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\noindent
\parbox{2in}{\bf Math 3110} 
\hfill {\Large \bf Test \#3} \hfill
\parbox{2in}{\bf \hfill April $10^{\mathrm{th}}$, 2015}

\vspace{0.1in}

\noindent {\large\bf Name:} \underline{\Large \quad\color{red}\sc\bf 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 $G = D_3 \oplus Q$ where $D_3 = \langle x,y \;|\; x^3=1,y^2=1,xyxy=1 \rangle = \{ 1,x,x^2,y,xy,x^2y\}$ is a dihedral group and $Q = \{ \pm 1, \pm i, \pm j, \pm k \}$ is the group of quaternions. 

\vspace{0.05in}

The order of $G$ is $|G|=$ \underline{\ $|D_{3}| \cdot |Q|=6 \cdot 8=48$}

\vspace{0.05in}

Does $G$ have an element of order 12? If so, give an example. If not, explain why not.

\vspace{0.05in}

Yes, $x$ has order 3 in $D_3$ and $i$ has order 4 in $Q$.  Therefore, $|(x,i)| = \mathrm{lcm}(|x|,|i|)=\mathrm{lcm}(3,4)=12$

\vspace{0.05in}

\item Let $G$ be a group with subgroups $H$ and $K$ such that $H \subseteq K \subseteq G$. In addition, suppose that $|H|=2$ and $|G|=30$. What are the possible orders of $K$?

By Lagrange's Theorem, $|H| \divides |K|$ and $|K| \divides |G|$.  This implies $2 \divides |K|$ (i.e. $|K|$ is even) and $|K| \divides 30$.  Because of this, the possible orders of K are \fbox{2,6,10, or 30}

\vspace{0.05in}

\end{enumerate}


\noindent {\large 2. (10 points)} Let $\varphi:G \to H$ be a homomorphism between two groups $G$ and $H$. Prove that $\mathrm{ker}(\varphi) \triangleleft G$.

\vspace{0.05in}

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


$K=Ker(\varphi)=\{g \in G \;|\;  \varphi (g)=e\}$

\vspace{0.1in}

We need to run the normal subgroup test\dots

\vspace{-0.05in}

\begin{itemize}

\item $\varphi(e)=e$.  This implies $e \in K$.  Therefore, $K$ is nonempty.

\item Suppose $a,b \in K$.  Then $\varphi(a) =e$ and $\varphi(b)=e$.  So $\varphi(ab)=\varphi(a) \varphi(b)=e \cdot e=e$.  Therefore, $ab \in K$.

\item Suppose $a \in K$.  Then $\varphi(a)=e$.  This implies $\varphi(a^{-1})=\varphi(a)^{-1}=e^{-1}=e$, so $a^{-1} \in K$.

\vspace{0.05in}

Therefore, $K$ is a subgroup of $G$.

\vspace{0.05in}

\item Suppose $g \in G$ and $x \in K$.  Then $\varphi(x)=e$ and so $\varphi(gxg^{-1}) = \varphi(g)\varphi(x)\varphi(g^{-1})=\varphi(g)\cdot e \cdot \varphi(g)^{-1}=e$ and so $gxg^{-1} \in K$

Thus, $K$ is a normal subgroup of $G$.

\end{itemize}

\noindent {\large 3. (25 points)} Quotients

\begin{enumerate}[(a)]
   \item Given: $H=\{(1),(12)(34),(13)(24),(14)(23) \}$ is a normal subgroup of $S_4$. 
   
\vspace{0.05in}

   The order of $\diagquotient{S_4}{H}$ is \underline{\quad $\dfrac{|S_{4}|}{|H|} =  \dfrac{4!}{ 4}  = \dfrac{24}{4} =6$ \quad}.

\vspace{0.05in}

   The identity of $\diagquotient{S_4}{H}$ is \underline{\quad $H=(1)H$ \quad}.
   \hfill
   $((1234)H)^{-1}=$ \underline{\quad $(1234)^{-1}H=(1432)H$ \quad [$=(1234)H$] \quad}.

\vspace{0.05in}

  The order of $(1234)H$ in $\diagquotient{S_4}{H}$ is \underline{\quad 2 \quad}. \hfill The size of the set $(1234)H$ is \underline{\quad 4 \quad}.

\vspace{0.05in}

Scratch work:

Notice that $((1234)H)^2 = (1234)^{2}H=(12)(34)H=H$ since $(12)(34)\in H$, which implies $|(1234)H|=2$ (order as an element of the quotient group)

Also, all cosets have the same size, so since $|H|=4$, $(1234)H$ also has 4 elements [here $|H|$ means the size of the set H, not the order of H as an element of the quotient.  In the quotient H is the identity so $|H|=1$]

Now, $((1234)H)^{-1} = (1234)^{-1}H=(1432)H = \{ (1432)(1), (1432)(12)(34), (1432)(13)(24), (1432)(14)(23) \}$\\ $= \{ (1432), (24), (1234), (13) \}$. Thus $((1234)H)^{-1} = (1234)H$ since both $(1234)$ and $(1432)$ belong to $(1432)H$.  Another way to see that $((1234)H)^{-1}=(1234)H$ is to notice that the order of $(1234)H$ is 2, so it's its own inverse!

\vspace{1.75in}   

 \item Consider $\diagquotient{\mathbb{Z}_{20}}{H}$ where $H=\langle 4 \rangle = \{0,4,8,12,16\}$. List all of the cosets (and their contents) of $H$ in $\mathbb{Z}_{20}$. Then make a Cayley table for this quotient group.
 
 Notice that $[\mathbb{Z}_{20} : H] = \dfrac{|\mathbb{Z}_{20}|}{|H|} = \dfrac{20}{5}=4$, so there should be 4 cosets.
 
 \begin{minipage}{3.5in} 
 \begin{itemize}

\item $H=\langle 4 \rangle = \{0,4,8,12,16\}$   ($=4+H=8+H$  etc.)

\item $1+H= \{1,5,9,13,17\}$     ($=5+H=9+H$  etc.)

\item $2+H=\{2,6,10,14,18\} $    ($=2+H=6+H$ etc.)

\item $3+H=\{3,7,11,15,19\}$    ($=3+H=7+H$  etc.)

 \end{itemize}
\end{minipage} \hfill
 \begin{minipage}{3.5in}  
Cayley Table:

\vspace{0.05in}

\begin{tabular}{c||cccc}
  & H & 1+H & 2+H & 3+H \\ \hline\hline
H & H & 1+H & 2+H & 3+H \\
1+H & 1+H & 2+H & 3+H & H \\
2+H & 2+H & 3+H & H & 1+H \\
3+H & 3+H & H & 1+H & 2+H
\end{tabular}
\end{minipage}

\vspace{0.05in}

An example of computing in the Cayley table above: $(2+H)+(3+H) = (2+3)+H= 5+H = 1+H$.

Also, notice (from the Cayley table) that $\mathbb{Z}_{20}/\langle 4 \rangle \cong \mathbb{Z}_4$.

\vspace{0.05in}

\end{enumerate}

\noindent {\large 4. (15 points)} Let $\varphi:\mathbb{Z}_6 \to \mathbb{Z}_{10}$ be defined by $\varphi(x)=5x$.  

\vspace{0.05in}

\begin{enumerate}[(a)]

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

It is important to note that yes, we {\bf do} need to prove that $\varphi$ is well defined, since $\mathbb{Z}_{6}$ consists of equivalence classes and our map, $\varphi(x)=5x$, has been ``defined'' in terms of representatives of such equivalence classes.

Suppose $x=y$ in $\mathbb{Z}_{6}$.  Then there is some $k \in \mathbb{Z}$ such that $x=y+6k$. This implies that $5x=5y+30k$, so $5x=5y+10(3k)$.  Therefore, $5x=5y$ in $\mathbb{Z}_{10}$  Thus $\varphi$ is well defined.

Note that $\varphi(x+y)=5(x+y)=5x+5y=\varphi(x)+\varphi(y)$.  Therefore, $\varphi$ is a homomorphism.   

\vspace{0.05in}

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

\vspace{0.05in}

Let's just compute $\varphi(x)$ for $x=0,1,\dots,5$: \qquad $0\mapsto0$, $1\mapsto5$, $2\mapsto0$, $3\mapsto5$, $4\mapsto0$, $5\mapsto5$.

So $\mathrm{Ker}(\varphi)=\{x \in \mathbb{Z}_6 \;|\; \varphi(x)=0 \} = \{0,2,4\}=\langle 2 \rangle$ and $\varphi(\mathbb{Z}_{6})=\{0,5\}=\langle 5 \rangle$.

\vspace{0.05in}

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

\vspace{0.05in}

The first isomorphism theorem tells us that $\diagquotient{\mathbb{Z}_{6}}{\mathrm{Ker}(\varphi)} \cong \varphi(\mathbb{Z}_{6})$, so $\diagquotient{\mathbb{Z}_{6}}{\langle 2 \rangle} \cong \langle 5\rangle$     $(\subset \mathbb{Z}_{10})$

\vspace{0.05in}

\end{enumerate}

\noindent {\large 5. (10 points)} Let $H,K \triangleleft G$ and $G=HK$. The second isomorphism theorem says that $\diagquotient{G}{K} \cong \diagquotient{H}{H \cap K}$. We proved this in class using the function $\varphi:H \to \diagquotient{G}{K}$ defined by $\varphi(x)=xK$.

\begin{enumerate}[(a)]

\item What needs to be proven about $\varphi$ to establish this theorem?

\vspace{0.05in}

We need to show that $\varphi$ is an onto homomorphism.  Then we need to show that $\mathrm{Ker}(\varphi)=H\cap K$.  Then, by applying the first isomorphism theorem, we get $\diagquotient{H}{\mathrm{Ker}(\varphi)} \cong \varphi(H)$, and so $\diagquotient{H}{H \cap K} \cong \diagquotient{G}{K}$

Note: We don't need to check if $\varphi$ is well-defined since $\varphi$ has elements of $H$ as inputs (not representatives of equivalence classes).

\vspace{0.05in}

\item Determine the kernel of $\varphi$.

\vspace{0.05in}

Recall that $K$ is the identity in $G/K$,
so $$\mathrm{Ker}(\varphi)= \{ x \in H \;|\;  \varphi(x)=K \}=\{x \in H \;|\; xK=K \} = \{x \in H \;|\; x \in K \} = H \cap K$$

\vspace{0.05in}

\end{enumerate}

\newpage
\noindent {\large 6. (25 points)} Finite Abelian Groups

\begin{enumerate}[(a)]
   \item List all of the non-isomorphic abelian groups of order $36=2^2 3^2$. Circle any that are cyclic.
   
\vspace{0.05in}

The exponent ``2'' can be partitioned in two ways: $2$ or $1+1$. So there are two abelian groups of order $2^2=4$ (up to isomorphism): $\mathbb{Z}_{4}$ and $\mathbb{Z}_2 \oplus \mathbb{Z}_2$. Likewise, there are two abelian groups of order $3^2=9$ (up to isomorphism): $\mathbb{Z}_{9}$ and $\mathbb{Z}_3 \oplus \mathbb{Z}_3$. Therefore, we can build up four non-isomorphic abelian groups from these pairs.

\vspace{0.05in}
  
   \fbox{$\mathbb{Z}_{4} \oplus \mathbb{Z}_{9} \cong \mathbb{Z}_{36}$}
   
   $\mathbb{Z}_{2} \oplus  \mathbb{Z}_{2} \oplus  \mathbb{Z}_{9} \cong \mathbb{Z}_{2} \oplus \mathbb{Z}_{18}$ \hfill [Not cyclic since the largest order appearing is $\mathrm{lcm}(2,2,9) = \mathrm{lcm}(2,18)=18 \not= 36$.]
   
   $\mathbb{Z}_{4}\oplus  \mathbb{Z}_{3} \oplus \mathbb{Z}_{3} \cong \mathbb{Z}_{3} \oplus \mathbb{Z}_{12} $ \hfill [Not cyclic since the largest order appearing is $\mathrm{lcm}(4,3,3) = \mathrm{lcm}(3,12)=12 \not= 36$.]
   
   $\mathbb{Z}_{2} \oplus \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus  \mathbb{Z}_{3} \cong \mathbb{Z}_{6} \oplus  \mathbb{Z}_{6} $ \hfill [Not cyclic since the largest order appearing is $\mathrm{lcm}(2,2,3,3) = \mathrm{lcm}(6,6)=6 \not= 36$.]

\vspace{0.05in}

   \item How many non-isomorphic abelian groups of order 25,508,082,600 are there?

   {\it Note:}  25,508,082,600 $=2^3 3^4 5^2 7^1 11^3 13^2$ and there are 5 non-isomorphic abelian groups of order $3^4=81$. \smiley
   
\vspace{0.05in}

   Let $p(k)$ be the number of partitions of $k$.  Then we have $p(1)=1$, $p(2)=2$, $p(3)=3$, and $p(4)=5$.  so the number of non-isomorphic abelian groups of order 25 is $p(3) \cdot p(4) \cdot p(2) \cdot p(1)\cdot p(3)\cdot p(2)=3 \cdot 5 \cdot 2 \cdot 1\cdot 3\cdot 2= \fbox{180}$. I think I'll skip listing them. \smiley

\vspace{0.05in}

   \item Are the groups $\mathbb{Z}_{6} \oplus \mathbb{Z}_{20} \oplus \mathbb{Z}_{9}$ and $\mathbb{Z}_{30} \oplus \mathbb{Z}_{36}$ isomorphic? Explain your answer.

\vspace{0.05in}

Yes.  $\mathbb{Z}_{n} \oplus \mathbb{Z}_{m} \cong \mathbb{Z}_{nm}$ if and only if $n$ and $m$ are relatively prime.  Since we can pull apart and put back together the various moduli to get from one group to the other, they are isomorphic.

We see that ${\color{red}\mathbb{Z}_{6}} \oplus {\color{blue}\mathbb{Z}_{20}} \oplus \mathbb{Z}_{9} \cong {\color{red}\mathbb{Z}_{2} \oplus \mathbb{Z}_{3}} \oplus {\color{blue} \mathbb{Z}_{4} \oplus \mathbb{Z}_{5}} \oplus \mathbb{Z}_{9} \cong {\color{green}\mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} }\oplus {\color{cyan}\mathbb{Z}_{4} \oplus \mathbb{Z}_{9}} \cong {\color{green}\mathbb{Z}_{30}} \oplus  {\color{cyan}\mathbb{Z}_{36}}$

\vspace{0.05in}

   \item Is the group $\mathbb{Z}_{2} \oplus \mathbb{Z}_{21} \oplus \mathbb{Z}_{15}$ cyclic? Explain you answer.
   
\vspace{0.05in}

   No, it is not cyclic.  If it were cyclic, we should be able to rewrite it as $\mathbb{Z}_{n}$ for some $n$ (actually $n=630$).  But we can't since 21 and 15 aren't relatively prime.
   
   Note that $\mathbb{Z}_{2} \oplus \mathbb{Z}_{21} \oplus \mathbb{Z}_{15} \cong  \mathbb{Z}_{2} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{7} \oplus \mathbb{Z}_{3} \oplus \mathbb{Z}_{5} \cong  \mathbb{Z}_{3} \oplus \mathbb{Z}_{210}$ (the largest element order is $\mathrm{lcm}(2,21,15)=\mathrm{lcm}(3,210)=210 \not= 630$, so not cyclic).

\end{enumerate}

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