.05in}

Scratch work:
\quad
Notice that $(xH)^1 = xH$, $(xH)^2 = x^2H$, and $(xH)^3=x^3H=H$ (since $x^3 \in H$), so the order of $xH$ (as an element of the quotient group $D_9/H$) is 3. 
Next, $xH = x\{1,x^3,x^6\} = \{x,x^4,x^7\}$, so the size of the set $xH$ is also 3.

\vspace{0.05in}

\item This theorem should be helpful: Every group of order $2p$ (where $p$ is an odd prime) is either cyclic (isomorphic to $\mathbb{Z}_{2p}$) or dihedral (isomorphic to $D_p$).

\vspace{0.05in}

$xH \, yH = xyH = xy\{1,x^3,x^6\} = \{xy,xyx^3,xyx^6 \} = \{xy,x^4y,x^7y \}$ \qquad [List the elements in the resulting coset.]

\vspace{0.05in}

$yH \, xH = yxH = yx\{1,x^3,x^6\} = \{yx,yx^4,yx^7\} = \{x^2y, x^5y, x^8y \}$  \qquad [List the elements in the resulting coset.]

\vspace{0.05in}

Therefore, $xHyH \not= yHxH$. This means that $\diagquotient{D_9}{H}$ is not abelian. So $\diagquotient{D_9}{H}$ is a non-abelian group of order $6=2 \cdot 3$. By the above theorem, it must be isomorphic to $D_3$.

\vspace{0.05in}

The above calculation tells us that $\diagquotient{D_9}{H}$ is isomorphic to \underline{\quad $D_3$ \quad}.

\vspace{-0.05in}

\end{enumerate}

 \noindent 
 {\large 4. (10 points)} Consider $\diagquotient{\mathbb{Z}_{12}}{H}$ where $H=\langle 3 \rangle = \{0,3,6,9\}$. List all of the cosets (and their contents) of $H$ in $\mathbb{Z}_{12}$. Then make a Cayley table for this quotient group.
 
\vspace{0.05in}

Keep in mind the operation in $\mathbb{Z}_{12}$ is addition modulo 12, so we will have ``additive'' cosets. The order of our quotient group is $\dfrac{|\mathbb{Z}_{12}|}{|H|} = \dfrac{12}{4} = 3$. Thus we should find 3 distinct cosets.

The cosets: \fbox{$0+H = H = \{ 0,3,6,9\}$, \quad $1+H = \{1+0,1+3,1+6,1+9\} = \{1,4,7,10\}$, \quad and \quad $2+H = \{2,5,8,11\}$}. 

Next, we need our Cayley table. Most of the computations are completely mundane such as: $H+(2+H) = (0+H)+(2+H)=(0+2)+H=2+H$ and $(1+H)+(1+H)=(1+1)+H=2+H$. Some are {\it slightly} more interesting such as $(1+H)+(2+H)=(1+2)+H=3+H=H$ since $3 \in H$. Also, $(2+H)+(2+H)=(2+2)+H = 4+H = 1+H$ since $4 \in 1+H$.

\vspace{-0.2in}

\begin{center}
\begin{tabular}{c||c|c|c|} 
& $H$ & $1+H$ & $2+H$ \Large\strut \\ \hline \hline
$H$ & $H$ & $1+H$ & $2+H$  \Large\strut\\ \hline 
$1+H$ & $1+H$ & $2+H$ & $H$  \Large\strut\\ \hline 
$2+H$ & $2+H$ & $H$ & $1+H$  \Large\strut\\ \hline 
\end{tabular}
\end{center}

Of course, we immediately recognize that this is the Cayley table of $\mathbb{Z}_3$ which makes sense since $\diagquotient{\mathbb{Z}_{12}}{H} = \diagquotient{\mathbb{Z}_{12}}{\langle 3 \rangle} \cong \mathbb{Z}_3$.

\vspace{0.05in}
 
\noindent {\large 5. (15 points)} Let $\varphi:\mathbb{Z}_8 \to \mathbb{Z}_{12}$ be defined by $\varphi(x)=3x$.  

\begin{enumerate}[(a)]

\item Show that $\varphi$ is a homomorphism. [Do we need to prove that $\varphi$ is well-defined?]

\vspace{0.05in}

Since $\varphi$ is defined in terms of a representative ``$x$'' of an equivalence class (i.e. all integers congruent to $x$ modulo 8), we {\bf do} need to check to see if $\varphi$ is well-defined. 

Suppose $x=y$ (mod 8). Then $x=y+8k$ for some $k \in \mathbb{Z}$. Therefore, $3x=3(y+8k)=3y+24k = 3y+12(2k)$. This means that $3x=3y$ (mod 12). Therefore, $\varphi(x)=\varphi(y)$ (mod 12) and so $\varphi$ is well-defined.

Next, notice that for all $a,b \in \mathbb{Z}_8$, $\varphi(a+b)=3(a+b)=3a+3b=\varphi(a)+\varphi(b)$. Thus $\varphi$ is operation preserving and so $\varphi$ is a homomorphism.

\vspace{0.05in}

\item Compute the kernel and image of $\varphi$.

\vspace{0.05in}

Let's record what $\varphi$ does (keep in mind that the outputs are computed ``mod 12''): \quad
$0 \mapsto 3\cdot 0 = 0$, \ \ 
$1 \mapsto 3\cdot 1 = 3$, \ \ 
$2 \mapsto 3\cdot 2 = 6$, \ \ 
$3 \mapsto 3\cdot 3 = 9$, \ \ 
$4 \mapsto 3\cdot 4 = 12 = 0$, \ \ 
$5 \mapsto 3\cdot 5 = 15 = 3$, \ \ 
$6 \mapsto 3\cdot 6 = 18 = 6$, \ \  and \ \  
$7 \mapsto 3\cdot 7 = 21 = 9$. 

\vspace{0.05in}

\fbox{$\mathrm{ker}(\varphi) = \{ x \in \mathbb{Z}_8 \;|\; \varphi(x)=0 \} = \{ 0,4 \} = \langle 4 \rangle$} (in $\mathbb{Z}_8$) and \fbox{$\mathrm{Im}(\varphi) = \varphi(\mathbb{Z}_8) =  \{ 0,3,6,9 \} = \langle 3 \rangle$} (in $\mathbb{Z}_{12}$).


\item When applied to this $\varphi$, what does the first isomorphism theorem tell us?

\vspace{0.05in}

$$\diagquotient{\mathbb{Z}_8}{\mathrm{ker}(\varphi)} \cong \mathrm{Im}(\varphi)  \qquad \Longrightarrow \qquad \diagquotient{\mathbb{Z}_8}{\langle 4 \rangle \mbox{ \tiny (in  $\mathbb{Z}_8$)}} \cong \langle 3 \rangle \mbox{ \tiny (in $\mathbb{Z}_{12}$)}$$

\vspace{0.05in}

\end{enumerate}


\noindent {\large 6 (10 points)} Let $\varphi:\mathrm{GL}_2(\mathbb{R}) \to \mathbb{R}_{\not=0}$ be defined by $\varphi(A) = \mathrm{det}(A)$. Show $\varphi$ is a homomorphism. What does the first isomorphism theorem tell us in this particular situation?

\vspace{0.05in}

There's nothing special about $n=2$, so I'll work with $\mathrm{GL}_n(\mathbb{R})$ for any $n \in \mathbb{Z}_{>0}$. Let $A,B \in \mathrm{GL}_n(\mathbb{R})$. Then $\varphi(AB) = \mathrm{det}(AB)=\mathrm{det}(A)\mathrm{det}(B) = \varphi(A)\varphi(B)$. Thus $\varphi$ is a homomorphism. [Of course, this assumes a basic, yet not that easy to prove, property of determinants.]

$\mathrm{ker}(\varphi) = \{ A \in \mathrm{GL}_n(\mathbb{R}) \;|\; \varphi(A)=1 \} = \{ A \in \mathrm{GL}_n \;|\; \mathrm{det}(A)=1 \} = \mathrm{SL}_n(\mathbb{R})$ (the ``special linear group''). By the way, this shows that $\mathrm{SL}_n(\mathbb{R}) \triangleleft \mathrm{GL}_n(\mathbb{R})$ (since kernels are normal).

Notice that $\varphi$ is onto. Consider $r \in \mathbb{R}_{\not=0}$. Then let $A = \begin{bmatrix} r & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix}$. Then $\mathrm{det}(A)=r \cdot 1 \cdots 1 = r \not=0$ so $A \in \mathrm{GL}_n(\mathbb{R})$ and $\varphi(A)=\mathrm{det}(A)=r$. Thus $r \in \mathrm{Im}(\varphi)$. So in our situation the first isomorphism theorem says\dots

$$\diagquotient{\mathrm{GL}_n(\mathbb{R})}{\mathrm{ker}(\varphi)} \cong \mathrm{Im}(\varphi)  \qquad \Longrightarrow \qquad \diagquotient{\mathrm{GL}_n(\mathbb{R})}{\mathrm{SL}_n(\mathbb{R})} \cong \mathbb{R}_{\not=0}$$

\vspace{0.05in}

\noindent {\large 7. (25 points)} Finite Abelian Groups

\begin{enumerate}[(a)]
   \item List all of the non-isomorphic abelian groups of order $225 =3^2 5^2$. Circle any that are cyclic.

\vspace{0.05in}

There are 2 ways to split up $3^2$: $3^2=9$ or $3\cdot 3$. The same is true for $5^2$: $5^2=25$ or $5 \cdot 5$. This leaves us with $2\cdot 2 = 4$ isomorphism classes:
\begin{itemize}
\item \fbox{$\mathbb{Z}_9 \oplus \mathbb{Z}_{25} \cong \mathbb{Z}_{225}$} \quad [These are cyclic.]
\item $\mathbb{Z}_9 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5 \cong \mathbb{Z}_5 \oplus \mathbb{Z}_{45}$
\item $\mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_{25} \cong \mathbb{Z}_3 \oplus \mathbb{Z}_{75}$
\item $\mathbb{Z}_3 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5 \cong \mathbb{Z}_{15} \oplus \mathbb{Z}_{15}$
\end{itemize}

\vspace{0.05in}

   \item How many non-isomorphic abelian groups of order 14,346,832,500 are there?

   {\it Note:}  14,346,832,500 $=2^2  \cdot3^2  \cdot5^4 \cdot 7^3  \cdot11 \cdot 13^2$ and there are 5 non-isomorphic abelian groups of order $625 = 5^4$. \smiley

\vspace{0.05in}

There are 2 ways to break up $2^2$. The same is true for $3^2$ and $13^2$. There are 3 ways to break up $7^3$. $11$ can't be split up. Finally, there are 5 ways to split up $5^4$ (as alluded to in the note above). This means that there are $p(2)p(2)p(4)p(3)p(1)p(2) = 2 \cdot 2 \cdot 5 \cdot 3 \cdot 1 \cdot 2 =$ \fbox{120 non-isomorphic abelian groups of order 14,346,832,500}.

\vspace{0.05in}

   \item Are the groups $\mathbb{Z}_{10} \oplus \mathbb{Z}_{20} \oplus \mathbb{Z}_{30}$ and $\mathbb{Z}_{100} \oplus \mathbb{Z}_{60}$ isomorphic? Explain your answer.

\vspace{0.05in}

Although not the quickest solution, we can determine if these groups are isomorphic by breaking them up as much as possible (keeping in mind that $\mathbb{Z}_{k\ell} \cong \mathbb{Z}_k \oplus \mathbb{Z}_{\ell}$ if and only if $k$ and $\ell$ are relatively prime). 

\vspace{-0.15in}

$$\mathbb{Z}_{10} \oplus \mathbb{Z}_{20} \oplus \mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \cong \mathbb{Z}_2 \oplus \mathbb{Z}_2 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_5  \oplus \mathbb{Z}_5$$

\vspace{-0.1in}

\force \hfill vs. \hfill \force

\vspace{-0.15in}

$$\mathbb{Z}_{100} \oplus \mathbb{Z}_{60} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_{25} \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \cong \mathbb{Z}_4 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_{25}$$

\vspace{-0.05in}

{\it Answer:} \fbox{\bf No}. Notice that we cannot put $\mathbb{Z}_2 \oplus \mathbb{Z}_2$ together to get $\mathbb{Z}_4$ and also we cannot put $\mathbb{Z}_5 \oplus \mathbb{Z}_5$ to get $\mathbb{Z}_{25}$. So these groups are not isomorphic.
 
\vspace{0.05in}

   \item Is the group $\mathbb{Z}_{14} \oplus \mathbb{Z}_{5} \oplus \mathbb{Z}_{81}$ cyclic? Explain you answer.

\vspace{0.05in}

{\it Answer:} \fbox{\bf Yes}. Notice that $14=2 \cdot 7$, $5=5$, and $81=3^4$. These numbers are pairwise relatively prime, therefore all of these groups can be combined: $\mathbb{Z}_{14} \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_{81} \cong \mathbb{Z}_{14 \cdot 5 \cdot 81} = \mathbb{Z}_{5,670}$ is cyclic.

\vspace{0.05in}
\end{enumerate}

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