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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill April $7^{\mathrm{th}}$, 2021}

\vspace{0.1in}

\noindent {\large\bf Name:} \underline{\color{red}\Large\sc\qquad Answer Key\qquad} \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 $G = Q \times A_4$ where $Q = \{ \pm 1, \pm i, \pm j, \pm k\}$ is the quaternion group and $A_4$ is the group of even permutations in $S_4$ (i.e., $A_4 = \{ (1), (123), (132), (124), (142), (134), (143), (234), (243), (12)(34), (13)(24), (14)(23) \}$).

\vspace{0.1in}

The order of $G$ is $|G|=$ \underline{\quad $|Q| \cdot |A_4| = 8 \cdot 4!/2 = 8 \cdot 12 =$ \fbox{96} \quad}

\vspace{0.1in}

What is the largest element order in $Q \times A_4$? {\bf Give an example} of such an element.

\vspace{0.05in}

The elements in $Q$ have orders 1, 2, and 4. The elements in $A_4$ have orders 1, 2, and 3. The largest least common multiple we can obtain from these choices is $\mathrm{lcm}(4,3)=12$.  The largest element order in $G$ is \fbox{12}. For example, \fbox{$(i,(123))$} has order $\mathrm{lcm}(|i|,|(123)|)=\mathrm{lcm}(4,3)=12$.

\vspace{0.05in}

\item Let $G$ be a group of order 50 with subgroups $H$, $K$, and $L$. In addition, suppose the order of $H$ is 5, the order of $L$ is 2, and that $H \subseteq K \subseteq G$. 

\vspace{0.05in}

By Lagrange's theorem, the orders of these subgroups must divide $|G|=50$.

\vspace{0.05in}

\noindent
What is/are the possible order(s) of $K$?

\vspace{0.05in}

Notice that we also have that $H$ is a subgroup of$K$ (since $H \subseteq K$), so the order of $H$ (i.e., $|H|=5$) must divide the order of $K$. Thus $|K|$ divides 50 and is a multiple of 5. Thus \fbox{$|K|=$ 5, 10, 25, or 50}.

\vspace{0.05in}

\noindent
What can we conclude about $H \cap L$?

\vspace{0.05in}

$H \cap L$ is a subgroup of both $H$ and $L$. Thus $|H \cap L|$ divides both $|H|=5$ and $|L|=2$. In particular, $|H \cap L|=1$. Therefore, \fbox{$H \cap L$ is the trivial subgroup} (i.e., $H \cap L = \{\text{the identity}\}$). 

\vspace{0.05in}

\end{enumerate}


\noindent {\large 2. (15 points)} Let $\varphi:G \to H$ be a homomorphism.\\ \force \qquad \qquad $\longrightarrow$ State the definition a homomorphism.\\ \force \qquad \qquad $\longrightarrow$ State the definition of the kernel: $\mathrm{ker}(\varphi)$.\\ \force \qquad \qquad $\longrightarrow$ Then prove that $\mathrm{ker}(\varphi)$ is a normal subgroup of $G$. \hfill
{\it Note:} Prove it is a subgroup \underline{and} that it is normal in $G$.

\vspace{0.05in}

The map $\varphi$ is a homomorphism if it is operation preserving: For all $a,b \in G$, we have \fbox{$\varphi(ab)=\varphi(a)\varphi(b)$}. Next, the kernel of $\varphi$ is \fbox{$\mathrm{ker}(\varphi) = \{ x \in G \;|\; \varphi(x)=1_H\}$} where $1_H$ is the identity of $H$ (the codomain). In words, the kernel is everything in domain that maps to the identity (of the codomain). Alternatively, it is the inverse image of the trivial subgroup of $H$ (i.e., $\mathrm{ker}(\varphi) = \varphi^{-1}\{1_H\}$).

\vspace{0.05in}

To show that $\mathrm{ker}(\varphi) \triangleleft G$, we go through the normal subgroup test:
\begin{itemize}
\item Obviously, $\mathrm{ker}(\varphi)$ is a subset of the domain $G$. Notice that $\varphi(1_G)=1_H$ where $1_G$ is the identity of $G$ since homomorphisms always map the identity of the domain to the identity of the codomain. Thus $1_G \in \mathrm{ker}(\varphi)$ (and so the kernel is non-empty).
\item Suppose $a,b \in \mathrm{ker}(\varphi)$. Then $\varphi(a)=1_H$ and $\varphi(b)=1_H$. Therefore, $\varphi(ab)=\varphi(a)\varphi(b)=1_H 1_H = 1_H$ and so $ab \in \mathrm{ker}(\varphi)$ (we have closure under the operation in $G$).
\item Suppose $a \in \mathrm{ker}(\varphi)$ so that $\varphi(a)=1_H$. Therefore, $\varphi(a^{-1})=\varphi(a)^{-1}=1_H^{-1}=1_H$ (since homomorphisms send inverses to inverses) and so $a^{-1} \in \mathrm{ker}(\varphi)$ (we have closure under inverses in $G$ and thus $\mathrm{ker}(\varphi)$ is a subgroup of $G$).
\item Suppose $a \in \mathrm{ker}(\varphi)$ and $g \in G$ so that $\varphi(a)=1_H$. Therefore, $\varphi(gag^{-1}) = \varphi(g)\varphi(a)\varphi(g^{-1}) = \varphi(g)1_H\varphi(g)^{-1} = \varphi(g)\varphi(g)^{-1} =1_H$. Thus $gag^{-1} \in \mathrm{ker}(\varphi)$ (we have closure under conjugation by elements of $G$).
\end{itemize}
Therefore, $\mathrm{ker}(\varphi)$ is a normal subgroup of $G$.

\vspace{0.05in}

\newpage

\noindent {\large 3. (15 points)} Consider $K = \{ 1,x^2,x^4 \}$ in $D_6 = \langle x,y \;|\; x^6=1,y^2=1,xyxy=1 \rangle = \{ 1,x,\dots,x^5, y,xy,\dots, x^5y\}$. 

{\it Note:} It can be shown that $K$ is a normal subgroup of $D_6$ (just accept this for now).

\vspace{-0.05in}

\begin{enumerate}[(a)]
   \item  Quick questions about $\diagquotient{D_6}{K}$.
   
\vspace{0.05in}

   The order of $\diagquotient{D_6}{K}$ is \underline{\quad $\dfrac{|D_6|}{|K|} = \dfrac{12}{3}=$ \fbox{4} \quad}.
   \hfill
   List the distinct elements of $\diagquotient{D_6}{K} = \Big\{$ \underline{\quad \fbox{$K,xK,yK,xyK$} \quad} $\Big\}$.
   %\\    \force \hfill {}[In terms of $K$, for example: ``$xK$''.] \  \  \force

\vspace{0.05in}   

   The identity of $\diagquotient{D_6}{K}$ is \underline{\quad \fbox{$K$} \quad}.
   \hfill
   $(xK)^{-1}=$ \underline{\quad $x^{-1}K = x^5K =$ \fbox{$xK$} \quad}.
   %\\    \force \hfill {}[Simplify please.] \force

\vspace{0.05in}

  The order of $xK$ in $\diagquotient{D_6}{K}$ is \underline{\quad \fbox{2} \quad}. \hfill List the contents of $xK = \Big\{$ \underline{\quad \fbox{$x,x^3,x^5$} \quad} $\Big\}$.

\vspace{-0.05in}

Scratch work:

\vspace{0.05in}

We have $K= \{1,x^2,x^4\}$ and so $xK = \{x \cdot 1, x\cdot x^2, x\cdot x^4\} = \{x,x^3,x^5\}$. We need to try something we haven't seen so far to get a new coset. Consider $yK = \{y \cdot 1, y \cdot x^2, y \cdot x^4 \} = \{ y,x^{-2}y,x^{-4}y\} = \{y,x^4y,x^2y\} = \{y,x^2,x^4y\}$. Since there are only a total of 4 cosets, the last coset must be what's left over $xyK = \{ xy,x^3y,x^5y\}$. Note that $x,x^5 \in xK$ so $x^5K=xK$. Next, $xK$ has order 2 since it is its own inverse. Alternatively, $(xK)^2=x^2K = K$ since $x^2 \in K$ so the order of $xK$ is 2.

%\vspace{0.05in}

\item Let $H = \{ 1, x^3, y, x^3y\}$. While $H$ is a subgroup of $D_6$, it is {\bf not} a normal subgroup of $D_6$. Prove it isn't normal.

\vspace{0.05in}

Notice that $xH = \{ x\cdot 1, x\cdot x^3, x\cdot y, x\cdot x^3y \} = \{ x,x^4,xy,x^4y\}$ (also, since there are only $|D_6|/|H|=12/4=3$ left cosets of $H$, the last coset is just the leftovers $x^2H = \{x^2,x^5,x^2y,x^5y\}$). On the other hand, $Hx = \{1\cdot x, x^3\cdot x, y\cdot x, x^3y \cdot x \} = \{ x,x^4,x^{-1}y,x^3x^{-1}y \} = \{x,x^4,x^5y,x^2y\} = \{x,x^4,x^2y,x^5y\}$ (also, the last right coset is the leftovers $Hx^2 = \{x^2,x^5,xy,x^4y\}$). Since $xH \not= Hx$, we have that $H$ is not a normal subgroup of $D_6$.

Alternatively, notice $y \in H$ but $xyx^{-1}=x^2y \not\in H$. Thus $H$ is not closed under conjugation, so it isn't normal in $D_6$.

\vspace{-0.05in}

\end{enumerate}

 \noindent 
 {\large 4. (10 points)} Consider $\diagquotient{\mathbb{Z}_{12}}{H}$ where $H=\langle 3 \rangle = \{0,3,6,9\}$.\\
 \force \qquad \qquad $\longrightarrow$ List all of the cosets of $H$ (and their contents) in $\mathbb{Z}_{12}$.\\ 
 \force \qquad \qquad $\longrightarrow$ Then make a Cayley table for this quotient group. 
 
\vspace{0.05in}
 
Notice that $\left|\diagquotient{\mathbb{Z}_{12}}{H}\right| = \dfrac{|\mathbb{Z}_{12}|}{|H|} = \dfrac{12}{4}=3$ so we should expect there are 3 cosets. 

The cosets are \fbox{$H = \{0,3,6,9\}$, $1+H=\{1,4,7,10\}$, and $2+H=\{2,5,8,11\}$} (we add since the operation in $\mathbb{Z}_{12}$ is addition mod 12). The Cayley table is easy enough to fill out. For example, $(2+H)+(2+H)=(2+2)+H=4+H=1+H$ since $4 \in 1+H$. Also, $(1+H)+(2+H)=(1+2)+H=3+H=H$ since $3 \in H$.

\vspace{0.05in}

\force\hfill
\begin{tabular}{c||c|c|c|} & $H$ & $1+H$ & $2+H$ \\ \hline \hline
$H$ & $H$ & $1+H$ & $2+H$ \\ \hline
$1+H$ & $1+H$ & $2+H$ & $H$ \\ \hline
$2+H$ & $2+H$ & $H$ & $1+H$ \\ \hline
\end{tabular} \qquad \qquad{\it Note:} $\diagquotient{\mathbb{Z}_{12}}{H} = \diagquotient{\mathbb{Z}_{12}}{\langle 3 \rangle} \cong \mathbb{Z}_3$ \qquad
\begin{tabular}{c||c|c|c|} & $0$ & $1$ & $2$ \\ \hline \hline
$0$ & $0$ & $1$ & $2$ \\ \hline
$1$ & $1$ & $2$ & $0$ \\ \hline
$2$ & $2$ & $0$ & $1$ \\ \hline
\end{tabular}
\hfill\force

\vspace{0.05in}

Of course, we could have concluded what our quotient group should look like from the very beginning since quotients of cyclic groups are cyclic and so $\mathbb{Z}_{12}/H$ must be a cyclic group of order 3 (i.e., isomorphic to $\mathbb{Z}_3$).

\vspace{0.05in}
 
 What is the order of $8+H$ in $\mathbb{Z}_{12}/H$?
 
\vspace{0.05in}
 
We could compute directly: $8+H \not= H$, $(8+H)+(8+H)=16+H=4+H \not= H$, and $(8+H)+(8+H)+(8+H)=(8+8+8)+H=0+H=H$ so the order of $8+H$ is \fbox{3}.
Alternatively, we could first swap out the representative 8 for a smaller one: $8+H=2+H$ and work with that. Or, we could skip the calculation and notice that since $8+H \not=H$, it is a non-identity element in a (cyclic) group of order 3. Thus its order must be 3.

\vspace{0.05in}

\noindent {\large 5. (15 points)} Something is terribly, horribly wrong!

\vspace{-0.1in}

\begin{enumerate}[(a)]
\item Let $H \triangleleft \mathbb{Z}_{100}$. Why is $\diagquotient{\mathbb{Z}_{100}}{H} \cong \mathbb{Z}_{5} \times \mathbb{Z}_5$ impossible?

\vspace{0.05in}

While $\mathbb{Z}_{100}/H$ could be a group of order 25 (since 25 divides 100), we know that quotients of cyclic groups must be cyclic and \fbox{$\mathbb{Z}_5 \times \mathbb{Z}_5$ is not cyclic} (the moduli are not relatively prime or in a related note the largest order in $\mathbb{Z}_5\times\mathbb{Z}_5$ is $\mathrm{lcm}(5,5)=5 \not=25$ so not cyclic). Alternatively, if the order of the quotient was 25, we must be quotienting by a subgroup of order 4 (there is only one such subgroup: $H=\langle 25 \rangle$) so $\mathbb{Z}_{100}/H=\mathbb{Z}_{100}/\langle 25\rangle \cong \mathbb{Z}_{25} \not\cong \mathbb{Z}_5 \times \mathbb{Z}_5$.

\vspace{0.05in}

\item Let $\varphi:A_4 \to Q$ be a homomorphism (where $A_4$ and $Q$ are the same as in problem 1(a)). Why can't $\varphi$ be onto?  

\vspace{0.05in}

We know that the order of the image (=range) of a homorphism must divide the order the the domain. Thus the order of $\varphi(A_4)$ must divide $|A_4|=4!/2=12$. However, $|Q|=8$. Briefly, this is impossible since \fbox{$|Q|=8$ does not divide $|A_4|=12$}.

{\it Note:} The first isomorphism theorem says that for a homomorphism $\varphi:G\to H$ one has $\diagquotient{G}{\mathrm{ker}(\varphi)} \cong \varphi(G)$ (domain mod kernel is isomorphic to the image). This implies that $[G:\mathrm{ker}(\varphi)]=|G/\mathrm{ker}(\varphi)| = |\varphi(G)|$ and so by Lagrange's theorem $|G| = |\mathrm{ker}(\varphi)| \cdot [G:\mathrm{ker}(\varphi)] = |\mathrm{ker}(\varphi)| \cdot |\varphi(G)|$ (thus the order of the image divides the order of the domain).

\vspace{0.05in}

\item Recall $H = \{1,x^3,y,x^3y\}$ (a subgroup of $D_6$) from problem 3(b). Suppose $\varphi:D_6 \to \mathbb{Z}$ is a homomorphism. Why can't $\mathrm{ker}(\varphi)=H$?

\vspace{0.05in}

In problem 3(b) we showed that $H$ is not a normal subgroup of $D_6$, but \fbox{kernels must be normal subgroups}.

\vspace{0.05in}

\end{enumerate}

\noindent {\large 6 (10 points)} Let $H$ be a subgroup of $G$ where $G$ is an Abelian group. First, prove that $H$ is in fact a normal subgroup. Then, prove that $\diagquotient{G}{H}$ is Abelian.

\vspace{0.05in}

Suppose that $a \in G$. Then $aH = \{ah \;|\; h \in H\} = \{ha \;|\; h \in H\} = Ha$ where $ah=ha$ since $G$ is Abelian. Therefore, since the left and right cosets of $H$ match, $H \triangleleft G$.

Suppose $aH,bH \in \diagquotient{G}{H}$. Then $aH \cdot bH=abH = baH = bH \cdot aH$ where $ab=ba$ since $G$ is Abelian. Therefore, coset multiplication is commutative and so $\diagquotient{G}{H}$ is an Abelian group.

\vspace{0.05in}

\noindent {\large 7. (20 points)} Finite Abelian Groups

\begin{enumerate}[(a)]
   \item List all of the non-isomorphic Abelian groups of order $100 =2^2 5^2$. Circle any that are cyclic.

\vspace{0.05in}

Both exponents are 2 and there are only two partitions of 2: $2=1+1$. I will present both the invariant factor and elementary divisor versions of each of our $2 \cdot 2=4$ distinct isomorphism classes (and I will list the cyclic one first).

\fbox{$\mathbb{Z}_4 \times \mathbb{Z}_{25} \cong \mathbb{Z}_{100}$}, $\mathbb{Z}_4\times \mathbb{Z}_5 \times \mathbb{Z}_5 \cong \mathbb{Z}_5 \times \mathbb{Z}_{20}$, $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{25} \cong \mathbb{Z}_2 \times \mathbb{Z}_{50}$, and finally $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_5 \times \mathbb{Z}_5 \cong \mathbb{Z}_{10} \times \mathbb{Z}_{10}$.

\vspace{0.05in}

   \item How many non-isomorphic Abelian groups of order 449,878,000 are there?

   {\it Note:}  449,878,000 $=2^4  \cdot 5^3  \cdot 11^3 \cdot 13^2$ and there are 5 non-isomorphic Abelian groups of order 16 $=2^4$. \smiley

\vspace{0.05in}

We have 4 isomorphism classes associated with $2^4$, 3 with $5^3$, 3 with $11^3$, and 2 with $13^2$ since there are 5 partitions of 4: $4=3+1=2+2=2+1+1=1+1+1+1$, 3 partitions of 3: $3=2+1=1+1+1$, and 2 partitions of 2: $2=1+1$. Therefore, there are a total of $5 \cdot 3 \cdot 3 \cdot 2=$ \fbox{90} non-isomorphic Abelian groups of order 449,878,000.

\vspace{0.05in}

   \item Are the groups $\mathbb{Z}_{6} \times \mathbb{Z}_{10} \times \mathbb{Z}_{10}$ and $\mathbb{Z}_{12} \times \mathbb{Z}_{50}$ isomorphic? Explain your answer.

\vspace{0.05in}

An easy way to resolve this question is to split each group into prime powers and see if the factors match up. First, $\mathbb{Z}_{6} \times \mathbb{Z}_{10} \times \mathbb{Z}_{10} \cong  \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{2} \times \mathbb{Z}_{5}$. On the other hand, $\mathbb{Z}_{12} \times \mathbb{Z}_{50} \cong \mathbb{Z}_{3} \times \mathbb{Z}_{4} \times \mathbb{Z}_{2} \times \mathbb{Z}_{25}$. Therefore, \fbox{these groups are {\bf not} isomorphic}. For example, $\mathbb{Z}_6 \times \mathbb{Z}_{10} \times \mathbb{Z}_{10}$ has no elements of order 4 or 25 whereas $\mathbb{Z}_{12} \times \mathbb{Z}_{50}$ does have elements of those orders.   


\vspace{0.05in}

   \item What is the largest order among elements of $\mathbb{Z}_{6} \times \mathbb{Z}_{9} \times \mathbb{Z}_{15}$? Explain you answer.

\vspace{0.05in}

The largest order would be the largest least common multiple resulting from a divisor of 6, a divisor of 9, and a divisor of 15. In particular, this is $\mathrm{lcm}(6,9,15) = \mathrm{lcm}(2\cdot 3,3^2,3\cdot 5) = 2\cdot 3^2 \cdot 5 =$ \fbox{90}. If we wanted an element of such an element $(1,1,1)$ does the trick.

Alternatively, we could split these groups into their invariant factors and then put together elementary divisors:
$$\mathbb{Z}_{6} \times \mathbb{Z}_{9} \times \mathbb{Z}_{15} \cong \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{3^2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{5} \cong \mathbb{Z}_{2} \times  \mathbb{Z}_{3} \times \mathbb{Z}_{2\cdot 3^2\cdot 5} = \mathbb{Z}_{3} \times \mathbb{Z}_{3} \times \mathbb{Z}_{90}.$$ 
Now 90 is visibly the largest possible order.     


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\end{enumerate}

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